Gr-12_M-1_NEET_IL-Ranker_Physics_Electrostatic Potential and Capacitance - V2

CHAPTER

2

ELECTROSTATIC POTENTIAL AND CAPACITANCE Chapter Outline 2.1

Electrostatic Potential

2.2

Potential Due to Discrete Charges

2.3

Potential Due to an Electric Dipole

2.4

Potential Due to Continuous Charge Distribution

2.5

Electrostatic Potential Energy

2.6

Potential Energy in an External Electric Field

2.7

Electrostatics of Conductors

2.8

Dielectrics and Polarisation

2.9

Capacitors and Capacitance

2.10 Effect of Dielectric on Capacitance 2.11 Combination of Capacitors 2.12 Energy Stored in a Capacitor 2.13 Capacitance of a Concentric Spherical Capacitor 2.14 Van de Graaff Generator* 2.15 Atmospheric Electricity The notion of potential energy was introduced in work power energy and oscillations. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, the work gets stored as potential energy of the body. *

This topic is not included in NEET 2025 syllabus.

When the external force is removed, the body moves such that it gains kinetic energy and looses an equal amount of potential energy.

The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces.

Coulomb force between two (stationary) charges is also a conservative force like the gravitational force. Both obey inverse-square law and differ mainly in the proportionality constants. We can define electrostatic potential energy of a charge in an electro-static field like the potential energy of a mass in a gravitational field. The potential energy of a charge at any point is work done by an external force in bringing the charge from infinity to that point We can define difference of electric potential energy between two points as the work required to be done by an external force in moving (with constant velocity) charge from one point to another for electric field of any arbitrary charge configuration.

It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other which is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path.

Capacitor is a device used to store electrical charge in the same way that a tank for storing liquid. The capacity of conductor to hold charge depends on the size, shape

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CHAPTER 2: Electrostatic Potential and Capacitance

and surrounding medium of conductor.

A combination of two conductors of any shape placed close to each other constitutes a capacitor condenser. Capacitors are useful devices in storing charge and electrostatic energy which are of special interest in physics and engineering applications.

In this chapter we will discuss about potential, potential energy of a system of charges, capacitance of a capacitor, energy stored in capacitor and effect of dielectrics on capacity and energy of capacitor.

2.1

ELECTROSTATIC POTENTIAL

Consider any general static charge configuration. The potential energy of a test charge is the work done on the charge in bringing it from infinity to a point in the field.

This work is obviously proportional to the test charge . So it is convenient to divide the work by the amount of charge so that the resulting quantity is independent of test charge. That means the work done or potential energy per unit test charge is characteristic of the electric field associated with the charge configuration.

This leads to the idea of electrostatic potential due to a given charge configuration.

Definition: “The electric potential at a point in an electric field is defined as the work done per unit positive test charge in bringing it from infinity to that point against the electric field.” If an amount of work W is done in bringing a test positive charge q0 from infinity to a point ‘P’ against the electric field, then the potential at that point is given by = V

W UP −U∞ = q0 q0

Definition of volt: The electric potential at a point is said to be one volt, if one joule of work is to be done in bringing one coulomb of positive charge from infinity to that point against the electric field.

1 joule 1 Volt = 1 coulomb Potential difference: Potential difference between two points in an electric field is defined as the work done per unit positive test charge in moving it from one point to the other against the electric field. The potential difference (p.d.) is measured in volt.

It is a scalar quantity.

Consider two points B and C in an electric field. Let UB and UC are potential energies , and VB and VC potentials at B and C respectively as shown in figure.

A

E

B

C

+q

+1C ∞

VC

VB

The work done per unit positive test charge in moving a test charge from infinity to C against the field is VC, which is the potential at C. Similarly the work done per unit positive test charge in moving it from infinity to B against the field E is VB which is the potential at B. Therefore, the work done per unit positive test charge from C to B against the field E is VB – VC (Evidently VB > VC). By definition the potential difference between C and B is VBC = VB − VC =

U B −UC q0

Conclusions: ■ VBC will be positive (VB > VC) if we bring a

Where UP and U∞ are electrostatic potential energies at the point P and infinity respectively.

positive test charge from C to B opposite

In SI, the unit of potential is joule/coulomb (JC–1) or volt (V). Electric potential is a scalar quantity.

■ In moving a positive charge from B to C we need not do any work. The field does work and the p.d. VCB = VC – VB will be negative.

to the direction of the field.

IL RANKER SERIES FOR NEET

■ The field produced by a positive charge is directed away from the charge. So, in bringing a positive charge from infinity to any point in this field, we have to do work and hence the potential at any point around a positive charge is positive.

■ A positive charge gives rise to positive potential. A negative charge gives rise to negative potential.

■ The potential difference VB – VC between two points does not depend on the path followed from C to B. This is because the electrostatic field is a conservative field.

■ Work done in moving a charge q through a potential difference V is W = qV.

++++++ ++++++

Electrons

2.1.1 Relation between Electrostatic Potential (V) and Electric Intensity (E) In the diagram shown Lines (1) and (2) are the intersections of two equipotential surfaces  with the plane of the figure. E is the electric field at point A normal to the surface. An external agent is slowly moving a test charge Q0 (point positive from A to B where  charge)  displacement AB = ∆r . Force applied by the external agent is F = Q0E. (2)

■ If a negatively charged conductor is grounded, electrons flow from it to the earth until the potential of the conductor is same as that of the earth, So the initial potential of the negatively charged conductor is lower than that of the earth.

Hence it is taken as negative potential.

–––––– ––––––

Electrons Current

If a positively charged conductor is grounded, electrons flow from the earth to the conductor, until the potential of the conductor is same as that of the earth, So the initial potential of the positively charged conductor is higher than that of the earth. Hence it is taken as positive potential.

89

Current

Key Insights: ■ The earth is a huge reservoir of charges.

Addition of any amount of charge to the earth or removal of any amount of charge from the earth does not change the electrical status of the earth. So the earth is considered as electrically neutral and the potential of earth is taken as zero. The potentials of all the bodies are measured with respect to the potential of the earth.

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B

(1) Qo

 E

A

a F V+Dv V

DW = work done by external agent = F.Dr.cosa = Q0E Drcos(p – q). DW = –Q0Ecosq . Dr . . .(1) Also, DW = [(V + DV)Q0 – DV. Q0 ]. . . (2) from (1) and (2), DV = –EDr.cosq ⇒ E cos θ = −

dV dr

But, Ecosq = Er = component electric field in the direction of displacement.

In general Er = −

dV dr

If potential v is a function of x only, Ex = −

dV dx

If v is a function of both x and y, then

90

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CHAPTER 2: Electrostatic Potential and Capacitance

2.1.2 Conservative Nature of Electric Field

δV δV and E y = Ex = − − δx δy  δV  δV  i− j.

Also, E = − δx δy

A conservative force is that force for which the work done around a closed path is zero and the work done by the force is independent of path.

Therefore,

y

i) The negative  sign indicates that the direction of E is always in the direction of decrease of electric potential.

i rf

ii) The magnitude of electric field at a point is the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

That means electric field at a point is equal to the negative gradient of the electrostatic potential at that point.

Unit of potential gradient: In SI, the unit of potential gradient is volt metre­–1 (Vm–1) iii) The spacing between equipotential surfaces enables to identify regions of strong and weak fields E= −

+q

Then

⇒

40 V

Key Insights: ■ If we choose the potential at initial point ‘i’ as Vi = 0, then potential V at final point f     f   is given by V = − ∫ E.dr where dr= rf − ri i

■ If A and B are two points in an electrostatic     B   field, VB − VA = − ∫ E.dr where dr =− rB rA A

■ In the direction of electric field, potential decreases.

■ For uniform electric field the relation between E and V is E = V/d Where d is distance between two points, having V potential difference.

x

Consider a closed path i l1 f l2 i in the electric field of point charge +q. Let a small test charge +q0 be moved over the closed path.

R EP<EQ<ER

Q 20 V 10 V

f

z

and

P 30 V

l2 rf

1 dV . So E ∝ (if dV is constant). ∆ r dr

Example:

l1

f   = − ∫ E.d  (along l1) = Vf –Vi

wi → f q0

w f →i q0

wi − f q0

i

  = − ∫ E.d  i

f

+

w f −i q0

(along l2) = Vi –Vf

= (V f − Vi ) + (Vi − V f ) = 0

In other words work done in moving a unit positive test charge over a closed path in electrostatic field is zero. It means electrostatic field is a conservative field and electrostatic forces are conservative.   ⇒ ∫ E.d  = 0 The line integral of electrostatic field over a closed path in an electric field is zero.

Key Insights: ■ As electrostatic force is conservative, all paths from i to f yield the same result, when the line integral is found.

IL RANKER SERIES FOR NEET 1. If electric potential V at any point (x, y, z) all in metres in space is given by V = 4x 2 volt. Calculate the electric field at the point (1 m, 0 m, 2 m).

dV dV d ⇒ Ex = − = − (4 x 2 ) = −8 x dr dx dx

dV d − = − (4 x 2 ) = 0 E y = dy dy dV d And, Ez = − = − (4 x 2 ) = 0 dz dz  So, E = iˆ E + ˆj E + kˆ E = −8 xiˆ x

y

z

i.e., it has magnitude 8 V/m and is directed along negative x-axis.

91

V 5 3

Sol: As electric field E is related to potential V through the relation

E= −

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1 0

2

4

6

d

(1) 2.5 Vm–1 (2) –2.5 Vm–1 (3) 0.4 Vm–1 (4) –0.4 Vm–1 4. A, B and C are three points in a uniform electric field. The electric potential is: A B

Try yourself: 1. Find the potential difference (V A –V B ) between points A and B in an electric field  ˆ ˆ ˆ NC −1 where rA =(iˆ − 2 ˆj + kˆ) m E =(2i+3j+4k)  and r = (2iˆ + ˆj − 2kˆ) m .

B

Ans: –1 volt

TEST YOURSELF 1. Two parallel plates are 0.03 m apart. The electric field intensity between them is 3000 newton per coulomb. Calculate the potential difference between the plates.

(1) 80 V (2) 70 V (3) 100 V (4) 90 V 2. The electric potential (V) in volts varies with x (in metre) according to the relation V = 5 + 4x2. The force experienced by a negative charge of 2μC located at x = 0.5m is (1) 2 × 10–6 N (2) 4 × 10–6 N (3) 6 × 10–6 N (4) 8 × 10–6 N 3. The variation of electric potential with distance d from a fixed point is as shown in the figure. The electric intensity at d = 5m is

C

(1) Maximum at B (2) Maximum at C (3) Same at all the three points A, B and C (4) Maximum at A 5. The electric potential decreases uniformly from 120 V to 80 V as one moves on the X-axis from x = –1 cm to x = +1 cm. The electric field at the origin (1) must be equal to 20 V/cm (2) may be equal to 20 V/cm (3) may be greater than 20 V/cm (4) may be less than 20 V/cm 6. A charge of 5C experiences a force of 5000N when it is moved in a uniform electric field.

The potential difference between two points separated by a distance of 1cm is (1) 10 V (2) 250 V (3) 1000 V (4) 1000 V (1) 4 (5) 3

Answer Key (2) 4 (3) 1 (6) 1

(4) 1

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CHAPTER 2: Electrostatic Potential and Capacitance

The electric potential at P is

ELECTRIC POTENTIAL DUE TO DISCRETE CHARGES

2.2

V=

The electric potential resulting from discrete charges is determined by summing the contributions of each individual charge’s potential.

2.2.1 Electric Potential Due to Point Charge Consider a point charge +q fixed at a point O in free space. Let us find electric potential at P due to charge +q. Let r be the position vector of P from O, i.e. OP = r. +q

P r

dr B A

+ 1 C at ∞

r

 Consider point having position vector r with respect to the point O. The electro static force on unit positive charge or the electric field intensity at that point A is  q  1 E= r' 4 πε0 ( r ' )2

1 q 4πε0 r

If q is positive, then potential at P is positive. On the other hand if q is negative, then potential at P is negative. Potential due to a charge at its location is not defined it is infinite.

5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0

–1/r –1/r2 E V

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Variation of potential V with r [ in units of (Q/4πε0)m–1] ( cyan curve) and field with r [ in units of (Q/4πε0) m–2] ( magenta curve) for a point charge Q.

2. Find the electric potential at a distance of 2 m from a 2 nC point charge.

1

Q

2 × 10−9

Sol. V = . =× 9 109 × If an unit positive test charge is moved 4 πε0 r 2 from A to B, then the −9  amount of work done = 9 Volt. 9 2 × 10 1 Q against the field is F .dr ' = 9 10 × Volt E ( −dr ' ) = − Edr ' V = . =× πε 4 2  0 r     [ s i n c e r d e c r e a s e s , dr =(r + dr ) − r i s= 9 Volt. = 9 Volt. negative] Try yourself: Total amount of work done in bringing unit r r r q 1 W == dr ' −∫ 2 ∫∞ dW −∫∞ E dr ' = ∞ 4 πε r ( ') 0

r

r q q  1 1 = − − − dr ' = 2 ∫ ∞ 4 πε0 4 πε0  r '  ∞ (r ')

=

q 1 1  q −=   4 πε0  r ∞  4 πε0 r

By definition, this work done in bringing unit test charge from infinity to the point P is called electric potential at P.

2. If the electric potential at a distance r from a point charge Q is 5 volt, what is the electric potential at a distance 2r from a point charge 3Q? Ans: 7.5 volt

positive test charge from infinity to r is

Volt

2.2.2 Potential at a Point Due to System of Charges Electric potential obeys superposition principle. The potential at any point P due to a group of point charges q 1,q 2,........q n is equal to algebraic sum of the potentials due to q1,q2,........qn at P.

IL RANKER SERIES FOR NEET qn

r1 r2

q2

rn

q4

r4

P r3

Let V1,V2,V3........Vn be the potentials at P due to the charges q1,q2..........qn respectively.

Now, total potential at P is qn  1  q1 q2  + + .... +  rn  4 π∈0  r1 r2

Points of Zero Potential Point of zero potential exists only for unlike charges both in between and outside the line joining the charges.

Two unlike charges are separated by a distance d. Two zero potential points are obtained on the line joining the two charges.

One point between two charges and other outside nearer to smaller charge.

–q2

A

+q1 x

(d-x)

d

v1 + v2 = 0 (or) v1 = – v2

q d −1 =2 x q1

d q2 = +1 x q1

d ∴x = q2 +1 q1

+q1 y

At the point B

d

q2 −1 q1

■ Points of zero potential exists only for unlike charges.

Key Insights: ■ In fact, there will be infinite zero potential points lying on the circle of diameter x + y x

y

–q2

q1

3. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 ..... on the x-axis and a charge –q if fixed at each of the points x = 2x0, x = 4x0 , x = 6x0 ...... , here xo is a positive constant. Then find the potential at the origin due to the above system of charges.

V =

=

=

If q1 < q2 in magnitude zero potential point is obtained outside nearer to q1.

B

y=

Sol: Potential at origin will be given by

q2 1 q1 1 = 4 π∈0 x 4 π∈0 d − x

d − x q2 = x q1

d q2 = –1 y q1

q d + 1 =2 y q1

Conclusion:

V = V1 + V2 +.........+ Vn = ⇒V

q2 1 q1 1 = 4 π∈0 ( y ) 4 π∈0 d + y q2 1 q1 1 = 4 π∈0 ( y ) 4 π∈0 d + y

q3

93

–q2 d

v1 = – v2

 q 1 1 1 1 + − +   − 4 π∈o  xo 2 xo 3xo 4 xo  q 1  1 1 1  ⋅ 1 − + − +   4 π∈o xo  2 3 4  q ln (2 ) 4 π∈o xo

Try yourself: 3. Two point charges +6 µC and –2 µC are kept a separation of 30 cm. Find the distance of the point from +6 µC charge where the potential is zero.

Ans: 22.5 cm

q1

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94

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CHAPTER 2: Electrostatic Potential and Capacitance

TEST YOURSELF 1. Electric potential at a point P due to a point charge of 5 × 10–9 C is 50V. The distance of P from the point charge is

  1 = 9 × 10+9 Nm 2 C −2   Assume, 4 πε0   (1) 0.9 cm (2) 90 cm (3) 3 cm (4) 9 cm 2. Charges 5µC, –2µC, 3µC and –9µC are placed at the corners A, B, C and D of a square ABCD of side 1m. The net electric potential at the centre of the square is (1) –27 kV (3) –90 kV

(2) −27 2 kV (4) zero

3. An infinite number of electric charges each equal to 2 nano coulombs in magnitude are placed along x-axis at x = 1 cm, x = 3 cm, x = 9 cm, x = 27cm ... and so on. In this setup if the consecutive charges have opposite sign, then the electric potential at x = 0 is (1) 1250 V (2) 1350 V (3) 2700 V (4) 2500 V 4. Two charges 8 μC each are placed at the corners A and B of an equilateral triangle of side 0.2 m in air. The electric potential at the third corner C is (1) 7.2 × 105 V (2) 1.8 × 105 V (3) 3.6 × 105 V (4) 3.6 × 104 V 5. Four identical charges each +Q are at the vertices of a square of side “L”. The electric potential at the center of the square is ‘V’.

If one of the charges is reversed in sign, the potential now at the center of the square will be (1) –V/2 (2) 3V/4 (3) V/2 (4) –3V/4 6. Two electric charges of 9µC and –3µC are placed 0.16 m apart in air. There will be a point P at which electric potential is zero on the line joining the two charges and in between them. The distance of P from 9µC charge is

(1) 0.14 m (3) 0.08 m

(2) 0.12 m (4) 0.06 m

7. Figure shows three circular arcs, each of radius R and total charge as indicated.

The net electric potential at the centre of curvature is: +Q 450 –2q

300

R +3Q

Q (2) 4 πε R

Q 2πε0 R 2Q (3) πε0 R (1)

0

(4)

(1) 2

Answer Key (2) 2 (3) 2

(5) 3

(6) 2

2.3

Q πε0 R (4) 1

(7) 1

POTENTIAL DUE TO AN ELECTRIC DIPOLE

An electric dipole consists of two equal and opposite charges separated by a very small distance. If ‘q’ is the charge and 2a is the length of the dipole, then electric dipole moment will be given by p = (2a)q. q

r1

p

r r2

2a p -q

Let AB be a dipole whose centre is at ‘O’ and ‘P’ be the point where the potential due to dipole is to be determined. The potential due to the dipole is the sum of potentials due to the charges q and -q = V

1 q q  −  4 πε0  r1 r2 

IL RANKER SERIES FOR NEET

Now, by geometry, r12 = r 2 + a 2 − 2ar cos θ

4. Find the electric potential at a point P on the axis of a short electric dipole of moment 2 nC–m as shown in figure.

P

r22 = r 2 + a 2 + 2ar cos θ

 2a cos θ a 2  2  2a cos θ  + 2  = r 1− r12 = r 2  1 −  r r r    

 2a cos θ a 2  2  2a cos θ  = r 1− + 2  = r 1−  r r r     2

 2a cos θ  Similarly,= r22 r 2  1 +  r   Using the Binomial theorem and retaining

2m

1 p 2 × 10−9 9 =× 9 10 × volt 2 4 πε0 r 2 (2)

Sol. VP =

= 4.5 Volt Try yourself: 4. An electric dipole of moment p is placed at the origin pointing towards positive x-axis.

A is a point in the first quadrant whose coordinates are (a, a). Find the electric potential at point A.

8 2πε0 a 2

We take r much greater than a(r>>a) and retain terms only upto the first order in a/r

terms up to the first order in a/r; we obtain.

−1

1 1  2a cos θ  2 1  a  ≅ 1+ ≅  1 − cos θ   r2 r  r r r   using the above eqs. and p = (2a)q.

p

−1

1 1  2a cos θ  2 1  a  ≅ 1−  ≅ r  1 + r cos θ  r1 r  r   

TEST YOURSELF 1. Equal and opposite charges q are placed at points A and B as shown in Fig and P1 and P2 are equidistant points from O. The ratio of potential at P1 and P2 is p1

q 2a cos θ p cos θ = 4 πε0 4 πε0 r 2 r2   In vector form, V = p.r 4 πε0 r 3 Special Cases:

= V

■ On the axial line : For a point on the axial line θ = 0° ∴Vaxial = p / 4 πε0 r 2

volts for a dipole.

■ Point on the equatorial line: For a point on the equatorial line θ = 90°.

∴Vequitorial = 0 volt Equatorial line is a line where the potential is zero at any point.

95

Ans:

Where r1 and r2 are the distances of the point P from q and -q, respectively.

|

b

–q

2a

+q

p2

b

(1) 1 (3) zero

(2) b : a (4) infinity

2. The potential due to the electric dipole (length of dipole = 2a) at a distance r from the dipole on the axial line (1)

1 2qa 4 πε0 r 2

(2) Zero

(3)

1 qa 4 πε0 r 2

(4)

1 4qa 4 πε0 r 2

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CHAPTER 2: Electrostatic Potential and Capacitance

3. The potential due to the electric dipole at a distance ‘r’ from the dipole on the equatorial line is (1)

1 2qa 4 πε0 r 2

(2) zero

(3)

1 qa 4 πε0 r 2

(4)

1 4qa 4 πε0 r 2

4. A short electric dipole has a dipole moment of 32 × 10–9 cm. he electric potential due to the dipole at a point at a distance of 0.18 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole

 1  = 9 × 109 Nm 2 / C 2   4 πε0 

axis is: 

(1) 800 V (3) zero

(2) 200 V (4) 400 V

ii) The electric field is always perpendicular to an equipotential surface. In other words electric field or lines of force are perpendicular to the equipotential surface. iii) No two equipotential surfaces intersect.

If they intersect like that, at the point of intersection field will have two different directions or at the same point there will be two different potentials which is impossible iv) At any point on the equipotential surface component of electric field parallel to the surface is zero.

In uniform field , the lines of force are straight and parallel and equipotential surfaces are planes perpendicular to the lines of force as shown in figure Equipotential surface

Answer Key (1) 3

2.4

(2) 1

(3) 2

(4) 1

POTENTIAL DUE TO CONTINUOUS CHARGE DISTRIBUTION

The electric potential arising from a continuous charge distribution is determined by integrating the contributions of infinitesimal charge elements over the distribution.

The equipotential surfaces are a family of concentric spheres for a uniformly charged sphere or for a point charge as shown in figure

Equipotential surface

2.4.1 Equipotential Surfaces We know that potential is same at every point on equipotential surface in an electric field.

In other words, the locus of all points which have the same electric potential is called equipotential surface.

An equipotential surface may be the surface of a material body or a surface drawn in an electric field. The important properties of equipotential surfaces are as given below. i) As the potential difference between any two points on the equipotential surface is zero, no work is done in taking a charge from one point to another.

v) Equipotential surfaces in electrostatics are similar to wavefronts in optics. The wavefronts in optics are the locus of all points which are in the same phase. Light rays are normal to the wavefronts. On the other hand the equipotential surfaces are perpendicular to the lines of force.

Key Insights: ■ In case of non-uniform electric field, the field lines are not straight, and in that case equipotential surfaces are curved but still perpendicular to the field.

IL RANKER SERIES FOR NEET

■ The potential difference is a property of two points and not of the charge q0 being moved.

