ELECTRIC CHARGES AND FORCES

25.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So, electrons have been removed from the glass rod to make it positively charged. (b) Because each electron has a charge of 1.60 x lo-'' C, the number of electrons removed is

25.2. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred into the other because they are relatively free to move. Protons, on the other hand, are tightly bound in the nuclei. So, electrons have been added to the plastic rod to make it negatively charged. (b) Because each electron has a charge of 1.60 x C, the number of electrons added is 20~10c -~ = 1.25 X 10" 1.60~ 10-l9 c

25.3. Model: Use the charge model. Solve: The mass of copper in a 2.0-mm-diameter copper ball is x

The number of moles in rhe ball is n=-= A The number of copper atoms in the ball is

1

m)' = 3.736 x

kg = 0.03736 g

0'03736 = 5.884 x l o 4 mol 63.5 g l m o l

N = nN, = (5.884 x l o 4 mo1)(6.02 x loz3mol-') = 3.542 x 10" We note that the number of electrons per atom is the atomic number, and both the atomic number (29) and the average atomic mass (63.5 g) are taken from the periodic table in the textbook. The number of electrons in the copper ball is thus 29 x 3.542 x IOz0 = 1.027 X 10". The number of electrons removed from the copper ball is

So, the fraction of electrons removed from the copper ball is

Assess: This is indeed a very small fraction of the available number of electrons in the copper ball.

25.4. Model: Use the charge model. Solve: Each oxygen molecule has 16 protons (8 per atom), and there are 6.02 x lo2' oxygen molecules in 1 .O mole of oxygen. Because one proton has a charge of +1.60 x C, the amount of positive charge in 1.0 mole of oxygen is 6.022 x 10" x 16 x 1.6 x C = 1 . 5 4 ~lo6 C

25-2

Chapter 25

25.5. Model: Use the charge model and the model of a conductor as a material through which electrons move. Visualize: Glass Charge transfer

*

Very fast

The charge carriers in a metal electroscope are the negative electrons. As the positive rod is brought near, electrons are attracted toward it and move to the top of the electroscope. The electroscope leaves now have a net positive charge, due to the missing electrons, and thus repel each other. At this point, the electroscope as a whole is still neutral (no net charge) but has been polarized. On contact, some of the electrons move to the positive rod to neutralize some (but not necessarily all) of the rodâ&#x20AC;&#x2122;s positive charge. After contact, the electroscope does have a net positive charge. When the rod is removed, the net positive charge on the electroscope quickly spreads to cover the entire electroscope. The net positive charge on the leaves causes them to continue to repel.

25.6. Model: Use the charge model. Solve: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between (i) two opposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon does charge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, the wall becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is a net attractive electric force between the wall and the balloon. This causes the balloon to stick to the wall, with a normal force balancing the attractive electric force and an upward frictional force balancing the very small weight of the balloon. (b)

*

Fpolar

25.7. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:

The first step shows two neutral metal spheres touching each other. In the second step, the negative rod repels the negative charges which will retreat as far as possible from the top of the left sphere. Note that the two spheres are touching and the net charge on these two spheres is still zero. While the rod is there on top of the left sphere, the right sphere is moved away from the left sphere. Because the right sphere has an excess negative charge by the same amount as the left sphere has an excess positive charge, the separated right sphere is negatively charged as shown in the third step. As the two spheres are moved apart further and the negatively charged rod is moved away from the spheres, the charges on the two spheres redistribute uniformly over the entire sphere surface. Thus, we have oppositely charged the two spheres.

Electric Charges and Forces

25-3

25.8. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:

Charging two neutral sphere with opposite but equal charges can be done through the following four steps. (i) Touch the two neutral metal spheres together. (E) Bring a charged rod (say, positive) close (but not touching) to one of the spheres (say, the left sphere). Note that the two spheres are still touching and the net charge on them is zero. The right sphere has an excess positive charge of exactly the same amount as the left sphereâ&#x20AC;&#x2122;s negative charge. (iii) Separate the spheres while the charged rod remains close to the left sphere. The charge separation remains on the spheres. (iv) Take the charged rod away from the two spheres. The separated charges redistribute uniformly over the metal sphere surfaces.

25.9. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:

Q&\

0-~ -

!- +- 7 J Q ~ 0~ e

Charging two neutral spheres with like charges of exactly equal magnitude can be achieved through the following six steps. (i) Bring a charged rod (say, negative) near a neutral metal sphere. (ii) Touch the neutral sphere with the negatively charged rod, so that the rod-sphere system has a net negative charge. (iii) Move the rod away from the sphere. The sphere is now negatively charged. (iv) Bring this negatively charged sphere close to the second neutral sphere. (v) Touch these two spheres. The excess negative charge is distributed evenly over the two spheres. (vi) Separate the spheres. The excess charge will have the same sign as the charge on the charging rod and will be evenly distributed between the two spheres.

Object polarizes a small piece of Paper

Object picks up a piece of paper

(b) A charged object causes the charges inside a molecule to move apart while still being a part of the molecule. This is because electrons, that is, the negative charges in an insulator such as a piece of paper cannot move freely. Thus, the negative part of each molecule is slightly closer to a positively charged object and the positive part of each molecule is slightly farther from the positively charged object. This charge polarization leads to a slightly greater positive-to-negative attractive force than the positive-to-positive repulsive force. Because there are many molecules in a piece of paper, the total force of attraction is enough