2.4.2 Electric Potential Due to a Linear Charge Distribution Consider a thin infinitely long straight line charge having a uniform linear charge density λ placed along YY1. Let P is a point at distance r from the line charge then magnitude of electric field at point P is given by λ 2π∈0 r   We know that V (r ) = − ∫ E.dr E=

Y

+ + + + + + + + + +

dS P

E Gaussian Surface

Y’

Here, E =

  λ and E.dr = Edr 2π∈0 r

λ So V ( r ) = − ∫ E dr = −∫ dr 2π∈0 r  −λ  = ∴V (r )  log e r  + C  2π∈0  where C is constant of integration and V(r) gives electric potential at a distance r from the linear charge distribution.

97

5. An infinitely long line of charge has linear charge density of 0.5 nC/m. A and B are two points at distances 2 m and 4 m from the line of charge. Find the potential difference (VA–VB) between A and B. A

A λλ Sol: E = E = , V, AV−AV−BV= = B− ∫B Edr ∫B−Edr 2πε r 2πε 0r 0 2

2 λ λ drdr −9 −9 4 4 9 9 − ∫ ∫ = × 10 × 0.5 × 10 = −= 9= ×910 2 ×20.5 × 10 l nl n 2πε r 2πε r 2 2 0 40 4 volt. = 6.23 volt. = 6.23

((

))

Try yourself: 5. A and B are two points at distances rA and rB (rB > rA) from an infinitely line of charge 1 having linear charge density nC/m. Find 9 rB . if potential difference between A and rA B is 2 volt.

Ans: 2.718

■ Electric potential and potential energy are always defined relative to a reference. In general we take zero reference at infinity.

The potential at a point P in an electric field is V if potential at infinity is taken as zero.

If potential at infinity is V0, the potential at P is (V + V0).

|

2.4.3 Electric Potential Due to Infinite Plane Sheet of Charge (Nonconducting) Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σ of the sheet. By symmetry, it follows that the electric field is perpendicular to the plane sheet of charge and is directed in outward direction. σ 2ε 0 Electrostatic potential due to an infinite plane sheet of charge at a perpendicular distance r from the sheet is given by   V (r ) = − ∫ E.dr = − ∫ Edr The electric field intensity is E =

 −σ  σ = −∫ dr = r +C  2 ∈0  2 ∈0  where C is constant of integration.

Electric potential due to an infinite plane conducting plate at a perpendicular distance r from the plate is given by   V (r ) = − ∫ E.dr = − ∫ Edr

98

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CHAPTER 2: Electrostatic Potential and Capacitance

  The potential V (r ) = − ∫ E.dr = − ∫ Edr

 −σ  σ V (r ) = − ∫ dr = r +C  ∈0  ∈0  where C is constant of integration 6. A and B are two points at distances 3 m and 4 m from a large charged thin sheet having surface charge density ‘s’. If the potential difference between A and B is 2 volt, find s.

Sol: VA − VB =

rA

 

σ

∫ E.dr = E ( r − r )= 2ε ( ∆r ) B

A

0

rB

2 × ε0 × ∆V 2 × 8.85 × 10−12 × 2 C/m 2 = 1 ∆r = 35.4 pC/m 2

= ⇒σ

Try yourself: 6. M and N are two large charged sheets having charge densities +2s and –s respectively.

Find potential difference between A and B. +2s –s A B d Ans: 3sd/2e0

N

M

2.4.4 Electric Potential Due to a Charged Spherical Shell (or Conducting Sphere) q + + + + R r +

+ +

+

O + +

+

+

P

E dS Charged spherical shell

Consider a thin spherical shell of radius R and having charge +q on the spherical shell.

Cases: (i) When point P lies outside the spherical shell. The electric field at the point is q 1 . 2 4 πε0 r

where C is constant of integration If r → ∞ , V (∞) → 0 and C = 0 V(r) =

1 q 4 π ε0 r

(r > R)

(ii) When point P lies on the surface of spherical shell then r = R. Electrostatic potential at P on the surface is 1 q V= 4 π∈0 R (iii) For points inside the charged spherical shell (r <R), the electric field E=0 dV so we can write − = 0 dr V is constant and is equal to that on the surface 1 q so, V = for r ≤ R 4 π∈0 R The variation of V with distance r from centre is as shown in the graph.

V 1 q 4π∈0 R

Gaussian Surface

E=

1 q 1 q = −∫ dr = +C 2 4 π∈0 r 4 π∈0 r

(for r> R)

E∝ 1 r R

r

Electric field is discontinuous across the surface of a spherical charged shell. σ It is zero inside and outside. Electric ε0 potential is continuous across the surface 7. Three concentric spherical metal shells A, B, C of radii a, b, c (c > b > a) have surface charge densities + σ,− σ and +σ respectively. i) Find the potentials of the three shells? ii) If the shells A and C are at the same potential, find the relation between a,b and c

IL RANKER SERIES FOR NEET potential at O. V+ =

c b

B C Sol: i) Charges on the three shells are qA = + σ4πa2 qB = + σ4πb2 qC = + σ4πc2

1  q A qB qC  = VA  + +  π∈ b c  4 0  a

σ = (a − b + c) ∈0

1  q A qB qC  = VB  + +  b c  4 π∈0  b

2 2  σ  a −b = +c   ∈0  b 

qC  q q 1  A B = VC  + +  c c  4 π∈0  c =

σ  a 2 − b2 + c 2    ∈0  c 

ii) If VA = VC, on substitution, we get a + b = c

8. A point charge q is at a distance r from the centre O of an uncharged spherical conducting layer, whose inner and outer radii are equal to a and b respectively. Find the potential at the point O if r<a.

Sol: Equal amount of charge will induce on the inner and outer surface of the conductor. Then each negative charge is at an equal distance a from O, and so potential due to this charge at O,

V− =

1 (−q) 4 π∈0 a + + –– –– + q – – –– q a ––– – –––– r + –– O H –––– + –– ––– b –––– + –– – –– + +

Similarly due to positive charge,

Potential at O due to charge q V = 0

1 q 4 π∈0 r

Thus total potential V = Vo + V– + V+

=

q 1 1 1  − + .

4 π∈0  r a b 

Try yourself: 7. Three concentric spherical shells of radii r1, r2 and r3 have charges q1, q2 and q3 respectively.

If the central sphere is connected to the earth by a conducting wire now, find the final potentials of the outermost shell.

r3

q1

q2

q3

r1 r2

q3 ( r3 − r2 )

a A

1 q 4 π∈0 b

4 πε0 r32

–σ +σ

99

Ans:

+σ

|

2.4.5 Electric Potential Due to a Uniformly Charged Non-conducting Solid Sphere Consider a charged sphere of radius R with total charge q uniformly distributed on it.

Cases: i) For points Outside the Sphere (r >R) The electric field at any point is q 1 E= . (for r > R) 4 πε0 r 2 The potential at any point outside the shell is   V (r ) = − ∫ E.dr = − ∫ Edr 1 q 1 q = −∫ dr = +C 2 4 π∈0 r 4 π∈0 r where C is constant of integration If r → ∞ , V (∞) → 0 and C = 0 V(r) =

1 q 4 π ε0 r

(r > R)

100 |

CHAPTER 2: Electrostatic Potential and Capacitance

ii) When point P lies on the surface of spherical shell then r = R

The variation of V with distance r from centre is as shown in the graph.

The electrostatic potential at P on the surface is

9. A non-conducting sphere has radius R. The sphere carries some charge distributed uniformly throughout its volume. If potential at its centre is V, what is the potential at a radius distance r = 2R?

V=

1 q 4 π∈0 R

iii) For points Inside the Sphere (r < R) The electric field is E =

1 qr 4 π∈0 R 3

+ + + E + + + + + + + + r + + + + + + + + R + +

Try yourself:

  dV = − E.dr = − Edr V

r

r

Vs

R

R

qr

1

− ∫ E dr = −∫ dr ∫ dV = 4 π∈ R 0

3

Application

1 q 1 q  r 2 R2  = −  −  4 π∈0 R 4 π∈0 R 3  2 2 

⇒ = V

8. A non-conducting solid sphere of radius R is uniformly charged throughout its volume.

If potential at a radial distance R/2 is equal to two times the potential at a distance r(> R), find r.

Ans: 16 R/11

r

1 q  r2  − V − Vs =   4 π∈0 R 3  2  R V−

2Vc 2V 3 = Vs ⇒ Vs = 2 3 3 R 2V R V = For r ≥ R, V ' =Vs = . r 3 2R 3

Sol: Vc =

1 q  3 r2  × −  4 π∈0 R  2 2R 2 

V  r2  ⇒V = s 3− 2  2  R  1 q where, Vs = 4 πε0 R

■ A metal sphere A of radius a is charged to potential V. It is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire.

If the charge on sphere of radius a is q, then V=

1 q 4 πε0 a B

At the centre r = 0 b

Potential at centre, 1 3q 3 1 q = 4 π∈0 2R 2 4 π∈0 R

= VC 3VS 2

⇒ q=

V VS =

VS R

1 q 4π∈0 R r

A a

( 4πε0a ) V

Now, when sphere A is enclosed by spherical conductor B and the two are connected by a wire, the charge flows from A to B and charge will reside on outer surface of B. So the potential of B will be,

IL RANKER SERIES FOR NEET

= VB

‘r’ from the common centre ‘O’ are given below.

1 q 1 4 πε0a a = = V V 4 πε 0 b 4 πε 0 b b

Now as sphere A is inside B so its potential,

+

a (V ) [<V as a< b] V= V= A B b ■ A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such that the surface densities are equal. Steps to find the potential at the common centre are given below.

+ +

If q 1 and q 2 are the charges on spheres of radii r and R respectively, then in accordance with conservation of charge q1 + q2 = Q

.... (1)

And according to given problem σ1= σ2, i.e.,

q1 r 2 q1 q2 or = = q2 R 2 4 πr 2 4 πR 2

.....(2)

So from Eqs. (1) and (2) QR 2 Qr 2 and = q q1 = 2 2 (r 2 + R 2 ) (r + R 2 )

....(3)

Now as potential inside a conducting sphere is equal to that at its surface, so potential at the common centre, V = V1 + V2 =

1  q1 q2  + 4 πε0  r R 

Substituting the values of q1 and q2 from Eq.(3) = V

| 101

1  Qr QR  + 2 2   2 2 4 πε0  (R + r ) (R + r ) 

1 Q( R + r ) = 2 2 4 πε0 (R + r ) ■ Two concentric spherical conducting shells of radii a,b ( > a) have charges q 1 and q2 respectively. Steps to find the electric intensity and electric potential at a distance

+

b

+ +

+

+ q2 +

+ + q1+ + O + a +r + + + +

+

+

+ + + + +

■ Cases: i) If r < a (i.e., for points inside the inner shell) = V

1  q 1 q2   +  4 π∈0  a b 

[ Potential at any point inside is equal to potential on the surface] ii) If r = a (i.e., for points on the surface of inner shell) = V

1  q1 q2  + 4 π∈0  a b 

[ The point lies on the surface of the inner shell & inside the outer shell] iii) If a < r < b (i.e., for points in between shells) = V

1  q1 q2  + 4 π∈0  r b 

[ The point lies outside inner shell & inside the outer shell] (iv) If r = b [ie for points on the surface of outer shell] = V

1  q1 q2  1  q1 + q2  +=   4 π∈0  b b  4 π∈0  b 

[ The point lies outside the inner sphere and on the surface & outer sphere] (v) If r > b [ie for points outside the outer shell] = V

1  q1 q2  1  q1 + q2  +=   4 π∈0  r r  4 π∈0  r 

102 |

CHAPTER 2: Electrostatic Potential and Capacitance (1) Zero

TEST YOURSELF 1. Two conducting spheres of radii R1 and R2 are at the same potential. The electric intensities on their surfaces are in the ratio of (1) 1 : 1 (2) R1 : R2 (3) R2 : R1 (4) R12 : R22

(2)

(r + R ) q 4 π∈0 r 2 + R 2 2

(3)

q 1 1  + 4 π∈0  r R 

2. Two concentric metallic shells are of radii r1 and r2 (r2 > r1). If charge given to outer sphere is q, and the inner sphere is ground.

Then the charge on the inner sphere q1 is

(4)

(r + R ) q 4 π∈0 r 2 + R 2

r2 r1

(1) Zero r1 (3) − q r2

(2) –q r1 q (4) r2

3. Surface charge density of a sphere of a radius 10 cm is 8.85 × 10–8 C/m2. Potential at the centre of the sphere is (1) 1000 V (2) 885 V (3) 10–3 V (4) 442.5 V 4. Two concentric spherical shells of radii R and 2R carry charges Q and 2Q respectively.

Change in electric potential on the outer shell when both are connected by a conducing   wire is  K = 1  4 πε0   3KQ (1) zero (2) 2R

KQ 2KQ (4) R R 5. A charge ‘q’ is distributed over two concentric hollow conducting sphere of radii r and R (> r) such that their surface charge densities are equal. The potential at their common centre is (3)

(

)

(

)

6. Two spheres A and B of radius 4 cm and 6 cm are given charges of 80 μC and 40 μC respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is (1) 20 μC from A to B (2) 16 μC from A to B (3) 32 μC from B to A (4) 32 μC from A to B 7. Three concentric metallic shells A, B & C of radii a, b & c (a < b < c) have surface densities σ, –σ & σ respectively. If the shells A and C are at the same potential then the relation between a, b and c is (1) c = a + b (2) a = b + c (3) b = a + c (4) none 8. Consider the situation of figure. The work done in talking a point charge from P to A is WA, from P to B WB and from P to C is WC q A P C

B

(1) WA < WB < WC (2) WA > WB > WC (3) WA = WB = WC (4) None of these

IL RANKER SERIES FOR NEET 9. Four equipotential curves in an electric field are shown in the figure. A, B and C are three points in the field. If electric intensity at A, B, C EA, EB are EC and respectively, then C

120 V 90 V

B

A

60 V

(1) EA = EB = EC (3) EA < EB < EC

30 V

(2) EA > EB > EC (4) EA > EB < EC

11. The diagrams below show regions of equipotentials A

20V 40V

B A 10V 30V

10V 30V

10V 30V B A

20V

40V B

B A 20V 40V

10V

30V

A positive charge is moved from A to B in each diagram (1) Maximum work is required to move q in figure (b) (2) Maximum work is required to move q in figure (c) (3) In all the four cases the work done is the same.

(4) Minimum work is required to move q in figure (a) (1) 3 (5) 4 (9) 3

Answer Key (2) 3 (3) 1 (6) 4 (7) 1 (10) 2 (11) 3

ELECTROSTATIC POTENTIAL ENERGY

We know that potential energy of a system of particles is defined only for conservative fields. Since the electric field is conservative, the electrostatic potential energy in electric field can be defined in two ways a) interaction energy of charged particles of a system

10. Electric field in a region is increasing in magnitude along x-direction. The equipotential surfaces associated are (1) Planes parallel to xy-plane (2) Planes parallel to yz-plane (3) Co-axial cylinders around x-axis (4) All of these

20V 40V

2.5

| 103

(4) 1 (8) 3

b) self energy of a charged body (we will discuss later)

2.5.1 Electrostatic Interaction Energy: The external work required to assemble the charged particles of a system by slowly bringing them from infinity to a given configuration is known as electro-static interaction energy of that system.

If charged particles are at infinite separation, potential energy of that system is taken as zero as there will be no interaction between them. When these charges are brought closer to form a given configuration, external work is required and energy is supplied to the system. If the force between the charges is repulsive, work is done by external agent and final potential energy of the system will be positive. If the force between the charges is attractive, work will be done by the system and final potential energy of the system is negative.

2.5.2 Electrostatic Potential Energy of a System of Two Charges Consider a system of two point charges q 1 and q2 initially at infinite separation. Let the charges are brought to the points A and B   having position vectors r1 and r2 respectively.

The work done by external agent in bringing the charges to finite separation is stored as interaction potential energy. q1 A(r1)

r12

q2 B(r2)

104 |

CHAPTER 2: Electrostatic Potential and Capacitance

When the charge q1 is brought from infinity to  the point A having position vector r1 , there is no need of doing work against electric field.

Now electric field is produced due to the charge q1.

The electric potential at the point B  having position vector r2 due to the charge q1 is V1 =

1 q1 , where r 12 is distance 4 π∈0 r12

 

between A and B and is equal to r1  r2

From the definition of potential, work done in bringing charge q2 from infinity to the point B is equal to q2 times the potential at B due to q1.

1 q1q2 .

4 π∈0 r12 Since the electrostatic force is conservative, this work is stored as potential energy of the system of two charges q1 and q2. work done on q2 is W =

1 q1q2 .

4 π∈0 r12 Suppose, the charges are initially at separation r1 and their separation is changed to r2. The work done by external agent is equal to change in electric potential energy of the system The potential energy U =

So, the work done or the change in potential energy is given by q q 1 1 W = U (r2 ) − U (r1 ) = 1 2  −  4 π∈0  r2 r1 

Key Insights: ■ The potential energy of the system of charges depends essentially on the separation between the charges and is independent of their location in space.

■ For a system of many charged particles q1, q2, q3 ........ qn, the interaction potential energy is given by

= U

 1  q1q2 q2q3 + + .....  4 π∈0  r12 r23 

■ Here, r12 indicates separation between q1 and q2. r23 indicates separation between q2 and q3. r31 indicates separation between q1 and q3 and so on. We have to consider all possible pairs while finding U. Here charges with their original signs must be used.

■ Potential energy of a system of charges may be positive, negative or even zero.

■ If the separation between two point charges is increased, electrostatic potential energy may increase or decrease.

■ Positive potential energy means that work can be obtained from the system. If two bodies having the same charges are free, they rush away from each other, releasing the stored potential energy as kinetic energy of the charges.

■ Negative potential energy means that external agent would have to do work to separate the charges.

■ As electrostatic force is conservative, work done by external agency is given by Wext = U(final) – U(initial) ■ Work done by the electric field is given by = W field U (initial ) − U ( final ) ⇒ dU = −dW field = dWext f   Wi → f U f − U i V f − Vi = = = − ∫ E.dr qo qo i

■ A positive charge has higher potential energy at higher potential point. A negative charge has lower potential energy at higher potential point.

■ When free to move a positive charge moves from higher potential point to lower potential point; similarly,a negative charge moves from lower potential point to higher potential point.

IL RANKER SERIES FOR NEET

| 105

10. Three point charges +Q, +2Q and –Q are placed at the vertices of an equilateral triangle A, B and C respectively. Length of each side is l. At the mid point D of the side BC, a fourth point charge +Q is placed. Find the potential energy of the system of charges.

2. The charges –q, Q, –q are placed along x-axis at x = 0, x = a and x = 2a respectively. If the potential energy of the system is zero, then Q : q is (1) 1 : 2 (2) 2 : 1 (3) 1 : 4 (4) 4 : 1

A (+Q)

3. Two charges q1 and q2 are placed 30 cm apart,

Sol:

as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40

(+2Q) B

cm from C to D. The change in the potential q3 k , where k is energy of the system is 4 πε0 q3 C

(+Q) D

C

3  2 ( +2Q ) + ( −Q )  1  +Q = ∴VD +   /2 /2  4 πε0  3 / 2 AD =

=

Q  3 +1   2πε0   3 

40 cm q1

(1) 8q2

Q2  3 + 1  ∴U = VD ⋅ Q =   2πε0   3 

Try yourself: 9. Two point charges are kept at a separation and potential energy stored in the system is 10 µJ. An external agent slowly increases the separation between the charges to two times its initial value. Find the work done by the external agent.

Ans: 5µJ

TEST YOURSELF 1. If 3 charges are placed at the vertices of an equilateral triangle of charge q each, what is the net potential energy, if the side of equilateral triangle is l cm?

1 q2 (1) 4 π∈0 

1 2q 2 (2) 4 π∈0 

1 3q 2 (3) 4 π∈0 

1 4q 2 (4) 4 π∈0 

q2 A

30 cm

(2) 8q1

D

B

(3) 6q2

(4) 6q1

4. 10C and – 10C are placed at y = 1 m and y = – 1m on y-axis 1c charge is placed on x-axis at x = +1. Now find the change in PE of system when 1 coulomb is displaced from x = +1m to x = –1 m keeping other two charges as fixed is (1) 109 J (2) 21 × 109 J (3) 10 × 109 J (4) zero 5. Two positive point charges of 12 μC and 5 μC, are placed 10 cm apart in air. The work needed to bring them 4 cm closer is (1) 2.4 J (2) 3.6 J (3) 4.8 J (4) 6.0 J 6. Two charges each ‘Q’ are released when the distance between is ‘d’. Then the velocity of each charge of mass ‘m’ each when the distance between them is ‘2d’ is (1) (3)

Q 8πε0 dm Q 4 πε0 dm

(2) (4)

Q 4 πε0 dm Q 2πε0 dm

106 |

CHAPTER 2: Electrostatic Potential and Capacitance

7. Three charges 1μC, 2μC 3μC are kept at vertices of an equilateral triangle of side 1m.

If they are brought nearer, so that they now form an equilateral triangle of side 0.5 m, then the work done in the process is nearly (1) 11 J (2) 1.1 J (3) 0.1 J (4) 19 J Answer Key (1) 3

(2) 3

(3) 1

(5) 2

(6) 2

(7) 3

2.6

(4) 4

POTENTIAL ENERGY IN AN EXTERNAL ELECTRIC FIELD

We shall discuss about the potential energy of a charge in an external field which is not produced by the given charges.

The electric field is produced by the sources external to the given charge(s). The external sources may be known or unknown. We assume that the charge q does not significantly affect the sources producing the external field, if the charge q is very small (or) the external sources are held fixed by other unspecified forces. Even if q is finite, its influence on the external sources may still be ignored when very strong sources far away at infinity produce a finite field E. We have to determine the potential energy of a given charge q in the external field but not the potential energy of the sources producing the external electric field.

The external electric field E and the corresponding external potential V may vary from point to point.

energy of charge q. If the point P has position  vector r relative to some origin, the potential  energy of q at r in an external field, is U = qV(r), where V(r) is the external potential at the point r.

The conventional unit of energy is joule.

But this unit is very large for computing nuclear energies, electron energies in atomic physics. So, a smaller and convenient unit called electron-volt is used in such cases.

“The amount of energy gained by an electron when accelerated through a potential difference of one volt is known as electronvolt.” ⇒ 1eV = 1.6 × 10–19 J

Key Insights: ■ Electron volt is a unit of energy. It is not a unit of potential difference or voltage.

■ Electron volt is not a standard SI unit of energy.

■ If an electron is decelerated through a potential difference of one volt, amount of energy lost by it is 1eV. ■ If an α–particle is accelerated by a p.d. of 2 V then the kinetic energy gained by it is 4 eV. ■ Electric field is conservative in an electric field work is path independent and work done in moving a point charge q between two fixed points having a potential difference V is equal to, WAB = –UAB = q(VB – VA)= qV And hence in moving a charged particle in an electric field work is always done unless the points are at same potential as shown in Fig. below.

2.6.1 Potential Energy of a Single Charge By definition, the potential V at a point P is the work done in bringing a unit positive charge from infinity to the point P. The work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential

I

B I A

II II

A (A)

+q

L

+Q (B)

B L

IL RANKER SERIES FOR NEET A I

II

+Q

B (C)

■ When a charged particle is accelerated by an electric field (uniform or non - uniform),

by Work energy theorem, i.e., DKE=W, we have 1 2 1 2 mv − mu = qV (but W = qV) or 2 2 v =

The work done on q2 against the field due to 1 q1q2 q1 is W2' = where r12 is the distance 4 πε0 r12 between q1 and q2.

The total work done in bringing the charge q2 against the two fields from infinity to the point B is = W2 q2V2 +

1 q1q2 4 πε0 r12

The total work done in assembling the configuration or the potential energy of the system is 1 q1q2 4 πε0 r12 The potential energy is characteristic of the present state of configuration, and not the way the state is achieved.

W = q1V1 + q2V2 +

 2 2qV  u + m   

And if the charged particle is initially at rest, i.e., u = 0 2qV v= m And if the field is uniform, i.e. E = (V/d) v=

| 107

2qEd m

2.6.2 Potential Energy of a System of Two Charges in an External Field Consider two charges q 1 and q 2 located at two points A and B having position vectors r1 and r2 respectively. Let V1 and V2 be the potentials due to external sources at the two points respectively.

The work done in bringing the charge q1 from infinity to the point A is W1 = q1V1.

In bringing charge q2, the work to be done not only against the external field but also against the field due to q1.

The work done in bringing the charge q2 from infinity to the point B is W2' = q2V2.

11. A particle of mass 40 mg and carrying a charge 5 × 10-9 C is moving directly towards fixed positive point charge of magnitude 10-8 C. When it is at a distance of 10 cm from the fixed point charge it has a velocity of 50 cm/s.

At what distance from the fixed point charge will the particle come momentarily to rest ?

Is the acceleration constant during motion?

Sol: If the particle comes to rest momentarily at a distance r from the fixed charge, then from ‘conservation of energy’ we have

1 2 1 Qq 1 Qq mu + = 2 4 πε0 a 4 πε0 r

Substituting the given data, we get

1 1 1 × 40 × 10−6 × × 2 2 2

1  = 9 × 109 × 10−8 × 5 × 10−9  − 10  r  

1 5 × 10−6 100 or,= − 10 = − 8 r 9 9 × 5 × 10 1 100 190 = + 10 = m i.e., r = 4.7 × 10–2 m r 9 9 F 1 As here, F = 1 qQ ; So, = a ∝ 2 2 m r 4 πε0 r i.e., acceleration is not constant during motion. or,

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Try yourself: 10. Charge q1 is fixed and another point charge q 2 is placed at a distance r 0 from q 1 on a frictionless horizontal surface. Find the velocity of q2 as a function of separation r between them (treat the charges as point charges and mass of q2 is m)

q1q2  1 1   −  2π∈0 m  r0 r 

Consider a dipole with charges +q and –q seperated by adistance 2a, placed in a uniform electric field E as shown in fig. qE

θ 2a sin θ qE

–q

The potential energy of dipole in an electric field is U =pE(cos 900 − cos θ) =− pE cos θ   In vector form U =− p ⋅ E

= Ans: v

2.6.3 Potential Energy of a Dipole in an External Field

+q

perpendicular to the field to the given direction.

E

If θ = 0°; τ = 0 and U = –pE If θ = 90°; τ = pE and U = 0

If θ = 180°; τ = 0 and U = pE   So, if p is parallel to E then, potential energy is minimum and torque on the dipole is zero, and the dipole will in stable equilibrium.   If p is anti parallel to E then, potential energy is maximum and again torque is zero, but it is in unstable equilibrium.

Key Insights:

■ Force on dipole in non–uniform electric field: The force on the dipole due to electric field is given by F = −∇U (Force = negative potential energy gradient).  If the electric field is along r , we can write This torque will tend to rotate the dipole    dU  d   F =− er =− (− p ⋅ E )er (unless p is parallel or antiparallel to E ). dr dr Suppose an external torque is applied in such  d    a manner that it just neutralises this torque ⇒= F p ⋅ E er dr and rotates it in the plane of paper from angle  θ1 to θ2 at an infinitesimal angular speed and  If p and E are along the same direction without angular acceleration. we can write The work done by the external torque is  d θ2 θ2 = ( pE cos 0°)e r F W= ∫ τd= θ ∫ pE sin θ dθ dr θ1 θ1 dE = p e r θ2 dr =− pE(cos θ)θ =pE(cos θ1 − cos θ2 ) 1 Here, e r is the unit vector along r .

This work is stored as the potential energy 12. An electric dipole of dipole moment p is kept of the system.

The dipole experiences no net force, but experiences a torque in a uniform electric field.    The torque is given by τ = p × E = pE sin θ ,   where p= q × 2a is electric dipole moment.

(

The potential energy of dipole is the work done in rotating the dipole from a direction

)

at a distance r from an infinite long charged wire of linear charge density λ as shown.

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IL RANKER SERIES FOR NEET

Potential energy of dipole P2 in the field of P1 is given by

Find the force acting on the dipole? + + + + + +

P

U = –P2E1cos180° =

r

Now, force between them is given by

Sol: Field intensity at a distance r from the line of charge is E =

−1 P1P2 4 π∈0 r 3

= F

λ 2π∈0 r

F

The potential energy of dipole is

 

U = − P .E = − PE

−dU d  −1 P1P2  =   dr dr  4 π∈0 r 3 

1 3 p1 p2 4 0 r 4

As the force

is positive, it is repulsive. p Similarly if 1 || − p2 , the force is attractive.

The force on the dipole is

dU dE = +p dr dr  −λ  − pλ = +p = 2 2  2π∈0 r  2π∈0 r

− F =

b) If the dipoles are on the same axis: P1

P2

E2 r

−2P1P2 4 π∈0 r 3 The force between the dipoles is given by

Here, the net force on dipole due to the wire will be attractive.

In this case, U P2 E= = 1 cos 0°

Try yourself: 11. A n e l e c t r i c d i p o l e o f m o m e n t  p = 2i − 3j + 4k µC − m is placed in a  uniform electric field E= 4i + j + 2k V/m

dU dU  2P1P2  − = − F= −  dr dr  4 π∈0 r 3 

(

)

(

)

. Find the potential energy stored in the system.

Ans: 13 µJ

2.6.4 Potential Energy and Force between Two Short Dipoles Consider two short dipoles separated by a distance r. There are two possibilities. a) If the dipoles are parallel to each other:

P1 E2

r

E1

P2 E1

F

1 6 p1 p2 4 0 r 4

As the force is negative, it is attractive.

2.6.5 Electrostatic Field Energy (Additional) Consider a charged particle of mass m and charge q placed in an electric field E. If that particle is released from rest, it starts moving due to the force applied by the electric field on it. As a result, that particle gains some kinetic energy. Here, the electric field is doing work to increase the kinetic energy of that particle. In other words, the electric field has some energy which enables it to do that work.

So, whenever electric field exists, it has energy associated with it. We can find energy density of the electric field as explained below.

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CHAPTER 2: Electrostatic Potential and Capacitance

Consider a charged conductor as shown.

We know that electric field just outside the surface of that conductor at any point is given σ

by E = ∈0 +

+

+

+

+

+

+

+ +

+

+

+

+

The charged body experiences an outward σ2 1 electric pressure given by P= = ∈0 E 2 2 ∈0 2 If we treat the surface of the charged body flexible, due to the outward pressure it expands as shown. Inside the body there will be no field.

We know that to expand the charged body work is done by electric forces in the body.

This work done in increasing the volume is equal to loss of field energy in the volume dV. Now we can write dW = PdV The field energy stored in the volume dV dU =P is given by dU = dW = PdV and dV dU = u , which is known as field energy dV density in the electric field. u= P=

2

σ 1 = ∈0 E 2 2 ∈0 2

If the electric field is uniform in a region, total field energy stored in a given volume V 1 is given by U= ∈0 E 2V 2 If the electric field is non uniform in a given region, total field energy stored is given by ∫ dU where dU=

1 ∈0 E 2dV 2

2.6.6 Self Energy (Additional) In the previous topic we have discussed about interaction potential energy of a system of point charges. Now let us discuss about self energy of a charged body. When a body is charged, all the charge on it must be brought from infinity (the reference taken by us) onto that body. In doing so, work has to be done against the electric field of that body. This work will be stored in that body in the form of potential energy which is known as self energy. “Self energy of a charged body is the total field energy associated with the electric field due to this body in its surrounding”. Let us consider two important cases for which we are going to find self energy.

2.6.7 Self Energy of a Charged Conducting Sphere Consider a conducting sphere of radius R charged with charge Q on it. In the process of charging, we have to bring charge to the sphere from infinity in steps of elemental charge each dq. While bringing the elemental charge the field produced by the charge on the sphere already accumulated opposes the charge element. At an instant sphere has a charge ‘q’ on it. Due to this charge, potential of the sphere is V=

1 q 4 π∈0 R

As we brought charge dq to its surface from infinity, work done by the external agency is 1 q dq 4 π∈0 R Total work done in charging the sphere with final charge Q in it is given by dW = V dq =

W=

∫ dW=

Q

1 q 1 Q2 ⇒ = dq W ∫ 4π∈0 R 4 π∈0 2R 0

So self energy of charged conducting sphere Q2 is U = 8π∈0 R

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II Method: We can find the self energy from energy density as explained below

2.6.8 Self Energy of a Uniformly Charged Nonconducting Sphere

We know that energy per unit volume in 1 2 an electric field is u= ∈0 E .

2

Consider a non-conducting sphere charged uniformly with a charge Q on it. We know that outside region of that sphere, every point is same as that of conducting sphere of radius R only. So, the field energy in the surrounding of this sphere from its surface to infinity can be given as

When the sphere has no charge, there was no electric field in its surroundings. But when the sphere is charged, there exists an electric field in its surroundings from its surface to infinity.

U out =

Electric field due to the charged sphere at outer point is given by E=

Q2 8π∈0 R

1 Q 4 π∈0 r 2

dr r R

Q R

dr

r

Consider an elemental spherical shell of radius r and width dr as shown. The volume enclosed in this shell is dV = 4 πr 2 dr The field energy stored in this volume is dU=

We know that at interior point of this sphere, E ≠ 0. So, field energy exists in the interior region. Consider an elemental shell of radius r and thickness dr as shown. Field energy in the volume of this shell is given as dU in=

2

1 ∈0 E 2dV 2

1  1 Qr  2 = ∈0   4 πr dr 3 2  4 π∈0 R  2

 Q2 1  Q 2 × π = dU = ∈0  r dr dr 4  2  4 π∈0 r 2  8π∈0 r 2 2

 Q  Q ∈0  × 4 πr 2dr = dr  2  4 π∈ r  8π∈0 r 2 0   2

= U

dU ∫=

Q2

  1 Qr Q 2r 4 for r < R  E = = dr 3 4 π∈0 R   8 ∈0 R6  and dV = 4 πr 2 dr 

 

Now, total field energy inside the sphere will be given as

Total field energy associated with the sphere can be calculated by integrating the above expression from r = R to r = ∞ (no electric field inside the sphere). So, total field energy in the surrounding of the sphere is ∞

1 ∈0 E 2dV 2

2

Q ∫ 8π∈ r 2 dr ⇒ U = 8π∈0 R 0 R

= U in

by

dU in ∫=

Q2 r 4 Q2 = dr ∫ 8 ∈ R6 40π∈0 R 0 0

R

So, total self energy of this sphere is given U = U out + U in =

3Q 2 20π∈0 R

CHAPTER 2: Electrostatic Potential and Capacitance

Key Insights:

Here V2 is potential at q1 due to charge q2 and V1 is potential at q2 due to charge q1.

13. Two uniformly charged conducting spheres of radii R1 and R2 are charged with charges Q1 and Q2 respectively separated by a distance r.

Find total electrostatic energy of this system.

Sol: Here= U U self + U interaction

Q12 Q22 = U self +

8π∈0 R1 8π∈0 R2 1 Q1Q2 U interaction = 4 π∈ r 0

2  Q1

Q22 2Q1Q2  1 = ⇒ + + U  

8π∈0  R1 R2 r  If the two spheres are non conducting, in this case,

3Q12 3Q22 U self = +

20π∈0 R1 20π∈0 R2 So, = U

1  3Q12 3Q22 Q1Q2  + +   r  4 π∈0  5R1 5R2

14. T wo concentric shells of radii R 1 and R 2 are charged uniformly with charges Q1 and Q2 respectively. Find the total electrostatic energy of the system.

Sol:= U U self + U interaction

Q12 Q22 = U self + 8π∈0 R1 8π∈0 R2

Try yourself: 12. A shell of radius R has a charge Q uniformly distributed over it. A point charge q is placed at the centre of the shell. Find the work done to increase the radius of the shell from R to 2R? Q

If q1 and q2 are two point charges separated by a distance r, interaction energy of that system is Ui = q1V2 = q2V1

1  Q12 Q22 2Q1Q2  +  +  8πε0  R1 R2 R2 

 

■ A point charge does not have any self energy. So for a pair of charges there will be interaction energy only.

So, = U

Q 

U total = ΣU self + ΣU interaction

1 Q1Q2 4π ∈0 R2

Ans: − 8π ∈ R  2 + q 

■ Total electrostatic energy of a system of charges is the sum of self energy of all charged bodies and interaction energy of all possible pairs of charged bodies.

U interaction =

0

112 |

TEST YOURSELF 1. 12 J work should be done in moving 0.01C charge between two points in an electric field. The potential difference between the points (1) 1200 V (2) 100 V (3) 10 V (4) 12 V 2. A molecule with a dipole moment p is placed in electric field of strength E. Initially the dipole is aligned parallel to the field. If the dipole is to be rotated to be anti-parallel to the field, the work required to be done by an external agency is (1) – 2pE (2) – pE (3) pE (4) 2pE 3. A cloud is at a potential of 8 × 106 volt relative to the ground. A charge of 80 coulomb is transferred in lightening stroke between the cloud and the ground. Assuming the potential of the cloud to remain constant, the energy dissipated is (1) 6.4 × 108 joule (2) 6.4 × 105 joule (3) 105 joule (4) 107 joule 4. A pellet carrying a charge of 0.5 coulomb is accelerated through a PD of 2000 volt. It attains a kinetic energy equal to: (1) 1000 erg (2) 1000 joule (3) 1000 kWh (4) 500 erg (1) 1

Answer Key (2) 4 (3) 1

(4) 2

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2.7

ELECTROSTATICS OF CONDUCTORS

Conductors contain mobile charge carriers, which are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal.

The free electrons form a kind of gas called ‘electron gas’. They collide with each other and with the ions, and move randomly in different directions. In an external electric field, they drift against the direction of the field. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. In electrolytic conductors, the charge carriers are both positive and negative ions. The movement of the charge carriers is affected by both the external electric field and the chemical forces.

1) Electrostatic field is zero inside a conductor: Consider a neutral or charged conductor in an external electrostatic field. In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. A conductor has free electrons.

As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have distributed themselves so that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor.

2) At the surface of a charged conductor, electrostatic field must be normal to the surface at every point: If electric field were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, the electric field should have no tangential component. Thus electrostatic field at the surface of a charged conductor

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must be normal to the surface at every point.

3) The interior of a conductor can have no excess charge in the static situation: A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation. This follows from the Gauss’s law. Consider any arbitrary volume element dV inside a conductor.

The electrostatic field is zero on the closed surface bounding the volume element dV. Thus the total electric flux through the surface is zero. This means there is no net charge at any point inside the conductor, and any excess charge must reside at the surface.

4) Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface: Since electric field is zero inside the con d uctor an d has n o tan gen ti al component on the surface, no work is done in moving a small test charge within the conductor and on its surface. So there is no potential difference between any two points inside or on the surface of the conductor. If the conductor is charged, electric field exists normal to the surface.

This means potential will be different for the surface and a point just outside the surface, for any arbitrary size, shape and charge configuration of the conductor.

5) Electric field at the surface of a charged conductor: Consider a charged conductor having surface charge density σ. Imagine a cylindrical Gaussian surface of cross sectional area dS at a point P on the surface. The electrostatic field is zero just inside the surface and the field is normal to the surface with magnitude E just outside.

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CHAPTER 2: Electrostatic Potential and Capacitance

  ∫ E.ds = 0

+ + + + + + + +++ σ + + ++ + + + + + + + ++ ++ + + + ++ + + + ++

For the dotted surface

dS E Gaussian Surface

The contribution to the total flux through the cylinder comes only from the outside (circular) cross-section and equals to EdS. The field E may be considered constant and E and dS are parallel or antiparallel.

The charge enclosed by the Gaussian surface is σdS.

By Gauss’s law E dS =

σ dS ε0

σ ε0 The electric field is normally outwards to the surface, if σ is positive and the field is normally inwards to the surface, if σ is negative.

The electric field E =

6) Electrostatic shielding: Consider a conductor without a cavity then there will be no electric field inside a charged conductor and all the charge resides on its outer surface only. Suppose that charged conductor has a cavity or cavities and there are no charges within the cavity or cavities, even then charge resides on the outer surface of the conductor.

There will be no charge on the walls of the cavity or cavities. This can be verified very easily using Gauss’s law by enclosing the cavity with a Gaussian surface. +

+

+

+

+ +

+

+ +

+ +

Cavity +

+

+ + +

+

+ +

q = 0 inside cavity.

From the above result, any cavity in a conductor remains shielded from outside electric influence i.e., the field inside the cavity is always zero. This is known as electrostatic shielding. The effect can be made use of in protecting sensitive instruments from outside electrical influence.

Key Insights: When metal conductor is kept in an electric field, ■ the field inside the conductor is zero ■ the charge resides only on the outer surface.

■ the field is normal to at every point on the surface ■ the potential is constant at everywhere on the surface and throughout the volume of conductor. +

+ + +

2.8

+

+

+

E=0 Q=0 φ=0 V = constant + + +

+ + + +

σ E0 = ∈0

Q

DIELECTRICS AND POLARISATION

The substances which do not possess free electrons or possess very less number of free electrons to constitute electric current are known as dielectric materials. In such materials the electrons will be tightly bound to the nucleus.

We know that the atom consists of central nucleus with +ve charge surrounded by revolving electrons. The centre of the +ve charge and the centre of the –ve charge are

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IL RANKER SERIES FOR NEET

supposed to be concentrated at a single point in certain molecules only. Such molecules are known as non polar molecules. Non polar molecules will have symmetrical structure of atoms with regards to electrical effect.

They have symmetric structure so that the electric dipole moment of the molecule is zero. Examples of such molecules are CO2, H2, O2, etc.

an electric field, the field gets modified. This effect can be felt by introducing a dielectric slab in an uniform electric field between two unlike charged parallel plates. +

E0

Ei

–

In the presence of electric field the positive charge centre get displaced in the direction of electric field and negative charge centre get displaced opposite to the field and each molecule get polarized.

E=0

E = E0 –

+

– –

–

+

No E-field

+

+ –

+

–

+

–

+

Applied E-field

The centres of positive charge and negative charge of a molecule do not coincide in some molecules. Such molecules are said to be a polar molecules. These polar molecules will have unsymmetric structure and possesses net dipole moment even in the absence of electric filed. Example of such molecules are H2O, HCl. Polar

The molecules of dielectric get polarized and every volume element DV of dielectric slab has a dipole moment in the direction of the field.

As a result, the induced charges will appear at the outer surfaces of dielectric. The induced positive and negative charges cancel each other inside the volume of dielectric.

Due to the induced surface charges at the surface of the dielectric, there will be an induced electric field Ei, developed in a direction opposite to E0.

Hence, the net electric field inside the dielectric is reduced to E = E 0 – E i. As a consequence, the potential difference V 0 (without the dielectric) will get reduced to V (with the dielectric). But, the charge on the plates remains the same as Q. The net electric field inside the dielectric is,

No E-field

Applied E-field

In the presence of electric field, each polar molecule experience a torque and the molecules will align into the direction of field and the effect of polar dielectric material in electric field is much more pronounced.

Whenever a dielectric material is placed in

E0 V and V = 0 K K The induced electric field E = E0 – Ei , also E =

Ei = E0 − E = E0 −

E0 K

1  Ei = E0  1 −  . It is opposite to E0.  K

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CHAPTER 2: Electrostatic Potential and Capacitance

Induced charge density σi =σ0  1 − 1   K 1  Induced charge= Qi Q0  1 −   K

2.8.1 Dielectric Polarisation Dielectric polarisation is defined as the dipole moment per unit volume of the dielectric When a polar or non polar dielectric slab is kept in an uniform electric field between two unlike charged parallel plates, the electric field applied across the plate will displace the centres of negative and positive charges and this displacement causes dipole effect in every atom. If we consider the entire dielectric substance the total dielectric will develop certain dipole moment with all dipoles put together. The dipole moment per unit volume is called polarisation.

 density on it. In vector form, σi = P .nˆ where n̂ is unit vector along outward normal to the surface.

2.8.2 Induced or Bound Charge Consider a parallel plate capacitor with charge q on its positive plate and –q on its negative plate. So, electric field between the plates will be E= 0

q σ = ∈0 A ∈0

When dielectric slab is introduced between the plates, polarisation takes place.

Let q' be the magnitude of induced charge.

Due to the induced charges there will be an induced electric field Ei. This field opposes the external   field E0. The net electric field will be = E E0 + Ei q

For linear isotropic dielectrics, polarisation is directly proportional to the electric field, p ∝ E or p = χe E where χe is a characteristic of dielectric and is known as electric susceptibility of the dielectric medium.

Let us consider a slab of cross-sectional area A and length ‘L’ (for convenience). Then the molecular charges developed on the edge faces of the dielectric are as shown in figure.

E0 +q1 -q

1

A L

The dipole moment of the total slab = Lq1 and volume of the slab = LA. By definition, polarisation

= p

Lq ' q ' = = σi LA A

Hence, dielectric polarisation (p) is numerically equal to induced surface charge

+ + + + + +

–q – –E – 0 – – – –

+ + Ei + + + +

–q1

q1

– – – – – – –

E = E0 – Ei   (as E0 and Ei are opposite in direction) where E= i But E =

σi q' = ∈0 A ∈0

E0 (K is dielectric constant) K

E0 = E0 − Ei K q q q' 1  = − ⇒ q ' = q 1 −  A ∈0 K A ∈0 A ∈0  K

2.8.3 Dielectric Strength If the applied electric field E0 becomes very high, in such a strong field the electrons of the atoms inside the dielectric get detached from their atoms. In such a situation the

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dielectric no longer can behave as an insulator.

It behaves just like a conductor. This is known as electrical breakdown or dielectric breakdown. The electric field strength at which the electrical (dielectric) breakdown occurs is called the dielectric strength of the material.

Application ■ If a metal sphere of radius r is charged to hold a maximum charge Q. The dielectric strength of surrounding medium is E. Then E =

1 Q = Q E(4 π∈0 r 2 ) 2 4 π∈0 r

15. A dielectric plate, having dielectric constant k = 4 is placed in an external uniform electric field of 100 V/m, with its plane normal to the field. Find the induced electric field inside the plate.  1 Sol: = Ein E0  1 −   k  1 = 100  1 −  V/m = 75 V/m  4

Try yourself: 13. A large charged non-conducting sheet has a surface charge density of 6 µC/m2. A large dielectric plate, having dielectric constant k = 4 is kept in front of the charged sheet with its plane parallel to the sheet. Find the intensity of electric field inside the dielectric plate.

Ans: 8.47 × 104 Volt/m

TEST YOURSELF 1. If the dielectric constant of substance is 4 k = , then its electric susceptibility is (ε0 3 permittivity of free space) (1)

ε0 3

(2) 3ε0

(3)

4ε0 3

(4)

2. Example to non-polar molecules (1) O2 & H2 (2) HCl and H2O (3) O2 & HCl (4) H2 & H2O 3. For metals the value of dielectric constant (K) is (1) One (2) Infinity (3) Zero (4) Two (1) 1

2.9

(The dielectric strength of air is 30,000 V/ cm or 3 × 106 Vm−1 ).

3ε0 4

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Answer Key (2) 1 (3) 2

CAPACITORS AND CAPACITANCE

Capacitor is a device which is used to store electrical charge in the same way that a tank is for storing liquid. Different containers with different volumes will have different capacities to hold a liquid, similarly different capacitors will have different capacities to hold electric charge. The capacity of a conductor to hold charge depends on the size, shape and surrounding medium of the conductor.

The capacity of a conductor to hold charge is called it’s capacitance or capacity. A given container will have fixed capacity of holding a liquid which does not depend on the quantity to be stored. Similarly capacity of a capacitor does not depend on the charge to be stored.

A combination of two conductors of any shape placed close to each other constitutes a capacitor or condenser. Capacitor can be of any shape, When a capacitor is charged, one of the two conductors will be given a positive charge (+q) and the other is given a negative charge (–q) of same magnitude. The charge on the positive plate is called the charge on the capacitor. The potential difference between the plates of a capacitor is known as potential of the capacitor (V=V+– V– ). Charge on the capacitor does not mean it is same as the total charge of two conductors (total charge is in fact zero). Figure (a) shows two conductors forming a capacitor and figure (b) shows the symbol used to represent a capacitor.

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CHAPTER 2: Electrostatic Potential and Capacitance

of capacitance is coulomb/volt which is called farad (Symbol is F) +Q

–Q

Farad : The capacity of conductor is said to be 1 farad, if 1 coulomb of charge is given to the conductor to raise its potential by one volt.

1C 1V Farad is a large unit for practical purpose.

So, its sub-multiples are used more frequently like i.e 1F =

a b

fixed

variable

fig (b)

In either symbol the vertical lines (straight or curved) represent the conductors and the horizontal lines represent wires connected to either conductor. One common way to charge a capacitor is to connect these two wires to opposite terminals of a battery. Once the charges Q and –Q are established on the conductors, the battery is disconnected. This gives a fixed potential difference Vab between the conductors (that is, the potential of the positively charged conductor with respect to the negatively charged conductor) that is just equal to the voltage of the battery.

2.9.1 Capacity of an Isolated Conductor If the charge on an isolated conductor is gradually increased, its potential also increases.

At any instant, the charge Q on the conductor is directly proportional to the potential V of the conductor. ⇒ Q ∝ V (or) Q = CV The proportionality constant C is called the capacitance of the conductor. It depends on the shape, size, and permittivity of the surrounding medium of the conductor. It does not depend on the charge or potential of the conductor.

Q So, capacitance, C = ----- (1) V i.e. “The ratio of charge on a conductor to its potential is known as capacitance or capacity of that conductor”. It is numerically equal to the charge to be given to a conductor to raise its potential by one unit. The S.I unit

1 micro farad (1µF)=10–6F 1 pico farad (1pF)=10–12F

2.9.2 Capacity of an Isolated Sphere Let us consider an isolated metal sphere of radius R having a positive charge of +Q. This forms a capacitor with earth as second conductor and air as the dielectric medium between the two conductors. We know that the potential of an isolated charged sphere is V =

Q .

4 πε0 R

By definition of capacitance, C=

Q ⇒ C = 4 π∈0 R V

+ + + ++ ++ + + + + + + + + + R + + + + + + + + + + + + + ++ +

Dielectric earth

Key Insights: ■ The capacity of a spherical conductor depends on dimensions of the conductor and nature of the medium in which conductor is placed.

■ The capacity of a spherical conductor is independent of charge on the conductor.

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■ In CGS system, the capacity of a spherical conductor is numerically equal to radius of the conductor.

■ For the earth, R = 6.4 × 106 m. The capacity of earth is C = 4 π∈0 R =

1 9 × 10

× 6.4 × 106 = 711µF 9

16. Two metal spheres of radii in the ratio 4 : 3 are kept in contact and a charge is given to them. Next they are separated wide apart so that there is no electric influence of one on another. Find the ratio of charges on them.

Sol: Charge q= CV=

( 4π∈0 R )V

Since they are in contact, potential ‘V’ is same.

q∝R

∴ q1 : q2 = R1 : R2 = 4 : 3

Try yourself: 14. Two metal spheres each of radius 10 cm, are carrying +10 µC and –10 µC charges.

The spheres are kept at a large separation so that one sphere is not influenced by the electric field of the other. Find the potential difference of the system.

Ans: 18 × 105 Volt

2.9.3 Capacity of a Capacitor It is a device to store large amount of charge at low potential (or) it is an arrangement of two conductors to increase the capacity of a conductor.

At any instant, the charge Q on the capacitor is directly proportional to the potential difference V between the plates of the capacitor. ⇒ Q ∝ V (or) Q = CV The proportionality constant C is called the capacitance of the capacitor. It depends on the shape, size, and the nature of medium between the plates of the capacitor. It does not depend on the charge or potential of the capacitor.

Q So capacitance C = ----- (1) V

| 119

i.e. “The ratio of charge on a capacitor to its potential is known as capacitance or capacity of that capacitor”.

2.9.4 Parallel Plate Capacitor A parallel plate capacitor consists of two parallel conducting plates separated by a dielectric material, capable of storing electrical energy when a voltage is applied across them.

Principle The simplest arrangement of a capacitor is, two parallel metal plates separated by air or a dielectric material as shown. The capacitance of a conductor can be increased by increasing its size. But for a given size, the other way of increasing the capacitance is by lowering the potential without altering the charge on the conductor. This is the principle of a capacitor.

A + + + + (a) + + + + + + + +

B

+ + + + + +

B – + – + – + – + – + – +

A

+ + + + + +

– – – – – –

B

Consider a metal plate A to which positive charge Q is given. The charge distributes on A as shown in figure (a). Its potential raises to a value V. If C is the capacitance of the conductor, Q = CV. At this potential V, the conductor can not hold charge more than Q. Now, as shown in figure (b), an uncharged metal plate B is kept parallel to A and closer to it. The positive charge on A induces an equal amount of negative charge on inner side of B. As the plate B is uncharged, an equal amount of positive charge is induced on the outer side of B. Here the induced negative charge on B lowers the potential of A, where as the induced positive charge on B increases the potential of A. But as the negative charge on B is nearer to the plate A, its effect will be more. Hence potential of A decreases.

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CHAPTER 2: Electrostatic Potential and Capacitance

Now, the outer side of plate B is earthed, as shown in figure (c). The induced positive charge on B gets neutralized due to earthing, while the induced negative charge on the inner side of B is held by the attraction of positive charge on A. The negative charge on B lowers the potential of A and so some more charge can be added to A to rise its potential to V. It means the capacitance or charge holding ability of plate A is increased. In this case the capacitance can be further increased by filling the space between the plates with a dielectric.

The principle of a capacitor is to increase the capacitance of a conductor by bringing an uncharged conductor nearer to it and earthing the outer surface of the uncharged conductor.

The capacitance of a capacitor depends on: ■ geometry of the plates, ■ separation between the plates, ■ the dielectric medium between the plates.

In a capacitor one plate need not necessarily be earthed. The two plates must have equal and opposite charges

Key Insights: ■ The method for the calculation of capacitance requires integration of the electric field between two conductors or the plates which are separated with a potential difference Vab a     − ∫ E.dr i.e. Vab = − ∫ E.dr or V+ − V− = b

q ⇒C = Vab

Capacity of Parallel Plate Capacitor Let us consider a parallel plate capacitor which consists of two parallel plates each of area A, separated by a distance ‘d’ as shown in figure.

+

+

+

+ +

–

–

–

– –

d

Let the charge on the inner surface of upper plate be +q and the bottom plate earthed. Then there will be –ve charge induced on the side which faces the positively charged plate. As the plates are uniform and parallel the distribution of charge will be uniform throughout its face (one face only). Hence the electric field will be uniform between the two plates.

However, the field will not be uniform at the outer edges of the plates. This is shown by the curved lines of force in figure. This is called ‘fringing’ and can be neglected when d < < A. q The surface charge density σ = .

A The electric field in the outer region of plates is σ σ E= − =0 2ε 0 2ε 0 But the electric field between the plates is σ σ σ q E= + = = 2ε0 2ε0 ε0 ε0 A Since the field is uniform, the electric field intensity between the plates is V , where V is potential difference d between the plates.

E=

From the above two equations, q q ε A V = ⇒ = 0 ε0 A d V d q But capacity C0 = V The capacity of air filled parallel plate ε A capacitor is C0 = 0 d

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Key Insights: ■ Unit of Permittivity C.d farad × m ∈ =0 = = farad/metre A m2 Unit of permittivity = farad/metre ■ For the values A = 1m2 and d = 1 mm,

| 121

17. Charges Q1 and Q2 are given to two identical metal plates. Find the charge distribution on the four surfaces of the plates. Find the potential difference of such arrangement shown in the figure?

(Area of each plate is A and separation between the plates is d. Ignore fringing effect) Q1

8.85 × 10−12 × 1 C = = 8.85 × 10−9 F .

10−3

Q2

If C = 1 F, d = 1 cm, A =

Cd 1× 10−2 =  109 m2 ∈o 8.85 × 10−12

Let us assume the charges on the four surfaces of the plates as shown. From the given data,

2.9.5 Charging a Capacitor C

Q1 = qa + qb; Q2 = qc + qd

+ – a b

S

Sol: Method 1:

electron motion

At P1, E = 0

qc

V

Consider a capacitor C connected in a circuit with a battery as shown. Here battery provides the energy and pulls the electrons from plate a and send to plate b as shown. As a result positive charge on plate a and negative charge on plate b start increasing. The potential difference between the plates increases until it equals the applied voltage V between the terminals of the battery. Finally plate a and the terminal of the battery connected to that plate will be at the same potential. There will be no longer an electric field in the conducting wires. Similarly plate b and the terminal of the battery connected to that plate also will be at the same potential and no electric field in the connecting wire between them. As there is no electric field, no further motion of electrons will be there in the connecting wires. Now we say that capacitor is fully charged with a potential difference V and charge Q = CV. The applied voltage ‘V’ is such that the electric field between plates becomes equal to dielectric break down then the capacitor cannot be further charged.

qd P1

qa

P2 qb

qa q q q − b − c − d = 0 2 A ∈0 2 A ∈0 2 A ∈0 2 A ∈0 qa – qb – qc – qd = 0 ..... (1)

Similarly at P2 also, E = 0

qa q q q + b + c − d = 0 2 A ∈0 2 A ∈0 2 A ∈0 2 A ∈0

qa + qb + qc– qd = 0 ..... (2) Add (1) and (2),

2(qa – qd) = 0 or qa = qd .... (3) subtract (1) from (2), 2(qb + qc) = 0 or qb = –qc ..... (4) But qa +qb = Q1 and qc + qd = Q2 ..... (5) From equations (3), (4) and (5)

Q − Q2 q= q= Q1 + Q2 ; q = −qc =1 a d b 2 2 The charge on inner surface of positive charged plate or the charge of capacitor is

Q1 − Q2 2

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CHAPTER 2: Electrostatic Potential and Capacitance

q (Q1 − Q2 ) where V is Capacitance C = = V 2V potential difference between plates of capacitor.

Capacity

C=

= V

∈0 A d

Method 2: The charge on the plates is distributed equally on both sides and charges are induced on the plates as shown.

Initial charges Q1 2

Q2 2

1. A body is charged to a certain potential.

When, an additional charges of 60nc is imparted to it, the rise in potential is found to be 15mv. Find the capacitance of the body (1) 6 μF (2) 8 μF (3) 4 μF (4) 2 μF 2. The capacitance (in F) of a spherical conductor of radius 1m is (1) 1.1 × 10–10 (2) 10–6 (3) 9 × 10–9 (4) 10–3

Q1 − Q2 (Q1 − Q2 )d = 2C 2 ∈0 A

Q1 2

TEST YOURSELF

Q2 2

3. The radius of the circular plates of a parallel plate condenser is ‘r’. Air is there as the dielectric. The distance between the plates if its capacitance is equal to that of an isolated sphere of radius ‘r’ is

r2 r r2 r2 (2) (3) (4) 4r ′ 4 r′ r′ 4. A capacitor has a capacitance of 8.5µF . How much charge must be removed to lower the potential difference of its plates by 50 volt (1) 426 µC (2) 425 µC (3) 500 µC (4) 425 C (1)

+

Q2 2

–

Q2 Q – 1 2 2

+

Q1 2

Induced charges The charge on inner surface of positive charger plate or the charge of capacitor is Q1 − Q2

2

q (Q1 − Q2 ) Capacitance, C = = V 2V

where V is potential difference between plates of capacitor.

Capacitance, C =

= V

∈0 A d

Answer Key (2) 1 (3) 1

(1) 3

2.10

(4) 2

EFFECT OF DIELECTRIC ON CAPACITANCE

Case I:

Q1 − Q2 (Q1 − Q2 )d = 2C 2 ∈0 A

Try yourself: 15. A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. Now a charge +q is given to its positive plate. Then find the potential difference across the capacitor.

Consider a case of parallel plate capacitor in which a medium of dielectric constant K is completely filled as shown in figure. Let Q be the charge on the capacitor plates having surface area A, seperated by a distance ‘d’. q + K

d –

Ans: V +

q 2C

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The dielectric is completely polarized by the electric field between the plates of capacitor and charges are induced on the surfaces of dielectric, normal to field. If σ0 is the charge density on the plates of capacitor and σi is induced charge density on the surface of dielectric then the electric field = E

σ0 − σi σ0 σ0 = = ε0 kε0 ε

Case II: Let us consider a case of parallel plate capacitor in which a medium of dielectric constant K is partially filled as shown in figure.

Then the field is uniform in air as well as in the medium, but fields will have different values. Let ‘t’ be the thickness of the medium whose relative permittivity is K. The remaining space of (d–t) thickness be occupied by air. q +++++++++ air d K t –––––––––

The p.d. between the two plates of the capacitor is

V = E= d

σ0 Q d d= kε0 A K ε0

The capacitance of the capacitor with the dielectric is

| 123

Since thickness of dielectric medium is t, its equivalent air distance is t/K. The effective distance between the plates of capacitor is d–t + t/K.

kε0 A εA Q Q k= ε0 A = ; V Qd d d Here ε = kε0 is called permittivity of the The capacity of capacitor is medium. ε0 A ε0 A = C = If the capacitor is completely filled with d - t + ( t / K )  d - t (1-1/ K ) dielectric of dielectric constant k, then its Gauss Method: Imagine a Gaussian surface capacity is enclosing the plate, as shown. εA kε0 A y = C = = kC0 x d d q

= C=

Key Insights: ε0 KA ε0 A = . d d/K While measuring capacity, the dielectric medium of thickness d is equivalent to d/K in air.

Definition of dielectric constant (In terms of capacity): It is defined as the ratio of capacity of a parallel plate capacitor completely filled with dielectric to the capacity of same parallel plate capacitor without dielectric.

C K= C0

+++++++++ x

d

1

y1 t

air

K –––––––––

C ■=

y

x +++++++++

y1

x1

If E0 is the field in air, then from Gauss law q

q

q

∫ E0ds =ε0 ⇒ E0 A =ε0 or E0 = Aε0 .... (a)

Similarly by considering a Gaussian surface through the medium, then by Gauss law,

124 |

CHAPTER 2: Electrostatic Potential and Capacitance

∫ E.ds =

If the plate separation is increased by x

q q ⇒ EA = ε0 K ε0 K

C' =

where E is a field in the medium q ∴E = Aε0 K

.... (b)

The p.d. between the two plates of the capacitor is

=

q q t (d − t ) + Aε0 K Aε0

q  t  (d − t ) +  Aε0  K q = V

= or, C

=

potential to its initial value the distance between plates should be increased by  1 t 1 −  .  k ■ On comparison we can conclude that capacitance of capacitor with multiple dielectric media (with no air gap) is given by

Aε0

  t  (d − t ) +  K     

Key Insights: ■ The capacity of an air filled parallel plate capacitor is C0. A dielectric slab of thickness t with dielectric constant k is filled between the plates as shown. ε A Initially, C0 = 0 after introducing a d dielectric, the capacity increases to C.

= C

When capacitor is partially filled with

the capacitor decreases. To bring back the

q q d − t + t / K  Aε0 

d

 1  1 ⇒ d = d + x − t  1 −  or x = t  1 −   k  k dielectric the effective distance between the  1 plates decreases by t  1 −  . So capacity  k increases and potential difference across

V = E0(d – t) + E.t V=

ε0 A But C1 = C0  1 d + x − t 1 −   k

t

ε0 A ε0 A = 1 t d −t + d − t (1 − ) k k

To have the capacitance of the arrangement C 0 again the plate separation must be increased by x.

C=

Aε0 Thickness of medium ∑ Re lative permittivity

i.e, C =

Aε0

 t1 t 2  + .....  +  K1 K 2 

K1 K2 K3

t1 t2 t3

■ A number of dielectric plates are inserted between the plates of capacitor with some air gap as shown.

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K1 K2

d

t1 t2

| 125

Try yourself: 16. An air cored parallel plate capacitor has capacitance C 0 and plate separaton d. A dielectric slab of dielectric constant k = 3 is placed in the space between the plates. If the new capacitance is 2C0, find the thickness of the slab.

d

The capacity of capacitor is given by ε0 A

  t1 t 2 + + .... d − (t1 + t 2 + ....) + K1 K 2  

18. A metal slab of thickness, equal to half the distance between the plates is introduced between the plates of a parallel plate capacitor as shown. Find its capacity.

d

q +++++++++ air

d/2

–––––––––

Sol: When capacitor is partially filled with dielectric capacity C=

ε0 A

d − t (1 − 1 )   k 

For metal slab of thickness t = d/2,

ε A C = 0 ( K = ∞ for metal slab) d −t

ε0 A εA = 2 0 d d d− 2

=

Ans: 3d/4

C=

TEST YOURSELF 1. The capacity of a parallel plate capacitor formed by the plates of same area A is 0.02µF with air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (K = 2) is introduced between the plates, the capacity is (1) 0.04 µF (2) 0.08 µF (3) 0.01 µF (4) 2 µF 2. A capacitor connected to a 10V battery collects a charge 40µC with air as dielectric and 100µC with oil as dielectric. The dielectric constant of oil is (1) 2 (2) 2.5 (3) 4 (4) 10 3. A parallel plate capacitor (condenser) has a certain capacitance (capacity). When 2/3rd of the distance between the plates is filled with a dielectric, the (capacity) capacitance is found to be 2.25 times the initial capacitance.

The dielectric constant of the dielectric (1) 1 (2) 3 (3) 7 (4) 6 4. The plates of a parallel plate capacitor are charged upto 200 Volts. A dielectric slab of thickness 4 mm is inserted between its plates. Then to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2 mm. The dielectric constant of the dielectric slab is (1) 1 (2) 4 (3) 5 (4) 6

126 |

CHAPTER 2: Electrostatic Potential and Capacitance

5. For charging the capacitance of given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as plates of capacitor. The 3 thickness of dielectric slab is d , where d is 4 the separation between the plates of parallel plate capacitor. The new capacitance (C‘) in terms of original capacitance (C0) is given by (1) C ′ =

4 KC0 K +3

(2) C ′ =

(4 + K ) C

(3) C ′ =

4C0 3+ K

(4) C ′ =

(3 + K ) C

3

0

4K

0

6. Consider a parallel-plate capacitor of capacitance 10 μF with air filled in the gap between the plates. Now, one half of the space between the plates is filled with a dielectric of dielectric constant 4 as shown in the figure. The capacitance of the capacitor changes to

K=4

(1) 25 μF (3) 40 μF

(2) 20 μF (4) 5 μF Answer Key

(1) 1

(2) 2

(5) 1

(6) 1

2.11

(3) 4

(4) 3

COMBINATION OF CAPACITORS

Several capacitors of capacitance C1, C2 ....

Cn can be combined to obtain a system with some effective capacitance C. The effective capacitance depends on the way in which the individual capacitors are combined.

2.11.1 Capacitors in Series In series combination, the capacitors are first arranged in a series order such that the second plate of first capacitor is connected to the first plate of second capacitor, the second plate of second capacitor is connected to first plate of third capacitor and so on. And finally the first plate of first capacitor and second plate of last capacitor are connected to opposite terminals of battery.

Let us consider three capacitors of capacities C1, C2 and C3 connected in series across a source of potential difference ‘V’ as shown in figure. +q

V1 C1

–q

+q

V2

–q

C2

+q

V3

–q

C3

V + –

At the moment, the system is connected to the source, left plate of first condenser acquires positive charge due to conduction. This in turn will produce negative charge of equal magnitude, on the left face of second plate of first condenser due to induction. The process continues for the remaining two condensers.

Hence, the charge acquired by all the three capacitors will be same.

As the capacitors are different, the potentials developed across them will be different. q = C 1V 1 = C 2V 2 = C 3V 3 = V1

q q q ; But = V1 +V2 + V3 = , V2 = , V3 C1 C2 C3

1 1 1  V= q  + +   C1 C2 C3 

.... (a)

If a single capacitor when connected across the same source draws the same charge, that capacitance is said to be the equivalent capacitance of the three capacitors. If CS is the equivalent capacitance.

IL RANKER SERIES FOR NEET

CS =

q q ⇒V = V CS

Let us consider three capacitors of capacities C1, C2 and C3 connected in parallel across a source ‘V’ as shown.

... (b)

Substituting (b) in (a)

The moment capacitors are connected, charge is drawn from the voltage source and this charge is drawn along three branches and thus gets shared. As all capacitors are connected in parallel, the potential across any of the capacitors is same. Here charge gets shared depending upon their capacitances for maintaining same potential.

q q q q = + + CS C1 C2 C3 ⇒

1 1 1 1 ... (c) = + + Cs C1 C2 C3

In general

1 1 =∑ CS Cn

... (d)

Key Insights: ■ The resultant capacity of series combination is smaller than the least capacity of the capacitors of the combination.

■ In series, ratio of charges on three capacitors is 1:1:1.

■ The ratio of potential differences across three capacitors is = V1 : V2 : V3

Q Q Q 1 1 1 = : : : : C1 C2 C3 C1 C2 C3

■ PD across first capacitor is V1 =

1 C1

 1 1 1  +   +  C1 C2 C3 

V similarly we can

find V2 and V3.

Capacitors are said to be connected in parallel if the two plates of any capacitor are connected one to positive terminal and the other to negative terminal of the source, then the connection is said to be parallel connection. + –

q1 C1

= V

q1 q2 q3 = = C1 C2 C3

∴ q1 + q2 + q3= C1V + C2V + C3V q = V(C1 + C2 + C3) q = C1 + C2 + C3 ... (a) V If a single capacitor when connected to the same source draws a charge q then that capacitor is said to be the effective or equivalent capacitor for the three parallel capacitors.

If the effective capacitance is CP, q ... (b) V from (a) and (b) CP = C1 + C2 + C3 CP =

In general, CP = ΣCn

2.11.2 Capacitors in Parallel

V

+ –

| 127

q2 C2

+ –

q3 C3

--- (4.6)

Key Insights: ■ The resultant capacity of parallel combination is greater than the largest capacity of the capacitors of the combination.

■ In parallel, ratio of PD on three capacitors is 1 : 1 : 1.

■ The ratio of charges on three capacitors is Q 1 : Q 2 : Q 3 = C 1V : C 2V : C 3V = C 1 : C 2 : C 3

128 |

CHAPTER 2: Electrostatic Potential and Capacitance

■ The charge on first capacitor is Q1 =

C1 Q C1 + C2 + C3

K2

K1

similarly we can find Q2 and Q3.

■ When n identical capacitors each of capacity C are first connected in series and next connected in parallel then the ratio of their effective capacities can be written as

A/2

A/2

K1

K1

d

d

Cs C = n2 : 1 = ; C p nC ; Cp n

= Cs

19. The equivalent capacity between A and B in the given circuit is A

8µF

B 12µF

12µF

8µF Sol: Here 12 µF and 12 µF are short circuited.

Capacity of the left half C1= K1

Take only 8 µF and 8 µF parallel combination.

C = 8 + 8 = 16 µF

ε0 A 2d

ε0 A 2d C1 and C2 may be supposed to be connected in parallel then effective capacity Capacity of the right half C2 = K2

C = C1 + C2 =

Hence they are not charged.

The equivalent circuit is as shown

ε0 A  K1 + K 2   d  2 

 K1 + K 2  = C0   2   where C0 is capacity of capacitor without dielectric.

K + K2 Effective dielectric constant K = 1 2 ■ When the space between the plates of a parallel plate condenser is completely filled with two slabs of dielectric constants K1 and K2 and each slab having area A and d thickness equal to as shown in the figure 2

Try yourself: 17. In the network three identical capacitors are connected as shown. Each of them can withstand to a maximum 100 V potential difference. What is the maximum voltage that can be applied across A and B so that no capacitor gets spoiled ?

C C B A C

Ans: 150 V

Application ■ When the space between the plates of a parallel plate condenser is completely filled with two slabs of dielectric constants K1 and K2 and each slab having area A/2 and thickness equal to distance of separation d as shown in the figure, the equivalent circuit is as shown.

K1 K2

The equivalent circuit is as shown A

d/2

K1

d/2

K1

A

C1C2 + C2

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2K1εo A d 2K ε A ■ Capacity of the lower half C2 = 2 o d ■ C1 and C2 may be supposed to be connected in series.

■ Effective capacity ■ Capacity of the upper half C1 =

C= 3 C= 4

K 3 εo A / 2 = K 3C d 2

C1 and C3 are in series, C2 and C4 are in series and the two combinations are in parallel.

C2C4 = CB = C2 + C4

CC

2 4 CB =  2 K1 K 2  = ε0 A  2 K1K 2  C1C2 C = C = 2 + C4  =  C0   C1 + C2 d  K1 + K 2  K + K 2   1

Here C0 is the capacity of the condenser with air medium.  2K1 K 2  ■ Effective dielectric constant K=    K1 + K 2  20. A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K1, K2, and , as shown in fig. If a single dielectric material is to be used to have the same effective capacitance as the above combination, then find its dielectric constant K.

d

A 2

A 2

K1

K2

 KK K K  C = C A + CB =  1 3 + 2 3  C0  K1 + K 3 K 2 + K 3 

The dielectric constant K is

K1K 3 K K = K + 2 3 K1 + K 3 K 2 + K 3

21. Four identical metal plates are located in air at equal distance d from one another. The area of each plate is A. Find the equivalent capacitance of the system between X and Y. 1 3

CA

K3 A 2

Y

Sol: Let us give numbers to the four plates. Here, X and Y are connected to the positive and negative terminals of the battery (say), then the charge distribution will be as shown.

A

d 2 d 2

X

2

––––– 1 2+++++ +++++ ––––– 3 +++++ 4–––––

The equivalent circuit is as shown in fig.

K1

K 2C0 .K 3Co  K 2 K 3  =  C ( K 2 + K 3 ) Co  K 2 + K 3  o

4

K3

A 2

K 2C0 .K 3Co  K 2 K 3  =  C ( K 2 + K 3 ) Co  K 2 + K 3  o

The equivalent capacity is given by

 2 K1 K 2  ε0 A  2 K1K 2   =  C0   d  K1 + K2   K1 + K 2 

Sol:

| 129

A 2 K2

CB

K4 A 2

ε0 A ε0 A 2 KC Let C0 = = ; C1 K= 1 1 d d 2 ε0 A 2 K C C2 K= = 2 2 d 2

X Y

+ –

Here, the arrangement can be represented as the grouping of three identical capacitors, each of capacity shown.

ε0 A . The arrangement will be as d 2 1

X (+)

3 4 2 3

Y (–)

130 |

CHAPTER 2: Electrostatic Potential and Capacitance

Now, the equivalent capacitance between X and

= Y is C XY

(C + C )C 2C 2 ∈0 A = = C +C +C 3 3d

Try yourself: 18. A parallel plate capacitor is constructed using three different dielectric materials as shown in the figure. Separation between the plates of the capacitor is d. Find the equivalent capacitance between the plates?

K

3K

2K d/2

39K  ε0 A    20  d 

2. Three condensers of capacity 4 mF, 2 mF and 3mF are connected such that 2 mF and 3 mF are in series and 4 mF is parallel to them.

The equivalent capacity of the combination is (1) 9 mF (2) 5.2 mF (3) 2.6 mF (4) 10 mF 3. The effective capacitance in µF in between A and B will be 12µF 2µF

(2) 4

(2) 10 µF (4) 135 µF

(3) 5

C

B

(4) 18

C

C

C

A B

(1) C/4

(2) 3C/4

(3) C/3

(4) 4C/3

7. Two identical capacitors are connected in series. Charge on each capacitor is q 0. A dielectric slab is now introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected.

The charge in each capacitor will now be (1)

2q0 1 1+ k

(2)

q0 1+

1 k

q0 2q0 (4) 1+ k 1+ k 8. Four plates of the same area of cross section A are joined as shown in figure. The distance between each plate is d. The equivalent capacity across A and B will be (3)

2µF

(1) 28/9

(1) 15 µF (3) 40 µF

6. The equivalent capacitance between A and B is

1. Three parallel connected conductors A, B and C have a total charge of 48µC. The ratio of their capacitances are 1 : 3 : 2. The charges on them individually are (1) 24 µC, 12 µC, 12 µC (2) 8 µC, 18 µC, 22 µC (3) 8 µC, 24 µC, 16 µC (4) 16 µC, 16 µC, 16 µC

A

B

A

d/2

TEST YOURSELF

2µF

5. In the given figure each capacitor is equal to 45µF then the equivalent capacity between A and B in the given circuit is

Plate - 2

Ans:

Plate - 1

4. Two identical parallel plate capacitors are joined in series to 100 V battery. Now a dielectric with K = 4 is introduced between the plates of second capacitor. The potential difference on capacitors are (1) 60 V, 40 V (2) 70 V, 30 V (3) 75 V, 25 V (4) 80 V, 20 V

IL RANKER SERIES FOR NEET Answer Key

A B

3 ∈0 A (2) d

2 ∈0 A (1) d

2 Cnet Vnet × (1) C V2

2 C Vnet × (2) Cnet V 2

2 C Vnet × (3) Cnet V

(4)

Cnet Vnet × C V

10. Two condensers C1 and C2 in a circuit are joined as shown in figure. The potential of a point A is V1 and that of B is V2. The potential of point D will be C1 C2 A B V1

(1)

V1 + V2 2

(2)

C1V1 − C2V2 C1 + C2

(3)

C1V1 + C2V2 C1 + C2

(4)

C2V1 + C1V2 C1 + C2

11. What would be the voltage across C3?

C1 C2

(C1 + C2 )V C1 + C2 + C3

C2V (3) C + C + C 1 2 3

(2) 2

(3) 3

(4) 4

(5) 4

(6) 4

(7) 1

(8) 2

(9) 1

(10) 3

(11) 1

ENERGY STORED IN A CAPACITOR

The energy of a capacitor is the amount of work done to charge it. Any charged conductor will store electric potential energy in the form of electrostatic field.

Consider two uncharged parallel conducting plates separated by a small distance so that ‘C’ is the capacitance of the system.

Imagine the process of transferring of charge from plate B to A until plate ‘A’ is positively charged to +Q and plate B is negatively charged to the charge –Q. During process, let ‘q’ be the charge at any instant at an intermediate state. The potential difference between the plates at that instant q is V = .

C Q

V2

D

C3

(1) 3

2.12

∈0 A 3 ∈0 A (3) (4) d 2d 9. Given a number of capacitors labelled as C, . Find the minimum number of capacitors needed to get an arrangement equivalent to Cnet, Vnet

(1)

| 131

+ + + + +

A

E

–Q – – – – – B

If ‘dq’ is the small charge transferred from B to A at the intermediate state, the work done against the field to increase charge of A from q to q + dq is q dq .

C Hence, the total work done in transferring the charge +Q to the plate A from B, we have to integrate the above function from 0 to Q. dW = Vdq =

V

(2) (4)

C1V C1 + C2 + C3 C3V C1 + C2 + C3

W

Q

Q

q 1  q2  dW ∫= dq W = ∫=   C C  2  0 0 0

CHAPTER 2: Electrostatic Potential and Capacitance

Q2 1 1 2 = CV= QV (Q = CV ) 2C 2 2 This work stored as potential energy in the form of electrostatic field. = W

= U

Q2 1 1 2 = CV= QV (or) 2C 2 2

U=

Q 2 ( Aσ ) d σ2 Ad = × = 2C 2 ε0 A 2ε 0

Application ■ If a battery is used to charge a capacitor, then work done by the battery is W and energy stored in the capacitor is U. Let Q be the charge delivered by the battery to the capacitor. Then work done by the battery is W = VQ and energy stored is

2

The remaining energy is

Key Insights: ■ When three capacitors are in series, the ratio of energies is = U1 : U 2 : U 3 2

2

1 QV ⇒ U = W 2 2

U =

Q2 Q2 Q2 1 1 1 = : : : : 2C1 2C2 2C3 C1 C2 C3

W W 1 = = QV will be lost 2 2 2 as heat energy and light energy.

W–U = W −

22. In the arrangement shown, find the energy stored in the 3µF capacitor.

3µF

6µF

2

Q Q Q 1 1 1 = : : : : 2C1 2C2 2C3 C1 C2 C3 ■ When three capacitors are in parallel, the ratio of energies is 1 1 1 U1 : U 2 : U 3 = C1V 2 : C2V 2 : C3V 2 2 2 2 = C1 : C2 : C3

■ Energy density (u) = energy/volume εA 2 Ed 1 CV 2 1 d ( ) 1 u= = = εE 2 Ad 2 Volume 2 2

1 1 u = εE 2 = kε0 E 2 2 2 It is true for electric field due to any configuration of charges. (Where K in the dielectric constant of medium between the plates) ■ Practically, it is not possible to transfer charge from one plate to another plate directly through the region between the plates. It is practically done by external source like battery.

10 v Sol:

3× 6 µF = 2 µF 3+6 ∴ Q =CeV =2 × 10 µC Ce =

= 20 µC Q 2 202 = 2C 2 × 3 = 66.67 µJ.

∴U =

Try yourself: 19. In the arrangement shown, initial charge on the 3 µF capacitor is 4 µC. Find the energy lost after switch S is closed.

3µF +

–

S 5µF

Ans: 5/3 µJ

1 : U2 : U3

132 |

2.12.1 Force between the Plates of a Capacitor Consider a parallel plate capacitor with plate area A. Let Q and –Q be the charges on the plates of capacitor. Let F be the force of

IL RANKER SERIES FOR NEET

attraction between the plates. Let E be the field between the capacitor plates. The expression for the force can be derived by energy method.

Let the distance between the plates be x.

| 133

23. Plates of a parallel plate capacitor are connected to the terminals of a battery. Force of attraction between the plates is F. Find the force between the plates, if the sepratation between them is doubled.

2

1 F'  d  1 ⇒ =  = 2 F d' 4 d F ⇒F'= 4

Sol: F ∝

|dx|

So electric field energy between the plates is U=

1 ∈0 E 2 ( Ax ) 2

⇒

dU 1 =∈0 E 2 A dx 2

Try yourself: 20. An isolated parallel plate capacitor has a charge Q. Its plate area is A and plate separation is d. Find the work done in increasing the plate separation to 2d.

Q 2d 2ε 0 A

x

Ans:

F

dU −1 By definition, F = − =∈0 E 2 A dx 2 (Conservative force)

2.12.2 Effect of Dielectric on Various Factors of Parallel Plate Capacitor

So the force of attraction between the plates 1 is F = ∈0 E 2 A 2

Consider an air filled capacitor of capacitance ∈ A C0. Such that C 0 = 0 . d

Key Insights: ■ For an isolated charged capacitor Q2 F= . This force does not depend 2 ∈0 A on the separation between the plates, and so the constant amount of force is needed to change the separation.

Now, a dielectric slab of dielectric constant K is inserted between the plates completely filling the gap.

Case 1: Introduction of dielectric without disconnecting charging battery: +

V0

–

+

V0

–

■ For a capacitor having constant potential difference across the plates the force = F

C 2V 2 ∈20 A2 V 2 = 2 ∈0 A d 2 2 ∈0 A

1 V2 ∈0 2 A ⇒ F= 2 d In this case force depends on the separation between the plates. Thus to change the separation variable force is needed.

C0

C

If the capacitor (air filled) is initially connected between the terminals of a battery of emf V0, charge on the capacitor is Q0 = C0V0. If the dielectric is inserted between the plates of the capacitor without disconnecting battery, then its potential remains same. (V = V0) only.

134 |

CHAPTER 2: Electrostatic Potential and Capacitance

i)

In this case capacity changes to

ii)

K ∈0 A ⇒C = KC 0 d potential V = V0.

Now, if a dielectric is introduced between the plates of capacitors still charge remains same.

The charge Q = Q0.

C=

iii) Now charge on the capacitor is Q = CV i.e., Q = KC0V0 = KQ0 ⇒ Charge on the capacitor increases to K times iv) Energy stored in the capacitor before 1 2

introducing the dielectric is U0 = C 0 V02 .

After introducing dielectric, energy 1 2 stored in the capacitor U = CV 2 1 = = KC 0 V02 KU 0 2 ⇒ Energy stored increases to K times v)

Electric field between the capacitor plates is V E = 0 (as V = V0) d After introducing dielectric, electric field between the plates

V V0 = ⇒ Electric field between the d d plates does not change. vi) Additional charge passing through battery is DQ = Q – Qo= Qo(K–1). viii) Additional work done by the battery W = DQ (Vo) = Q0V0 (K–1). = E

Case 2: Introduction of dielectric after disconnecting charging battery: If the air filled capacitor is initially charged to potential V0 and disconnected from the battery, then charge on it does not change.

V0

i) In this case the capacity changes to K ∈0 A C= d ⇒ C = KCo ii) Charge Q = Q0 = C0 V0 Q iii) Now potential of the capacitor is V = C Q0 V0 is= V = KC0 K ⇒ Potential of the capacitor reduces to 1/K times. iv) Energy stored in the capacitor, before Q2 introducing dielectric is U 0 = 0 2C0 Energy stored in the capacitor, after Q02 Q2 U = introducing dielectric is = 2C 2KC0

Q02 U0 = ⇒Energy stored in 2KC0 K 1 the capacitor reduces to times.

K v) Electric field between the plates of the V capacitor is E0 = 0 initially. d After introducing dielectric slab = U i.e.,

E0 K 1 ⇒ Electric field decreases to times K = E

V V0 = d Kd

i.e., E =

24. A parallel plate capacitor of plate area A and separation ‘d’ its charged to a p.d (V) and then battery is disconnected. A slab of dielectric constant (K) is inserted to fill the space between the plates. The work done on the system in the process of inserting the slab is ε A Sol: Let charge q = CV and C = 0 d 2

+

C

–

Initial energy of the system U = q I 2C

IL RANKER SERIES FOR NEET

| 135

After inserting slab, C1 = KC Since charge is q2 constant Final PE U f = 2KC

increased after disconnecting the charging battery:

Work done W = Ui – Uf

i) Capacity decreases

∈0 A   C =  d  

ii) Charge is constant

(Q = Q0)

iii) Potential increases

Q  V = C   

iv) Energy increases

 Q2  = U   2C  

= =

q2  1  c 2V 2  1  −= −1 1 2C  K  2c  K  ε0 AV 2  1  −1 2d  K 

Try yourself: 21. An air cored parallel plate capacitor has capacitance Co. A dielectric slab of dielectric constant K is now introduced in the capacitor such that it completely fills the space between the plates. The plates of the capacitor are now connected to the terminals of a battery of emf V. Now, an external agent slowly removes the dielectric slab from the capacitor. Find the work done by the external agent.

Ans: 1 1 ( K − 1) C0V 2 2

Case 3: Effect of distance between the plates on various factors of parallel plate capacitor: 1) If the capacitor (air filled) is connected between the terminals of a cell of emf V 0 and the distance between the plates is increased without disconnecting the charging battery:

v) Electric field is constant  σ Q  E = =  ∈0 ∈0 A  

2.12.3 Redistribution of Charge Two capacitors of capacities C1 and C2 are charged to potentials V 1 and V 2 separately and they are connected so that charge flows.

Here, charge flows from higher potential to lower potential till both capacitors get the same potential.

1) Two capacitors are connected in parallel such that positive plate of one capacitor is connected to positive plate of other capacitor.

– +

– +

Let V be the common potential

ii) Potential is constant

∈0 A   C =  d   (V = V0 )

iii) Charge decreases

(Q = CV)

(C1 + C2) V = C1V1 + C2 V2

i) Capacity decreases

1  2  U = 2 CV    V  v) Electric field decreases  E =  d  iv) Energy decreases

2) If the capacitor (air filled) is connected between the terminals of a cell of emf V0 and the distance between the plates is

Then Q = Q1 + Q2 (charge conservation)

V=

C1V1 + C2V2 C1 + C2

In this case, there will be loss in energy of the system ∆U = U i − U f 1 1 where = Uf C1V 2 + C2V 2 2 2

136 |

CHAPTER 2: Electrostatic Potential and Capacitance

= Ui

1 1 C1V12 + C2V22 2 2

In case of spherical conductors, C = 4πεor

1 C1C2 (V1 − V2 )2 2 (C1 + C2 )

Common potential:

= ∆U

so, Q1' : Q2' = r1: r2 By law of conservation of charge

Q 11 + Q 21 = Q 1 + Q 2 2) If positive plate of one capacitor is connected to negative plate of other C1V + C2V = C1V1 + C2V2 capacitor, common potential is given by C V − C2V2 C1V1 + C2V2 r1V1 + r2V2 V= 1 1 = V = C1 + C2 C1 + C2 r1 + r2 Here, charge flow takes place if V1≠V2 In this case, the loss of energy = ∆U

1 C1C2 (V1 + V2 )2 2 (C1 + C2 )

Application ■ Redistribution of charges when two conductors are connected by conducting wire: When two charged bodies are connected by a conducting wire then charge flows from a conductor at higher potential to that at lower potential until their potentials are equal.

Effective capacity: Q Q1 + Q21 Q11 Q21 C= = 1 = + =C1 + C2 V V V V When charge flows through the conducting wire then energy is lost on account of Joule’s heating effect, i.e., electrical energy is converted into heat energy. Hence the stored energy of the conductors some what decreases.

Loss in energy of the system: ∆ = U =

1 1 1 C1V12 + C2V22 − CV 2 2 2 2 1  C1C2  2   (V1 − V2 ) 2  C1 + C2 

Let the amounts of charge on two 1  4 πε0r1r2  conductors A and B are Q1 and Q2 their 2 =   (V1 − V2 ) capacities are C1 and C2 and their potentials 2  r1 + r2  are V1 and V2 respectively, then Q1 = C1V1 and Q2 = C2V2 If V1 = V2, then neither the charge flows nor the energy is lost.

Let the amount of charge after the conductors are connected, are Q1' and Q2' respectively, then Q1' = C1V ; Q2' = C2V Q1

++ +++ + + + C1 ++ A + + + + +++ + + +

V1

Q2

++ + +

+ ++

B

+ ++ ++ +

+ + + C2 + + ++

V2

25. Parallel plate capacitor of capacitance C is charged to a potential difference V. Another capacitor of capacitance 2C is charged to a potential difference 2V. The two capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to negative terminal of the other.

The final energy of the configuration is

Charges are redistributed in the ratio of their capacities.

Sol: Common potential

∴Q1' : Q2' = C1 : C2 (since V is same)

= VC

Q '1 − Q '2 CV − ( 2C ) 2V = = V C1 + C2 3C

IL RANKER SERIES FOR NEET Final energy =

3 2 Uf = CV 2

1 1 (C1 + C2 )VC2 = ( 3C )V 2 2 2

22. Two conductors of capacity 8.4 mF each charged to potential 500 V and –500 V are joined by a conducting wire. If the mass of the wire is 500 g and specific heat of the material is 0.1 cal/g/°C, find the rise in the temperature of the wire.

Ans: 10°C

2.12.4 Uses of Capacitors Capacitors are very useful devices to store charges and electrostatic energy which are of special interest in physics and engineering applications.

1. Capacitors are used to establish desired, uniform and strong electric fields in small space and they serve as useful devices for storing electrical energy.

2. Capacitors are widely used in tuning circuits of radio and T.V. receivers.

3. A capacitor blocks direct current and allows alternating currents. Capacitors are used in filter circuits.

4. Capacitors are used in the generation and detection of oscillating electric fields.

5. To reduce voltage fluctuations in electric power supplies, to transmit pulsed signals and to provide time delays, capacitors are essential.

Application ■ Let n identical charged liquid drops each having radius r, charge q and capacity C are combined to form a big drop.

1) radius, R = n1/3r 2) Charge, Q = nq

3) Capacity, C' = 4π∈oR

1/3 π ∈o n1/3r n= C ′ = 4= 4 π ∈o r n1/3C

4) Potential,

Try yourself:

For big drop,

| 137

Q C'

= V ' =

nq n2/3 q = = n2/3V 1/3 C n C

5) Energy, 2 2 n5/3 q2 Q2 n q U = = = = n5/3U 1/3 2C ' 2n C 2C 6) Surface charge density,

n1/3 q σ=' = = = n1/3σ 2 2/3 2 2 4 πR 4 πn r 4 πr

Q

S.No.

Quantity

nq

For each charged small drop

For the big drop

1.

Radius

r

R = n1/3r

2.

Charge

q

Q=n×q

3.

Capacity

C

C' = n1/3 × C

4.

Potential

V

V' = n2/3 × V

5.

Energy

U

U' = n5/3 U

6.

Surface charge

s

s' = n1/3 s

TEST YOURSELF 1. A parallel plate condenser has plates of area 200 cm2 and separation 0.05 cm. The space between plates have been filled with a dielectric having K = 8 and then charged to 300 volts. The stored energy (1) 1.6 × 10–5 J (2) 2 × 10–6 J (3) 12.4 × 10–5 J (4) 1.6 × 64 × 10–5 J 2. A 4 µF capacitor is charged by a 200 V battery.

It is then disconnected from the supply and is connected to another uncharged 2 µF capacitor. During the process loss of energy (in J) is (1) Zero (2) 5.33 × 10–2 (3) 4 ×10–2 (4) 2.67 × 10–2

138 |

CHAPTER 2: Electrostatic Potential and Capacitance

3. Two condensers, one of capacity C and the other of capacity C/2, are connected to a V volt battery, as shown. The work done in charging fully both the condensers i

V

C 2

C

(1) 2 CV2

(2)

(3) 1.2 J

1 4CV 2

1 3 (4) 2 2CV 2 4CV 4. Two spheres of radii 12 cm and 16 cm have equal charge. The ratio of their energies is (1) 3 : 4 (2) 4 : 3 (3) 1 : 2 (4) 2 : 1 (3)

5. Three condensers of same capacity connected in series has effective capacity 2 mF. If they are connected in parallel and charged using a battery of emf 12 V, the total energy stored in the combination is (1) 1296 mJ (2) 648 mJ (3) 162 mJ (4) 48 mJ 6. The work done in increasing the P.D. across the plates of a capacitor from 4V to 6V is W. The further work done in increasing the P.D. from 6V to 8V is (1)

4W 3

(2)

3W 7W (3) 4 5

(4)

5W 7

7. A series combination of n 1 capacitors, each of value C 1 is charged by a source of potential difference 4V. When another parallel combination of n2 capacitors, each of value C2, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has.

The value of C2, in terms of C1 is then

2C1 (1) n1n2 (3) 2

n2 C1 n1

n (2) 16 2 C1 n1

(4)

16C1 n1n2

8. The capacity of a parallel plate condenser with air as dielectric is 2 µF. The space between the plates is filled with dielectric slab with K = 5. It is charged to a potential of 200V and disconnected from cell. Work done in removing the slab from the condenser completely (1) 0.8 J (2) 0.6 J (4) 1.6 J

9. A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by ∆T, the potential difference V across the capacitance is (1) (3)

2mC ∆T s ms∆T C

(2) (4)

mC ∆T s 2ms∆T C

(1) 3

Answer Key (2) 4 (3) 3

(4) 2

(5) 1

(6) 3

(8) 1

(7) 4

(9) 4

2.13

CAPACITANCE OF A CONCENTRIC SPHERICAL CAPACITOR

Based on the charge distribution and connection of shells two cases will arise. In one case charge may be given to the inner shell and outer shell is earthed. In the second case charge may be given to outer shell and inner shell is earthed.

When charge is given to the inner shell and outer shell is earthed: Consider two concentric conducting shells of radii a and b (b > a). Let q be the charge given to the inner shell and its outer surface is earthed. Electric field inside the inner shell is zero. So there will be electric field only between the two shells.

IL RANKER SERIES FOR NEET

Induced charge on the inner shell −a Q' = .Q b Potential difference between shells or

b+ q a -q

potential of outer = shell, V2 Electric field at a distance r from the centre of the shells (a < r < b) is E=

1 q 4 π∈0 r 2

Now potential differences between the two shells is a

V= V+ − V− = − ∫ Edr b

⇒ V=

b

∫ Edr= a

| 139

b

q q 1 1 dr = − 4 π∈0 ∫a r 2 4 π∈0  a b 

Capacitance of the spherical capacitor q C= V  ab  ⇒ C = 4 π∈0    b−a 

=

1 Q Q '  + 4 πε0  b b 

Q b − a  1 Q  a  Q   −   = 4 πε0  b  b  b  4 πε0  b2 

Capacitance of the system, C ' =

C '= 4 πε0 .

b2 b−a

Q V2

26. Three concentric thin spherical shells are of radii r1, r2, r3 (r1 < r2 < r3). The first and third are connected by a conducting wire through a small hole in the second and the second is connected to earth through a small hole in the third. Find the capacity of the condenser so formed.

Sol: Let Q be the charge given to the outermost shell 3, some of the charge will go to the outer surface of the inner-most shell because they are connected together by a wire.

(Q-Q1+Q2)

If there is a medium of dielectric constant K between the two spheres, then we can write

-Q2 -Q1

 ab  C = 4 π∈0 K    b − a  If outer shell is given a charge + Q while inner shell is earthed: The potential at the surface of inner sphere is zero. So if Q' is the charge induced on inner sphere, then, = V1

1 Q ' Q  += 0 4 πε0  a b 

a

b

Charge Q1 on the inner shell will induce a charge – Q1 on the inner surface of the shell 2 and +Q1 on its outer surface. Shell 3 also induces the charge on shell 2., so that the charges on surfaces of 2 and 3 facing each other are Q2 and –Q2. Here we can observe three different capacitors between shells 1 –2, 2 – 3 and 3 –∞ denoted by C1, C2, C3

 r r   rr  C1 = 4 π∈0  1 2  ,C2 = 4 π∈0  2 3   r2 − r1 

 r3 − r2 

140 |

CHAPTER 2: Electrostatic Potential and Capacitance earthed. Then the equivalent capacitance of

and C3 = 4 π∈0 r3

the system is -

Effective capacity, C = C1 + C2 + C3

- +- + +- - A + - 2 - + + A1 ++ -+ r1 + - ++ r2 + + - + - ++ + ++- - - -

 rr  rr C = 4 πε0  1 2 + 2 3 + r3   r2 − r1 r3 − r2 

Try yourself: 23. The capacity of a spherical capacitor is C1

when inner sphere is charged and outer shell is earthed. C2 is capacity when outer shell is

charged and inner sphere is earthed. If a and b are radii of inner sphere and outer shell, C what will be the value of 1 ?

C2 Ans: a/b

2.13.1 Cylindrical Capacitor It consists of two concentric cylinders of radii a and b (a < b), inner cylinder is given a charge +Q while outer cylinder is earthed.

If the common length of the cylinders is, 2 πε0  then C = b log e   a b

a

Q

(1)

(3) 4πε0r2

(1)

r1 and r2 (r2 > r1) are placed concentrically

in air. A1 is given a charge +Q while A2 is

(4) 4πε0r1

1 µF 8

(2)

1 µF 4

1 1 µF µF (4) 2 16 3. A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be (1) Zero (2) 4πε0a (3)

(3) 4πε0b

2.14

1. Two spherical conductors A1 and A2 of radii

(2) 4πε0(r1 + r2)

2. A spherical drop of capacitance 1μF is broken into eight drops of equal radius. Capacitance of each small drop is:

(1) 1

TEST YOURSELF

4 π∈0 r1r2 r2 − r1

 b  (4) 4 πε0 a    b−a 

Answer Key (2) 3 (3) 3

VAN de GRAAFF GENERATOR

Van de Graaff generator is used to develop very high charges and intense electric field or very high voltages of order of a few million volts.

The resulting large electric fields are used to accelerate charged particles like electrons, protons and ions to high energies.

Principle: Whenever a charge is given to a

IL RANKER SERIES FOR NEET

metal body it will spread on the outer surface of it. If we put a charged metal body inside the hollow metal body and the two are connected by a wire, whole of the charge of the inner body will flow to the outer surface of the hollow body. No matter how large the charge is on the inner body, it completely transfers to the outer body. This principle is used in Van de Graaff generator.

Theory: Consider a spherical conductor 1 of radius r 1 holding charge q 1 uniformly distributed on it. It is kept inside a hollow conductor 2 of radius r2 which is uncharged Now, electric potential of inner sphere is = V1

1  q1 0  1 q1  +=  4 π∈0  r1 r2  4 π∈0 r1 r2

q1

q1

r2

r1

r1

Fig-(I) and (II) Electric potential of outer shell is = V2

1  q1 0  1 q1  +=  4 π∈0  r2 r2  4 π∈0 r2

Now, potential difference between the two conductors is − V2 V1=

q1  1 1   −  4 π∈0  r1 r2 

From this, we can conclude that potential difference in this case depends on q1 only. It does not depend on any charge on the outer shell. Let us check the same as given below. If q2 is the charge on the outer shell, then = V1

1  q1 q2   +  4 π∈0  r1 r2 

and, V2 =

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1  q1 + q2    4 π∈0  r2 

Now V1= − V2

q1  1 1   −  4 π∈0  r1 r2 

Here the potential difference (V1 – V2) will remain the same for any value of q2.

If the two spheres are connected as shown in 2nd figure above, the charge flows from higher potential to lower potential. Ultimately all the charge resides on the surface of outer sphere.

Construction and working: The figure shows various parts of Van de Graaff generator. It consists of a large spherical metal shell S mounted on two insulating supports. A belt made of insulating material is run at high speed over pulleys P1 and P2. The pulley P1 is in the base of the machine which is run by a motor. The pulley P2 is at the centre of the spherical shell. The metal combs C1 and C2 with sharp ends are mounted on the generator as shown. Comb C1 is called emitter comb and it is held near the lower end of the belt. This is given a high positive potential about 104 V with respect to the ground. The other comb C2 is called collector comb and it is placed near the upper end of the belt such that its pointed ends touch the belt. The other end is in contact with the inner surface of the metal spheres.

As C1 is given high positive potential with respect to the earth, it builds positive ions.

These particle charges are sprayed on the belt which carries them to the top. As the positive charges reach C2, they are transferred to the metal sphere. This charge immediately moves to the outer surface of the sphere. Now, the uncharged belt runs down to take positive charge from C1. This process continues and more and more charge builds up on the outer surface of the spherical shell. As the charge on the shell increases, its potential also increases. When the shell is charged to a sufficient potential, the leakage of charge

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CHAPTER 2: Electrostatic Potential and Capacitance

due to ionization of surrounding air begins.

So there is a limiting value of potential (about 6 to 8 millions volts) to which the shell can be charged.

Collector comb(C2) + + + ++ + + + + + + + + Idler pulley (P ) + + 2 + + + + + + + Metal sphere S + Belt Insulating support + + Emitter comb + Driven pulley (P1) (C1) + Motor EHT 2.15

ATMOSPHERIC ELECTRICITY

The electrical behaviour of atmosphere is highly complex. However, following are the electrical properties of atmosphere, on the basis of experimentally observed facts: ■ The conductivity of atmosphere increases with altitude.

■ At ground level, there is a downward vertical electric field of strength 100 V/m.

The strength of field decreases with altitude;

being negligible at a height of 50 km.

■ There us a potential difference of about 4 × 105 V between earth and 50 km high layer of atmosphere; the earth being negative. Most of the potential drop occurs at low altitude where the field strength is high.

■ Since the electric field is acting vertically downward, there is a negative surface density of charge all over the earth.

■ The earth and 50 km layer of atmosphere form a capacitor of capacitance 0.1 F. We can find the potential difference and electric field. The experimentally observed values are 4 × 105 V and 100 V/m respectively.

This discrepancy is explained by the fact that, as we go up, conductivity of air increases.

■ There are about 40,000 thunderstorms per day all over the earth. Each storm lasts for about an hour.

■ During each thunderstorm, charged ions separate due to some complex process. The positive charges are carried upward to a height of about 6 km from ground level and negative charges collect at 2–3 km above the ground. Thus the top of thunderstorm has a positive charge and the bottom a negative one.

CHAPTER REVIEW Electric Potential ■ If W is the work done in bringing a test charge Qo from infinity to a point, then the potential at that point, is V=W/ Qo ■ Electric potential (V) is a scalar quantity but potential gradient (dV/dr) is a vector quantity.

■ Electric potential at a distance r from a 1 Q point charge Q, is V = 4 πE0 r

■ Electric potential at a point due to a system of point charges Q1,Q2,Q3......is ■ = V

Q  1  Q1 Q2 Q3 + + .... n   + r3 rn  4ε0  r1 r2

■ Work done by an external agent in slowly moving a point charge from point A to point B is W = Q (VB–VA) ■ If V A and V B are potential at A and B respectively then

IL RANKER SERIES FOR NEET B   VB − VA = − ∫ E.dl  A  where E is the electric field and dl is

σ∝

1 r2 +s + + +

displacement vector.

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+

+ + + + + + r

+

+ + + +

■ If in a region, potential V = f(x), then  dV i E= dx

■ At point P, electric potential

■ If in a region, V = f(x, y), then  δV  δV  − E= i− j. δx δy

■ When a dipole is placed in an electric

■ Electric potential at a point on the axis of a thin charged ring of radius R, at a distance 1 Q x from its centre is V = 4 π∈o R 2 + x 2

Electrostatic Potential Energy ■ Electrostatic potential energy of a system of two point charges Q1 and Q2 separated 1 Q1Q2 by a distance r is, U = 4 π∈o r ■ Total electrostatic potential energy due to a system of point charges, = V

 1  Q1Q2 Q2Q3 Q3Q4 + + + ....   4 π∈o  r12 r23 r34 

■ Electric potential due to a charged conducting sphere of radius R, carrying a charge Q is and = V

1 Q 4 π∈o r

and V =

[when, r ≥ R]

1 Q 4 π∈o R

[when, r ≤ R]

■ On the surface of a charged metallic object, surface charge density (s C/m 2) at any point is inversely proportional to the radius of curvature of the surface at the point i.e.,

Vp

+ +

+ + + + +

1 p cos 4 0 r 2

field making an angle θ with the field, potential energy stored in the system is

U p.E pE cos when θ is the angle

between p and E ■ Work done in rotating a dipole in an electric field is W = Uf – Ui = –pEcosb – (–pEcosa) = pE (cosa – cosb) Vs r2 3 2 r R 2 R R Vs r1 R r 1 Q 2 where, Vs

.R 4 0 R 3 0

■ Electric potential V

Capacitors and Capacitance ■ Capacitor is a device which stores electrostatic energy.

■ Two metallic objects carrying equal and opposite charge, separated by a distance, form a capacitor.

■ Capacitance of a capacitor depends on plate area, plate separation , their relative orientation and the intervening medium.

■ Capacitance of a capacitor does not depend on charge on plate or potential difference between the plates.

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CHAPTER 2: Electrostatic Potential and Capacitance

■ Capacitance of an air cored parallel plate capacitor is C 0=ε 0 A/d and if the space between the plates is filled with a dielectric material of dielectric constant k, then C = ε0 kA/d.

Effect of Dielectric on Capacitance ■ If a dielectric slab of thickness t (t < d) is introduced between the plates of a parallel plate capacitor, then

supplied by the battery is QV. So efficiency of charging is 50% ■ If n number of capacitors are connected in series, then 1/Ceq = 1/C1+1/C2+…+1/Cn ■ If n number of capacitors are connected in parallel combination, then Ceq= C1+C2+…+Cn ■ An external agent is introducing a capacitor of dielectric constant k into a capacitor of

ε0 A C=  1 d − t 1 −   k

capacitance Co whose plates are connected to the terminals of a battery of emf V0.

■ If a metal slab of thickness t is introduced

c0

between the plates, then, C=

ε0 A (d − t )

V0

■ If a number of dielectric slabs of thickness t1, t2…..tn and dielectric constants k1, k2…..

■ From conversation of energy, Wb+ We = (Vf –Vi) + Lost energy, where

kn are inserted between the plates, each

W b= work done by battery = DQ × V 0,

parallel to plate surface, then,

DQ = (k – 1)C0V0

εo A

C= d − t1 (1 −

  1 1 1  ) − t 2  1 −  .... − t n  1 −  k1 k k 2  n   

■ Electric force on each plate of a parallel plate capacitor is, = F

Q 1 εo AV = ε 2 o A 2 d2 2

u i=

1 2

o

V 02, U f =

We = work done by external agent.

■ After the switch k is closed, Common potential V =

2

(C1V1 + C2V2 ) (C1 + C2 )

c 1, v 1

■ Energy stored in a capacitor 2

1 1 Q = = QV CV 2 2 2 2C ■ When an uncharged capacitor is charged = U

1 k C oV o2, 2

by a battery of emf V, then energy stored Q2 1 in the capacitor is (or) QV but energy 2C 2

K

+

-

+

-

c 2, v 2 = Energy lost

1 C1C2 (V1 − V2 )2 2 C1 + C2

IL RANKER SERIES FOR NEET

■ After the switch k is closed, Common potential V = C1,

V1

(C1V1 − C2V2 ) (C1 + C2 )

earthed C = 4 πε0

b2 . (b − a ) b +Q

+ – K

– + C2,

Loss= of energy

V2

1 C1C2 (V1 + V2 )2 2 C1 + C2

For spherecial capacitor shown in figure C = 4 πε0

a

ab . (b − a )

■ Cylindrical capacitor inner radius a, C=

2πε0   n (b / a ) b

–Q a

+Q

b

ℓ ■ Outer shell of the spherical capacitor is given charge +Q and the inner shell is

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CHAPTER 2: Electrostatic Potential and Capacitance

Exercises NEET DRILL Electrostatic Potential 1. Two parallel plates separated by a distance of 5mm are kept at a potential difference of 50V. A particle of mass 10–15 kg and charge 10–11C enters in it with a velocity 107m/s.

The acceleration of the particle will be (1) 108 m/s2 (2) 5 × 105 m/s2 (3) 105 m/s2 (4) 2 × 103 m/s2 2. The potential at the origin is zero due E 20iˆ + 30 ˆj NC –1 .The to electric field = potential at point P(2m, 2m) is (1) 1 V (2) –100 V (3) 10 V (4) 2 V

(1) 6 5N

(2) 30 N

(3) 24 N

(4) 4 11

4. F i n d ( V A – V B ) i n a n e l e c t r i c f i e l d  E = 2iˆ + 3 ˆj + 4kˆ NC −1 where   rA = iˆ − 2 ˆj + kˆ m and rB = 2iˆ + ˆj − 2kˆ m.

)

)

(

(1) – 2 V (3) – 4 V

)

(2) – 1 V (4) – 6 V

5. A field of 100 Vm –1 is directed at 30° to positive x-axis. Find (VA–VB) if OA = 2m and OB = 4m

)

(

)

(4) 200 2 + 3 V

Electrostatic Potential Due to Discrete Charges 6. An electric charge 10–3 µC is placed at the origin (0, 0) of X-Y coordinate system. Two points A and B are situated at

( 2, 2 ) and

(2, 0) respectively. The potential difference between the points A and B will be (1) 9 V (2) zero (3) 2 V (4) 4.5 V the vertices of an equilateral triangle of 10 cm side. The potential at the mid point in between –q, –q, if q = 5µC. (1) 103 V (3) –12.8 × 105 V

(2) 1 V (4) 10 V

8. At y = 1 cm, y = 3 cm y = 9 cm, y = 27 cm ... and so on, an infinite number of charges equal to 5C are placed. At x = 1 cm, x = 2 cm, x = 4 cm, x = 8 cm .... and so on, an infinite number of charges equal to – 5C are placed.

Find the electric potential at origin in volts

 1  K =  4 πε0   (1) 250 K (3) zero

(2) –250 K (4) 100 K

9. Two point charges 4µC and 9µC are separated by 50 cm. The potential at the

B 30° A

(

(2) +100 2 + 3 V

7. Three charges +q, –q and –q are kept at

3. In a region, the potential is represented by V(x, y, z) = 6x – 8xy + 6yz, where V is in volts and x, y, z are in metres. The electric force (in N) experienced by a charge of 2 coulomb situated at point (1,1,1) is

( (

( 3 − 2 )V (3) 100 ( 2 − 3 ) V (1) 100

0

point between them where the field has zero strength is (1) 4.5 × 105 V (3) 9 × 104 V

(2) 9 × 105 V (4) zero

IL RANKER SERIES FOR NEET 10. Point charges q1 = 2 μC and q2 = –1 μC are kept at points x = 0 and x = 6 respectively.

Electric potential will be zero at the points: (1) x = 1 and x = 5 (2) x = 2 and x = 9 (3) x = 4 and x = 12 (4) x = –2 and x = 2 11. Three charges 2q, –q, –q are located at the vertices of an equilateral triangle. At the centre of the triangle (1) the field is zero but potential is non-zero (2) the potential is zero but field is non-zero (3) both field and potential are non-zero (4) both field and potential are zero 12. Two point charges +5 μC and −2 μC are kept at a distance of 1m in free space. The distance between the two zero potential points on the line joining the charges is (1)

2 m 7

(2)

2 m 3

20 22 m m (4) 21 21 13. Two point charges – q and + q are located at points (0, 0, –α) and (0, 0, α) respectively.

The electric potential at a point (0, 0, z), where z > α is qα q (1) (2) 2 4 πε0 α 4 πε0 z 2qα

(

4 πε0 z 2 − α 2

)

(4)

2qα

(

4 πε0 z 2 + α 2

)

Potential Due to an Electric Dipole 14. An electric dipole is placed as shown in figure. The electric potential (in 102v) at p 1   due to the dipole is  K = 4 πε 0  

3 cm –q

5 cm 3 cm

+q

8 qK 5

(2)

8 qK 3

5 qK 8  15. A dipole of dipole moment p = 2iˆ − 3 ˆj + 4kˆ is placed at point A(2, –3, 1). The electric potential due to this dipole at the point B(4, –1, 0) is equal to _______× 109 volts. (All the parameters specified here are in S.I. units) (1) 4 (2) 2 (3) 1 (4) 6 (3)

3 qK 8

(4)

16. A dipole of dipole moment is kept at the centre of a circle of radius r as shown in the figure. The radius of the circle is very large in comparison to the distance between the two charges of the dipole. A & B are two points on the axis and C & D are two points on the equatorial line of the dipole. If VA, VB, VC and VD are potentials at A, B, C and D respectively, then which of the following is CORRECT? C

(3)

(3)

(1)

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p

B

A P D

2kp r2 kp (2) VA − VB = 0, VA − VD = 2 r 2kp kp (3) VA −= VB , VA −= VD 2 r r2 −kp 2kp , VA −= VB VD (4) VA −= 2 r r2 17. Two dipoles each of moment 5 × 10–12 Cm form a cross with their axis (– to +) along the coordinate axes. The potential at a point 20cm away in a direction making an angle of 30° with axis is (1) 1.12 V (2) 2.12 V (3) 2.4 V (4) 1.536 V (1) VA − VB = 0, VA − VD =

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CHAPTER 2: Electrostatic Potential and Capacitance

Potential Due to Continuous Charge Distribution 18. A hollow metal sphere of radius R is given ‘+Q ‘ charge to its outer surface. The electric potential at a distance (R/3) from the centre of the sphere will be

1 Q 3 Q (1) (2) 4 πε0 3R 4 πε0 R 1 Q 1 Q (3) (4) 4 πε0 9R 4 πε0 R 19. The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0 x 10–15 m (1) 80 volt (2) 8 × 106 volt (3) 9 volt (4) 9 × 105 volt 20. Initially the spheres A & B are at potential VA and VB. The potential of A when sphere B is earthed is, B

A

(1) VA (3) VA – VB

(2) VB (4) VA + VB

21. There are three concentric thin spherical shells A, B and C of radii R, 2R, and 3R. Shells A and C are given charges q and 2q and shell B is earthed. Then which of the given is correct?

4 q 3 (2) charge on outer surface of shell B is 4 − q 3 2 (3) charge on outer surface of shell C is q 3 (4) all the above

q potential at the point O if r < a is 4 πε0 times 1 1 1 (1)  − +  r a b

1 1 1 (2)  − +  a r b

1 1 1 (3)  − +  b c r 

1 1 1 (4)  − +  a b r 

23. Two isolated metallic solid spheres of radii R and 2R are charged such that both have same charge density σ. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ′ σ. The radio is: σ 9 4 5 5 (1) (2) (3) (4) 4 3 6 3 24. Two conducting spheres of radii R1 and R2 are at the same potential. The electric intensities on their surfaces are in the ratio of (1) 1 : 1 (2) R1 : R2 (3) R2 : R1 (4) R12 : R22 25. In the electric field of a point charge Q a certain charge is carried from point A to B, C, D and E. Then the work done is (Q is at the centre of the circle)

A

(1) charge on inner surface of shell C is

22. A point charge q is at a distance r from the centre O of an uncharged spherical conducting layer, whose inner and outer radii are equal to a and b respectively. The

o +q B

E C

D

(1) least along the path AB (2) least along the path AD (3) zero along any of the paths AB, AC, AD and AE (4) least along AE

IL RANKER SERIES FOR NEET 26. Some equipotential surfaces are shown is figure. The electric field strength is

10V y

20V

5cm

+q X(cm)

10cm 15cm

1+ 2 (3) –2q

(2) 100 V/m along y- axis

V/m at an angle 120° with x- axis

3 27. What is the angle between electric field and equipotential surface?

(1) 90° always (2) 0° always (3) 0° to 90° (4) 0° to 180°

28. Figure shows equipotential surfaces concentric at ‘O’ the magnitude of electric field at distance r (in meter) measured from

20CM o

30CM

10cm

(2)

−2q

2+ 2 (4) +q

of 2a and a third charge –2q is placed at the mid-point. The potential energy of the system is

q2 (1) 8πε0 a (3) −

7q 2 8πε0 a

6q 2 (2) 8πε0 a (4) −

9q 2 8πε0 a

31. Two equal point charges are fixed at x = –a and x = +a on the axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of system, when it is displaced by a small distance x along the X–axis is approximately proportional to (1) x (2) x2 (3) x3 (4) 1/x 32. Identical charges (–q) are placed at each corner of a cube of side b, then the electrostatic potential energy of charge (+q) placed at the centre of the cube will be

60V 30V 20V

(1)

9 Vm −1 2 r

(

)

(2)

6 Vm −1 r2

)

(1)

4 2q 2 π∈0

(3)

2 Vm −1 r2

)

(4)

16 Vm −1 r2

(2)

8 2q 2 π∈0 b

(

+q

a

30. Two equal charges q are placed at a distance

(3) 400 V/m at an angle 120° with x- axis

400

−q

(1)

(1) 100 V/m along x- axis

(4)

Q

30V

30°

0

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(

(

)

Electrostatic Potential Energy 29. Three charges Q, +q and +q are placed at the

(3) −

vertices of a right-angle isosceles triangle as shown in Fig. The net electrostatic energy of the configuration is zero if Q is equal to:

(4) −

4q 2 3π∈0 b 8 2q 2 4 π∈0 b

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CHAPTER 2: Electrostatic Potential and Capacitance

33. Two identical particles of charge q each are connected by a mass less spring of force constant K. They are placed over a smooth horizontal surface. They are released when the separation between them is r and spring is in its natural length. If maximum extension of the spring is r, the value of is (neglect gravitational effect)

K

q

K

q

(1)

q 4r

1 πε0 r

(2)

q 2r

(3)

2q 1 r πε0 r

(4)

q 1 r πε0 r

1 πε0 r

34. Two identical thin rings, each of radius R metres, are coaxially placed at a distance R metres apart. If Q1 coulomb and Q2 coulomb are respectively, the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is:

Q1

2 R

A

B

R R

(3)

(4)

(2) 5.7 × 10–21 J

(3) 9.21 × 10–23 J

(4) 5.96 × 1024 J

36. A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q was given a speed 2v, the closest distance of approach would be (1) r

(2) 2r

(3) r/2

(4) r/4

Potential Energy in an External Electric Field 37. Two opposite and equal charges 4 × 10 –8 coulomb when placed 2 × 10–2 cm away, form a dipole. If this dipole is placed in an external electric field 4 × 108 newton/coulomb, the value of maximum torque and the work done in rotating it through 180° will be (1) 64 × 10–4 Nm and 64 × 10–4 J (2) 32 × 10–4 Nm and 32 × 10–4 J (3) 64 × 10–4 Nm and 32 × 10–4 J 38. An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively.

(1) 2q.E and minimum

(1) Zero (2)

(1) 9.2 × 10–21 J

(4) 32 × 10–4 Nm and 64 × 10–4 J

Q2

q

35. A spherical drop of radius 10 –6 m has absorbed 40 electrons. The energy required to give an additional electron to it is

(2) q.E and p.E

q ( Q1 − Q2 )

( 2 − 1)

2 4 πε0 R q 2 ( Q1 + Q2 ) 4 πε0 R q ( Q1 + Q2 )

( 2 + 1)

2 4 πε0 R

(3) Zero and minimum (4) q.E and maximum 39. A body of mass 1 gm and carrying a charge 10–8 C passes from two points P and Q. P and Q are at electric potentials 600 V and 0 V respectively. The velocity of the body at Q is 20 cm s–1. Its velocity in ms–1 at P is (1)

0.028

(2)

0.056

(3)

0.56

(4)

5.6

IL RANKER SERIES FOR NEET 40. A charge of –2 C and mass 0.5 kg at infinity is given a speed of 2 m/s, just to reach a point where the potential is 4 V. The work done by the agent is (1) –8 J (2) +8 J (3) –9 J (4) –7 J 41. A region contains a uniform electric field  = E 10iˆ + 30 ˆj Vm −1 . A and B are two

(

)

points in the field at (1,2,0) m and (2,1,3) m, respectively. The work done when a charge of 0.8 C moves from A to B in a parabolic path is (1) 8 J

(2) 16 J (3) 40 J

(4) 80 J

Dielectric and Polarisation 42. In a uniform electric field of 24000 V/m, a dielectric slab of thickness 10 cm is kept such that the electric field lines are perpendicular to the surface of the slab. If the dielectric constant of the material of the slab is K = 6, the intensity of the induced electric field inside the dielectric is (1) 4000 V/m (2) zero (3) 20000 V/m (4) 28000 V/m 43. Example of polar molecules (1) O2 & H2 (2) HCl and H2O (3) O2 & HCl (4) H2 & H2O 44. Dielectric constant of pure water is 81. Its permittivity will be (1) 7.16 × 10–10 MKS units (2) 8.86 × 10–12 MKS units (3) 1.02 × 10–13 MKS units (4) Cannot be calculated

Capacitors and Capacitance 45. The plates of a parallel plate condenser are being moved away with velocity v. If the plate separation at any instant of time is d then the rate of change of capacity with time is proportional to (1) 1/d (2) 1/d2 (3) d2 (4) d 46. 64 identical mercury drops combine to form one bigger drop. The capacitance of the

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bigger drop, as compared to that of a smaller drop will be (1) 8 times (2) 64 times (3) 4 times (4) 16 times 47. The time in seconds required to produce a potential difference of 20 V across a capacitor of 1000 μF when it is charged at the steady rate of 200 μC /s is (1) 50 (2) 100 (3) 150 (4) 200 48. Capacitance of a capacitor made by a thin metal foil is 2 µF . If the foil is folded onto itself with a paper of thickness 0.15 mm and dielectric constant of paper is 205, width of paper is 40 mm then length of foil will be (1) 0.34 m (2) 1.33 m (3) 13.4 m (4) 4.13 m

Effect of Dielectric on Capacitance 49. A parallel plate condenser of capacity 5 µF is kept connected to a battery of emf 10 V. If the space between the plates is filled with a medium of dielectric constant 12, then the additional charge taken from the battery is (1) 400 µC (2) 450 µC (3) 500 µC (4) 550 µC 50. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1= 3 and thickness d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3. Capacitance of the capacitor is now (1) 1.8 pF (2) 45 pF (3) 40.5 pF (4) 20.25 pF 51. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to (1) 6 × 10–7 C/m2 (2) 3 × 10–1 C/m2 (3) 3 × 104 C/m2 (4) 6 × 104 C/m2

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CHAPTER 2: Electrostatic Potential and Capacitance

52. A dielectric slab of length l, width b, thickness d and dielectric constant K fills the space inside a parallel plate capacitor. At t = 0, the slab begins to be pulled out slowly with speed v. At time t, the capacity of the capacitor is

ε 0b  Kl − ( K − 1) vt  d  ε 0b Kl + K + vt  (2) d  ε 0b ( Kl + 1) vt  (3) d  ε 0b l + ( K − 1) vt  (4) d  (1)

56. Three capacitors each of capacitance C and of breakdown voltage V are joined in series.

The capacitance and break down voltage of the combination will be (1)

C V , 3 3

(2) 3C ,

V 3

C ,3V (4) 3C, 3V 3 57. Four capacitors of equal capacitance have an equivalent capacitance C1 when connected in series and an equivalent capacitance C2 C1 when connected in parallel. The ratio C2 is 1 1 1 1 (1) (2) (3) (4) 12 4 8 16 (3)

53. The plates in a parallel plate capacitor are separated by a distance d with air as the medium between the plates. In order to increase the capacity by 66% a dielectric slab of dielectric constant 5 is introduced between the plates. What is the thickness of the dielectric slab.

d (1) 4

system became 24 μF. The value of K will be: (1) 1.5 (2) 2.5 (3) 1.2 (4) 3

d (2) 2

5d (3) (4) d 8 54. A parallel plate capacitor of capacitance C (without dielectrics) is filled by dielectric slabs as shown in figure. Then the new capacitance of the capacitor is

58. The equivalent capacity between A and B in the given circuit is (C1 = 4 F, C2 = 12 F, C3 = 8 F C4 = 4 F, C5 = 8 F)

C5 A

C1

C2

C3

B

C4 K3

d

6

K1=2

K2=4

w

w

(1) 3.9 C

(2) 4 C

(3) 2 C

d

16 8 µF (4) µF 3 3 59. An infinite number of identical capacitors each of capacity 1 mF are connected as shown in the figure. The equivalent capacity between A and B (1) 24 µF (2) 36 µF (3)

(4) 3 C

Combination of Capacitors 55. Two capacitors, each having capacitance 40 μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the

8 capacitors 16 capacitors A

B

(1) 1/2 mF (2) 1 mF (3) 2 mF (4) 4 mF

IL RANKER SERIES FOR NEET 60. A number of capacitors, each of capacitance 1µF and each one of which gets punctured if a potential difference just exceeding 500 V is applied are provided. Then an arrangement suitable for giving a capacitor of capacitance 3 µF across which 2000 V may be applied requires at least (1) 4 component capacitors (2) 96 component capacitors (3) 48 component capacitors (4) 3 component capacitors 61. A capacitor 1mF withstands a maximum voltage of 6 kV while another capacitor 2 mF withstands a maximum voltage of 4 kV. If the capacitors are connected in series, the system will withstand a maximum voltage of (1) 2 kV (2) 4 kV (3) 6 kV (4) 9 kV

(1) 5 µF

(2) 3 µF (3) 4 µF

2C

C a

(1)

4C 3

C

2C

3C

3C

3C

(2)

C 3

(3)

b

2C 3

e

e0

C

d

d

d

C1

C1

ε0 A d εε0 A (3) d(ε + ε0 )

C

C

(1) 1.25 µF (2) 6 µF (3) 3 µF

(4) 12 µF

63. The capacity of each condenser in the following fig. is ‘C’. Then the equivalent capacitance across A and B is

A

B (1) C/4 (3) 4 C/3

(2) 3 C/4 (4) 3 C

64. If the equivalent capacity between A and B in the circuit is 12 µF, the capacity C is

6µF 6µF A

8µF

8µF

(2)

Energy Stored in a Capacitor 67. If the charge on a body is increased by an amount of 2 C, the energy stored in it increases by 21%. The original charge on the body.

(1) 1 C (2) 10 C (3) 20 C (4) 15 C 68. A capacitor of capacitance 2 µF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is

1 S

B C

5C 3

εε0 A d εε0 A (4) d(2ε + ε0 )

(1)

B

(4)

66. For configuration of media of permittivity between ε0 , ε, ε0 parallel plates each of area A, as shown in the figure the equivalent capacitance is

e0

C

(4) 8 µF

65. Find the effective capacitance between the terminals a and b shown in figure.

C

62. If in the infinite series circuit, C = 9 µF and C1 = 6 µF then the capacity across AB is

A

| 153

2

V 2µF (1) 20%

(2) 75% (3) 80%

8µF (4) 0%

154 |

CHAPTER 2: Electrostatic Potential and Capacitance

69. Consider the situation shown in the figure.

The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is: q + + + s + + + A B (1) zero

q 2

(2)

(3) q

(4) 2q

70. Two parallel plate condensers A and B have capacities 1 µF and 5 µF are charged separately to same potential of 100 V. Now the positive plate of A is connected to the negative plate of B and the negative plate of A to the positive of plate of B then (1) final charge on A and B are

200µC 1000µC and 3 3 (2) the loss of energy is 1.67 × 10–2 J (3) both 1 and 2 are correct (4) both 1 and 2 are wrong 71. Two circuits (a) and (b) have charged capacitors of capacitance C, 2C and 3C with open switches. Charges on each of the capacitor are as shown in the figures. On closing the switches, S S 2Q

2Q

Q

3C L

(a)

R

L

(b)

72. A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure.

When key ‘K’ is closed, the total energy stored across the combination is E1. Now key ‘K’ is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E2. The ratio E1 will be: E2 K ( )

V

C

(1)

1 10

(2)

2 5

(3)

5 13

(4)

5 26

73. A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel.

The total electrostatic energy of resulting system: (1) Increases by a factor of 2 (2) Increases by a factor of 4 (3) Decreases by a factor of 2 (4) Remains the same

Q 2C

2C

C

(4) Charges flow from L to R in both (a) and (b)

R

(1) No charge flows in (a) but charge flows from L to R in (b) (2) Charge flows from L to R in (a) but charge flows from R to L in (b) (3) Charges flow from R to L in (a) and from L to R in (b)

74. Two capacitors of capacity 4 µF and 6 µF are connected in series and a battery is connected to the combination. The energy stored is E1. If they are connected in parallel and if the same battery is connected to this combination the energy stored is E2. The ratio E1:E2 is (1) 4 : 9

(2) 9 : 14

(3) 6 : 25

(4) 7 : 12

IL RANKER SERIES FOR NEET 75. One plate of a capacitor is connected to a spring as shown. Area of both the plate is A and separation is d when capacitor is uncharged. When capacitor is charged, in steady state separation between the plates is 0.8 d. The force constant of the spring is

o

E ε AE 3 (1) 0 2d 3 ε AE 2 (2) 65 0 32 2d 3 ε AE 2 (3) 125 0 16 2d 3 (4) 125 4

ε0 AE 2 d3

(4)

(2n + 1) (n + 1)

77. Capacity of a spherical capacitor is C1 when

inner sphere is charged and outer sphere is earthed and C2 capacitor when inner sphere is earthed and outer sphere is charged. Then C1 is (a is radius of inner sphere and b is C2 radius of outer sphere)

a b a +b b (3) (4) a −b a 78. Three conducting spheres A, B and C are as shown in figure. The radii of the spheres are a, b and c respectively. A and B are connected by a conducting wire. The capacity of the system is (1) 1

(2)

C B A

Capacitance of a Concentric Spherical Capacitor 76. The capacity of an isolated copper sphere increases ‘n’ times, when it is enclosed by an earthed concentric thin copper shell. The ratio of the radius of the outer shell to that of the inner sphere is

a b

2

n (1) (n − 1) n (2) (n − 1) (3)

(2n) (n + 1)

| 155

C

(1) 4πε0(a + b + c)

 bc  (2) 4 πε0    c −b   1 1 1 (3) 4 πε0  + +  a b c abc   (4) 4 πε0   ab bc ca + +  

THEORY-BASED QUESTIONS Single Correct MCQ

(2) maximum at centre

1. Some charge is being given to a conductor, then its potential is

(3) same throughout the conductor

(1) maximum at surface

(4) maximum somewhere between surface and centre

156 |

CHAPTER 2: Electrostatic Potential and Capacitance

2. The value of electric potential at any point

 1  due to any electric dipole is  k =  4 πε0     p×r (1) k ⋅ 2 r   p×r (2) k ⋅ 3 r   p ⋅r (3) k ⋅ 2 r  p.r (4) k. 3 r 3. Out of the two following statements

(4) No change in direction perpendicular to field 6. In bringing an electron towards another electron, the electrostatic potential energy of the system (1) decreases (2) increases (3) remains same (4) becomes zero 7. At each corner of an equilateral triangle identical charges are placed. Then (1) at the centre of the triangle the resultant electric intensity is zero

A) Two like charges can only produce a null point.

(2) at the centre of the triangle the net electric potential is zero

B) Due to two opposite charges infinite number of zero potential points are formed

(3) the electrostatic potential energy of the

(1) A is correct B is wrong (2) A is wrong and B is correct (3) Both A and B are wrong (4) Both A and B are Correct 4. Out of the following two statements.

A) As we move in the direction of the field potential goes on decreasing

system is zero (4) the resultant electric intensity at any corner is zero 8. Out of the following statements A) three charge system can not have zero mutual potential energy B) the mutual potential energy of a system of charges is only due to positive charges

B) If a charged body is moved within the field work must be done by field.

(1) A is wrong and B is correct

(1) A is correct and B is wrong

(2) A is correct and B is wrong

(2) A is wrong and B is correct (3) Both A and B are correct (4) Both A and B are wrong 5. An electron enters into high potential region V2 from lower potential region V1 then its velocity (1) Will increase (2) Will change in direction of field (3) No change in direction of field

(3) Both A and B are correct (4) Both A and B are wrong 9. Identify the correct increasing order of the gain in kinetic energies in the following cases a) Alpha particle accelerated through a P.D. of ‘2V’ b) Proton accelerated through a P.D of 2V

IL RANKER SERIES FOR NEET c) Deuteron accelerated through a P.D. of 3V d) Electron accelerated through a P.D. of 5V (1) b, c, a, d

(2) c, d, a, b

(3) d, b, a, c

(4) a, c, b, d

10. Read the following statements a) Non polar molecules have uniform charge distribution b) Polar molecules have non - uniform charge distribution c) Molecules are already polarized without electric field in polar molecules d) Molecules are not already polarized without electric field in Non–polar molecules

(2)

IV

III

II

I

(3)

III

IV

I

II

(4)

I

II

III

IV

| 157

12. A capacitor works in (1) A.C. circuits (2) D.C. circuits (3) Both the circuits (4) Neither in A.C. nor in D.C. circuit  13. The electric field ( E ) between two parallel plates of a capacitor will be uniform if (1) the plate separation (d) is equal to area of the plate (A) (2) the plate separation (d) is greater when compared to area of the plate (A) (3) the plate separation (d) is less when compared to area of plate (A) (4) 2 (or) 3 14. Two conductors when connected by a wire, charge flows if they have

(1) only a & b are correct (2) only c & d are correct

(1) different charges

(3) only ‘c’ is wrong

(2) different potentials

(4) all are correct

(3) different capacities

11.

(4) different charge densities List–I

List–II

15. Figure shows two capacitors connected in series and joined to a cell. The graph shows the variation in potential as one moves from left to right on the branch containing capacitors.

(A) proton and electron

(I) gains same velocity in an electric field for same time

(B) proton and positron

(II) gains same KE in an electric field for same time

(C) Deuteron and α-particle

(III) experience same force in electric field

C1

(D) electron and positron

(IV) gains same KE when accelerated by same potential difference.

(1) C1 > C2

(4) data insufficient to conclude the answer

(1)

(A)

(B)

(C)

(D)

IV

III

I

II

C2

(2) C1 = C2 (3) C1 < C2

158 |

CHAPTER 2: Electrostatic Potential and Capacitance

16. Two thin dielectric slabs of dielectric constants K1 and K2 (K1 < K2) are inserted between the plates of a parallel plate capacitor, as shown in the figure alongside.

The variation of electric filed E between the plates with distance d as measured from plate P is correctly shown by P 1 Q + + + + +

17. An insulator plate is passed between the plates of a capacitor. Then current in external circuit

A

B

E

(1) always flows from A to B (2) always flows from B to A (3) first flows from A to B and then from B to A (4) first flows from B to A and then from A to B

0

18. A parallel plate air capacitor is fully charged and then the battery is removed. A dielectric slab is now put between the plates. Which of the following statements is correct?

K1

(1)

d

K2

(1) The charge on the plates decreases

(2)

(2) The charge on the plates does not change, but the potential difference increases

E

0

(3)

d

E

0

(3) The charge on the plates does not change, the potential difference between the plates decreases and the energy stored also decreases (4) The charge on the plates does not change, the potential difference between the plates increases and the energy stored also increases 19. In a parallel-plate capacitor, the region between the plates is filled by a dielectric slab. The capacitor is connected to a cell and the slab is taken out

d

(1) Some charge is drawn from the cell (2) Some charge is returned to the cell

E

(3) The potential difference across the capacitor is reduced

(4)

0 d

(4) No work is done by an external agent in taking the slab out

IL RANKER SERIES FOR NEET 20. A parallel plate condenser is charged and disconnected from the battery. If the plates of the capacitor are moved further apart by means of insulating handles: (1) The charge in the capacitor is zero (2) The capacitance becomes infinite (3) The charge in the capacitor increases (4) The voltage across the plates increases 21. A parallel plate capacitor (of capacity C) is charged using a battery of potential V. It is disconnected from the battery and the plates of the capacitor are moved away from each other. Now it is reconnected to the same battery. Then (1) Charge flows from the capacitor to the battery (2) Charge flows from the battery to the capacitor (3) There is no flow of charge (4) The direction of flow of charge depends on the values of V and C 22. Three identical condensers are connected in different combinations using all three each time arrange the following cases in the increasing order of effective capacity. i) all in series ii) all in parallel iii) two in series and one in parallel iv) two in parallel and one in series (1) i, iv, iii, ii

(2) i, ii, iii, iv

(3) ii, iii, i, iv

(4) iii, ii, iv, i

23. Which of the following statements are correct? a) When capacitors are connected in parallel the effective capacitance is less than the individual capacitances b) The capacitances of a parallel plate capacitor can be increased by decreasing the separation plates c) When capacitors are connected in series the effective capacitance is less than the least of the individual capacities

| 159

d) In a parallel plate capacitor the electrostatic energy is stored on the plates (1) a and b (2) a and c (3) c and d (4) b and c 24. The electric field intensity at a point between the plates of a charged isolated capacitor is a) proportional to the square of the charge on the plates b) proportional to the charge on the plates c) inversely proportional to the distance between the plates d) independent of the distance between the plates (1) a, b and c are correct (2) b and d are correct (3) b, c and d are correct (4) a, b, c and d are correct 25. An uncharged capacitor is connected to a battery. On charging the capacitor (1) All the energy supplied by the battery is stored in the capacitor (2) Half the energy supplied by the battery is stored in the capacitor (3) The energy stored depends upon the capacity of the capacitor only (4) The energy stored depends upon the time for which the capacitor is charged 26. A capacitor ‘C’ is charged to a potential V by a battery. The e.m.f of the battery is V. It is then disconnected from the battery and again connected with its polarity reversed to the battery. Then a) The work done by the battery is CV2 b) The total charge that passes through battery is 2 CV c) The initial and final energy of the capacitor is same d) The work done by the battery is 2 CV2 (1) only a, c, d are true (2) only a, b, d are true (3) only a, d, b are true (4) only b, c, d are true

160 |

CHAPTER 2: Electrostatic Potential and Capacitance

Assertion & Reason Type Questions Directions for following questions (Q.No. 27-40)

−dV ,–ve sign dr indicates potential decreases in the opposite direction of the electric field.

(R) : In the formula E =

In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as

32. (A) : A cavity inside a conductor is shielded from outside electrical influences.

(1) if both (A) and (R) are true and (R) is the correct explanation of (A),

(R) : Generally Iron is used for electric and magnetic shielding.

(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),

33. (A) : Suppose if we put charges inside cavity, the exterior of the conductor is not shielded from the fields by the inside charges.

(3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.

In light of the given statements, choose the correct answer from the options given below

(R) : Electric lines of force can pass through a charged conductor.

27. (A) : When a proton with certain energy moves from low potential to high potential then its KE decreases.

34. (A) : Electric field is continuous across surface of a spherical charged shell, where as electric potential is discontinuous across the surface.

(R) : The direction of electric field is opposite to the potential gradient.

28. (A) : When charges are shared between two bodies, there occurs no loss of charge, but there does occur a loss of electrical energy.

(R) : In case of sharing of charges law of conservation of energy fails.

29. (A) : Two adjacent spherical conductors, carrying the same positive charge can have a potential difference between them.

(R) : The potential to which a conductor is raised depends upon the charge.

30. (A) : In bringing an electron towards a proton electrostatic potential energy of the system decreases.

(R) : Potential due to proton is positive.

31. (A) : Electric field is in the direction in which the potential decreases steepest.

(R) : The electric field is not zero and the electric potential is zero inside the charged spherical shell 35. (A) : Electric displacement of dielectric medium filled between the plates of parallel plate condenser is D = ε0E + P Here E is electric field in the medium and P is the dielectric polarisation.

(R) : Electric susceptibility of dielectric medium is zero [K = dielectric constant, ∈0 = permitivity of free space.

36. (A) : Two capacitors are connected in parallel to a battery. If a dielectric medium is inserted between the plates of one of the capacitors then the energy stored in the system will increase.

(R) : On inserting dielectric medium between the plates of a capacitor, its capacity increases

IL RANKER SERIES FOR NEET

37. (A) : If a dielectric is placed in an external field the field inside dielectric will be less than applied field.

| 161

39. (A) : Potential energy of a system of point charges may be zero.

(R) : Electric field induce dipole moment opposite to field direction.

(R) : Potential energy of individual point charges in a system of point charges may be zero.

38. (A) : Potential at the surface of a solid metallic sphere may be non-zero, even if net charge on the surface is zero.

40. (A) : Force between the plates of an isolated charged capacitor does not depend on distance between the plates.

(R) : If potential at the center of a solid metallic sphere is non-zero, potential at the surface of the sphere is non-zero.

(R) : Force between the plates can not be calculated using coulomb’s law.

BRAIN TEASERS 1. Charge Q is uniformly distributed on a dielectric rod AB of length 2l. The potential at P shown in the figure is equal to

(2) (3)

q (1) 4 πε 2l 0

q (2) 4 πε l In(2) 0

q (3) 4 πε 2l In(3)

(4)

0

2q In(2) 4 πε0 l

 2. An electric field given by = E yiˆ + xjˆ N / C .

The work done in moving a 1C charge from   r = 2iˆ + 2 ˆj mtor = 4iˆ + ˆj m is

(

A

(

)

B

(1) +4 J (3) +8 J

(

)

)

(2) –4 J (4) zero

3. The gap between the plates of a parallel–plate capacitor is filled with isotropic dielectric whose relative permittivity K varies linearly from K 1 to K 2 (K 2 > K 1) in the direction perpendicular to the plates. The area of each plate equals A, the separation between the plates is equal to d. Then the capacitance of the capacitor will be given by (1)

( K 2 − K1 ) ∈O A K  d ln  2   K1 

(4)

( K 2 − K1 ) ∈O A d

( K 2 − K1 ) ∈O A 2d 2 ( K 2 − K1 ) ∈O A

K  d ln  2   K1  4. Two square plates of side ‘a’ are arranged as shown in the figure. The minimum separation between plates is ‘d’ and one of the plate is inclined at small angle ∝ with plane parallel to another plate. The capacitance of capacitor is (given ∝ is very small) α d a ∈0 a 2  αa  1− (1) d  2d 

∈0 a 2  αa  1− (2) d  d 

∈0 a 2  αa  1+ (3) d  2d 

∈0 a 2  αa  1+ (4) d  4d 

162 |

CHAPTER 2: Electrostatic Potential and Capacitance

5. Consider the situation shown in figure. The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf V. All surfaces are frictionless. The value of M for which the dielectric slab will stay in equilibrium is L

d

K

densities σ, –σ and σ respectively. If VA, VB and VC denote the potentials of the three shell. For c = a + b, we have (1) VC = VA ≠ VB (2) VC = VB ≠ VA (3) VC ≠ VA ≠ VB (4) VC = VB = VA 7. Find the equivalent capacitance across AB (all capacitances in µF)

M ε0bV 2 ( K − 1) (1) 4dg

(2)

ε0bV 2 ( K − 1) 2dg

ε0bV 2 ( K − 1) dg

(4)

ε0bV 2 ( K − 1) 8dg

(3)

6. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge

10

17 A 13

5

1 1

6 20 µF 3

(2) 9 F

(3) 48 F

(4) None

(1)

FLASHBACK (Previous NEET Questions) 1. The value of electric potential at a distance of 9 cm from the point charge 4 × 10–7 C is   1 = 9 × 109 N m 2 C -2  .  Given 4 πε0  

(1) 4 × 102 V (3) 4.4 × 105 V

(2024–R)

(2) 44.4 V (4) 4 × 104 V

2. When a particle with charge+q is thrown with an initial velocity v towards another stationary charge +Q, it is repelled back after reaching the nearest distance r from +Q. The closest distance that it can reach if it is thrown with initial velocity 2v, is (2022-O) (1) (r/2) (2) (r/16) (3) (r/8) (4) (r/4) 3. The effective capacitances of two capacitors are 3 µF and 16 µF, when they are connected in series and parallel, respectively. The capacitance of two capacitors are (2022-R) (1) 10 µF, 6 µF (2) 8 µF, 8 µF (3) 12 µF, 4 µF (4) 1.8 µF, 1.2 µF

4. Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

(2021) (1) 1980 V (2) 660 V (3) 1320 V (4) 1520 V 5. A parallel plate capacitor having cross– sectional area A and separation d has air in between the plates. Now, an insulating slab of same area but thickness d/2 is inserted between the plates as shown in the figure, having dielectric constant K(=4). The ratio of new capacitance of its original capacitance will be (2020-II) d

K

d/2

(1) 8 : 5

(2) 6 : 5 (3) 4 : 1

(4) 2 : 1

IL RANKER SERIES FOR NEET 6. A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel.

The total electrostatic energy of resulting system(2017)

d

A/3

A/3

A/3

k1

k2

k3

d/2

k4

(1) decreases by a factor of 2 (2) remains the same

| 163

A (1) k =k 1 + k 2 + k 3 + 3k 4

(3) increases by a factor of 2 (4) increases by a factor of 4 7. A parallel plate capacitor is to be designed, using a dielectric of dielectric constant 5, so as to have a dielectric strength of 109 Vm-1. If the voltage rating of the capacitor is 12 kV, the minimum area of each plate required to have a capacitance of 80 pF is (2017-R) (1) 21.7 x 10-6 m2 (2) 25.0 x 10-5 m2 (3) 12.5 x 10-5 m2 (4) 10.5 x 10-6 m2 8. A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4, as shown in the figure. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by (2016-II)

(2) k=

2 k + k 2 + k 3 + 2k 4 3 1

(

)

2 3 1 = (3) k k + k + k + k 1 2 3 4 1

1

1

1

1

2

3

3

(4) k = k + k + k + 2k

4

9. A parallel plate air capacitor has capacity C, distance of separation between plates is d, and potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is (2015-R) (1)

CV 2 d

(2)

C2 V 2 2d2

(3)

C2 V 2 2d

(4)

CV 2 2d

CHAPTER TEST 1. ABC  is an equilateral triangle of side 2 m. If E = 10 NC–1 then VA – VB is

A

B (1) 10V (3) 20V

C (2) – 10V (4) – 20V

2. Electric potential is given by V = 6x – 8xy2 – 8y + 6yz – 4z2, then electric force acting on 2 C point charge placed at origin will be (1) 2 N (2) 6 N (3) 8 N (4) 20 N

3. Eight charges having the values as shown in fig are arranged symmetrically on a circle of radium 0.1 m in air. Potential at center ‘O’ will be +5µC -5µC -7µC +15µC

+7µC -5µC

+7µC +11µC

(1) 252 × 10 V (2) 252 × 104 V (3) 252 × 106 V (4) zero 10

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CHAPTER 2: Electrostatic Potential and Capacitance

4. Eight identical point charges each charge q are placed at eight corners of cube of side ‘l’. Then potential at centre of cube is (1) Zero

(3)

1 16q 2πε0 3l

(2)

1 8q 4 πε0 3l

(4)

1 16q 4 πε0 3l

5. If the potential at certain distance from a point charge is 600 V and the intensity of electric field is 300 NC–1 The magnitude of the charge in coulomb is ___ (1) 1.33 × 10–7

(2) 2 × 10–6

(3) 2 × 10–7

(4) 33 × 10–6

6. Two electric charges 12 μC and –6 μC are placed 20 cm apart in air. There will be a point P on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of P from –6 μC charge is (1) 0.10 m

(2) 0.15 m

(3) 0.20 m

(4) 0.25 m

7. If the electric potential at a certain distance from a point charge is 900 V and electric intensity is 225 Vm–1. The charge is (1) 4 × 10–7 C

(2) 2 × 10–7 C

(3) 4 × 10–3 C

(4) 2 × 10–5 C

8. Choose the wrong statement (1) On the perpendicular bisector of electric dipole, intensity of electric field (E) and potential (V) are E = 0,V ≠ 0 (2) Electric lines of force are always perpendicular to equipotential surface (3) When the proton and an electron moves away from each other then the electrostatic potential energy increases (4) The dielectric constant of same material will have different values for direct current and alternating current

9.

List–I (A) Two like charges are brought nearer

List–II (P) the force between them decreases

(B) Two unlike charges are brought nearer

(Q) potential energy of the system increases

(C) When a third charge of some nature is placed equidistant from two like charges

(R) mutual forces are not affected

(D) When a dielectric medium is introduced between two charges

(S) potential energy of the system decreases

(1) (2) (3) (4)

(A) IV II IV III

(B) II IV II I

(C) III II I II

(D) I I III IV

10. A proton, a deuteron and an α− particle are accelerated through the same p.d. of V volt.

The velocities acquire by them are in the ratio (1) 1 : 1 : 2

(2) 1 : 2 : 1

2 :1:1  p placed in 11. An electric dipole of moment  a uniform electric field E has minimum potential  energy when the angle between  p and E is π (1) zero (2) 2 (3) 1 : 1 : 1

(4)

(3) π

(4)

3π 2

IL RANKER SERIES FOR NEET 12. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as reason (R).

Assertion (A) : When an uncharged conducting plate is brought near a charged conducting plate, the capacity of the system increases as potential decreases.

Reason (R) : Capacity of a conductor depends on its potential.

In the light of the given statements, choose the correct answer from the options given.

(1) Both (A) and (R) are true and (R) is the correct explanation of (A) (2) Both (A) and (R) are true but (R) is not the correct explanation of (A) (3) (A) is true but (R) is false (4) Both (A) and (R) are false 13. A parallel plate capacitor is filled with a dielectric of dielectric constant (relative permittivity) 5 between its plates and is charged to acquire an energy E. Then it is isolated (disconnected from the battery) and the dielectric is replaced by another dielectric of dielectric constant (relative permittivity) 2. The new energy stored in the capacitor is (1) E (2) 2.5 E (3) 5 E (4) 10 E 14. The force acting upon a charged particle kept between the plates of a charged capacitor is F. If one of the plates is removed then the force acting on the same particle will be (1) F (2) zero (3) 2F (4) F/2

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will have lower charge than the hollow one.

Reason

(R) : Capacitance of metallic spheres depend on the radii of spheres.

In the light of the above statements choose the correct answer from the options given below.

(1) A is false but R is true.

(2) Both A and R are true and R is the correct explanation of A (3) A is true but R is false (4) Both A and R are true but R is not the correct explanation of A 16. What is the area of the plates of a 3 F parallel plate capacitor, if the separation between the plates is 5 mm (1) 1.694 × 109 m2 (2) 4.529 × 109 m2 (3) 9.281 × 109 m2 (4) 12.981 × 109 m2 17. From the initial point A(1, 2 , 0), to the final point D(3, 2, 0), a charge of 2 C is moved through the points B(2, 2, 0 ) and C (3, 1, 0) along the path ABCD. The electric field in that region is 4 i N/C. The work done by the field is given by (1) 8 J (2) – 8 J (3) 16 J (4) –4 J 18. An electric dipole is along an uniform electric field. For a deflection of 60º, work to be done by an agent is 2 × 10−19 J. Then the extra work to be done by an agent if it is to be deflected by 30º further is (1) 2.5 × 10–19 J (2) 2 × 10–19 J –19 (3) 4 × 10 J (4) 2 × 10–16 J

15. Given below are two statements: One is labeled as Assertion A and the other is labeled Reason R.

19. A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is (1) C (2) C/2 (3) zero (4) 2 C

Assertion (A) : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere

20. The charge on a capacitor is 30 µC. It is connected to another capacitor of half of its capacity in parallel then the charge on the first capacitor is (1) 30 µC (2) 9 µC (3) 10 µC (4) 20 µC

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CHAPTER 2: Electrostatic Potential and Capacitance

21. A 20 F capacitor is charged to 5 V and isolated. It is then connected in parallel with an uncharged 30 F capacitor. The decrease in the energy of the system will be (1) 25 J (2) 100 J (3) 125 J (4) 150 J 22. There are two identical capacitors, the first one is uncharged and filled with a dielectric constant K while the other one is charged to potential V having air between its plates.

If two capacitors are joined end to end, the common potential will be

V (1) K −1

KV (2) V +1

V KV (4) K +1 K −1 23. Equivalent capacitance between A and B as shown in the circuit will be (3)

4µF

4µF

25. A dipole is placed in an electric field as shown. In which direction will it move?

(1) towards the right as its potential energy will increase (2) towards the left as its potential energy will increase (3) towards the right as its potential energy will decrease (4) towards the left as its potential energy will decrease 26. In a hollow conducting charged spherical shell,the potential (V) changes with respect to distance (r) from centre

2µF

V A

2µF

2µF

(1) 1 μF (3) 4 μF

(1)

B

(2) 5 μF (4) 3 μF

r

24. The work done in taking a unit positive charge from C to A is WA and from C to B is WB. Then

A

d

Q

d

V (2)

r

B

θ

V (3)

r C

r

(1) WA > WB (2) WA < WB (3) WA = WB (4) WA + WB = 0

V (4)

r

IL RANKER SERIES FOR NEET 27. Plots A and B represents variation of charge with applied P.d. across the combination of two capacitors (series and parallel). The values of capacitance of two capacitors are Q(µC) B 500

A

80

31. A charged sphere is connected to a similar uncharged sphere. Then the percentage loss of energy is (1) 50% (2) 25% (3) 20% (4) 10% 32. The closest distance of approach of an α– particle travelling with a velocity V towards a stationary nucleus is ‘d’. For the closest distance to become d/3 towards a stationary nucleus of double the charge, the velocity of projection of the α-particle has to be

V Q (4) 56 4 33. In a certain region of space with volume 0.2 m3, the electric potential is found to be 5V throughout. The magnitude of electric field in this region is: (1) 0.5 N/C (2) 1 N/C (3) 5 N/C (4) zero (1) 6 V

V(Volt)

10

(1) 20 μF, 30 μF

(2) 40 μF, 10 μF

(3) 20 μF, 40 μF

(4) 30 μF, 50 μF

28. Surface charge density of a sphere of a radius 20cm is 8.85 × 10–8 C/m2. The potential at the centre of the sphere is (ε0 = 8.85 × 10–12 C2 /N-m2) (1) 1000 V

(2) 2000 V

(3) 885 V

(4) 442.5 V

29. The variation of electric potential with distance r from a fixed point is as shown in the figure. The electric intensity at r = 1 m, 3 m and 5.5 m are respectively (in V/m) (volt) V 4 0

2

3

4 5.5

r (m)

(1) –2, 1, 8/3

(2) –2, 0, 8/3

(3) 2, 0, –8/3

(4) 8, –1, –4/3

30. Two spheres of radii in the ratio 2 : 3 have charges 2q and 3q. The force between them is F for certain separation. Now they are touched each other and placed at the same separation. The force between them becomes 5F (1) F (2) 2F (3) 5F (4) 3

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(2)

6V (3)

34. The capacity of a condenser is 4 × 10−6 farad and its potential is 100 volts. The energy released on discharging it fully will be (1) 0.02 joule

(2) 0.04 joule

(3) 0.025 jolue

(4) 0.05 jolue

35. Mark the correct statement (V is electric potential and E is electrostatic field) (1) If E is zero at a certain point, then V should be zero at that point (2) If E is not zero at a certain point, then Vshould not be zero at that point (3) If V is zero at a certain point, then E should be zero at that point (4) If V is zero at a certain point, then E may or may not be zero 36. In Millikan’s oil drop experiment, an oil drop carrying a charge Q is held stationary by a potential difference of 2400 V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 V. What is the charge on the second drop?

(1)

Q 4

(2)

Q 2

(3) Q

(4)

3Q 2

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CHAPTER 2: Electrostatic Potential and Capacitance

0.8m

37. Two charges Q1 C and Q2 C are shown in fig.

A third charge Q3 coulomb is moved from point R to S along a circular path with P as centre. Change in potential energy is R Q 3

Q1 P

Q2 0.6m

T

(1)

Q1Q2Q3 4 πε0

(2)

Q1Q2 πε0

(3)

Q2Q3 πε0

(4)

4Q2Q3 πε0

S

38. If radius of a hollow metallic sphere is ‘R’.

If the P.D between it’s surface and a point at a distance 3R from it’s centre is V then the electric field intensity at a distance 3R from it’s centre is

V V V V (2) (3) (4) 3R 4R 2R 6R 39. A solid conducting sphere having a charge ‘Q’ is surrounded by an uncharged concentric conducting hollow spherical shell. Let the P.D between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q. The new P.D between the same two surfaces is (1) V (2) 2 V (3) 4 V (4) –2 V (1)

40. The voltage of clouds is 4 × 106 volt with respect to ground. In a lightning strike lasting 100 ms, a charge of 4 coulombs is delivered to the ground. The power of the lightening strike is (1) 160 MW (2) 80 MW (3) 20 MW (4) 500 MW 41. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as reason (R).

Assertion (A) : Non–polar material do not have any permanent dipole moment Reason

(R) : When a non polar material is placed in a electric field.

The centre of the positive charge distribution of it’s individual atom or molecule coincides with the centre of the negative charge distribution.

In the light of above statements, choose the most appropriate answer from the options given below.

(1) Both (A) and (R) are correct and (R) is the correct explanation of (A).

(2) Both (A) and (R) are correct and (R) is not the correct explanation of (A).

(3) (A) is correct but (R) is not correct.

(4) (A) is not correct but (R) is correct.

42. Given below are two statements. One is labelled as Assertion (A) and the other is labelled as reason (R).

Assertion(A) : A capacitor is connected to an ideal battery. A dielectric slab is slowly inserted into the gap between the plates of capacitor. Work required in the process is positive.

Reason (R) : Force due to capacitor will try to pull dielectric into the capacitor. To slowly insert external force needs to be outward i.e. opposite to displacement.

In the light of above statements, choose the most appropriate answer from the options given below.

(1) Assertion is false but Reason is true.

(2) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

IL RANKER SERIES FOR NEET (3) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(4) Assertion is true but Reason is false.

43. When a number of liquid drops each of surface charge density s and energy E combine, a large drop is formed. If the charge density of the large drop is 3s, its energy is (1) 81 E (2) 3 E (3) 27 E (4) 243 E 44. Two identical metal plates are given positive charges Q1 and Q2(< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C,

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the potential difference between them is (1) (Q1 + Q2)/(2C) (2) (Q1 + Q2)/(C) (3) (Q1 – Q2)/(C) (4) (Q1 – Q2)/(2C) 45. An air filled parallel plate capacitor has a capacitance of 10 –12 F. The separation of plates is doubled and wax is inserted between them, which makes capacitance 2 × 10–12 F. The dielectric constant of wax is (1) 2.0 (2) 3.13 (3) 4.0 (4) 8.0

ANSWER KEY NEET Drill (1) 1

(2) 2

(3) 4

(4) 2

(5) 2

(6) 2

(7) 3

(8) 2

(9) 1

(10) 3

(11) 2

(12) 4

(13) 3

(14) 3

(15) 2

(16) 3

(17) 4

(18) 3

(19) 2

(20) 3

(21) 4

(22) 1

(23) 3

(24) 3

(25) 3

(26) 3

(27) 1

(28) 2

(29) 2

(30) 3

(31) 2

(32) 3

(33) 2

(34) 2

(35) 1

(36) 4

(37) 4

(38) 3

(39) 1

(40) 3

(41) 2

(42) 3

(43) 2

(44) 1

(45) 2

(46) 3

(47) 2

(48) 4

(49) 4

(50) 3

(51) 1

(52) 1

(53) 2

(54) 1

(55) 1

(56) 3

(57) 2

(58) 3

(59) 3

(60) 3

(61) 4

(62) 3

(63) 3

(64) 1

(65) 1

(66) 4

(67) 3

(68) 3

(69) 1

(70) 3

(71) 3

(72) 3

(73) 3

(74) 3

(75) 4

(76) 2

(77) 2

(78) 2

Theory-based Question (1) 3

(2) 4

(3) 2

(4) 1

(5) 1

(6) 2

(7) 1

(8) 4

(9) 1

(10) 4

(11) 1

(12) 3

(13) 3

(14) 2

(15) 3

(16) 3

(17) 4

(18) 3

(19) 2

(20) 4

(21) 1

(22) 1

(23) 4

(24) 2

(25) 2

(26) 4

(27) 1

(28) 3

(29) 2

(30) 2

(31) 3

(32) 2

(33) 3

(34) 4

(35) 3

(36) 1

(37) 3

(38) 2

(39) 2

(40) 3

(3) 1

(4) 1

(5) 2

(6) 1

(7) 2

Flashback (Previous NEET Questions) (1) 4 (2) 4 (3) 3 (4) 1

(5) 1

(6) 1

(7) 1

(8) 3

(9) 4

Chapter Test (1) 2 (2) 4

(5) 1

(6) 3

(7) 1

(8) 1

(9) 2

Brain Teasers (1) 3

(2) 4

(3) 2

(4) 4

(10) 4

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CHAPTER 2: Electrostatic Potential and Capacitance

(11) 1

(12) 3

(13) 2

(14) 4

(15) 1

(16) 1

(17) 3

(18) 2

(19) 4

(20) 4

(21) 4

(22) 4

(23) 2

(24) 3

(25) 3

(26) 2

(27) 2

(28) 2

(29) 2

(30) 1

(31) 1

(32) 2

(33) 4

(34) 1

(35) 4

(36) 2

(37) 3

(38) 4

(39) 1

(40) 1

(41) 3

(42) 1

(43) 4

(44) 4

(45) 3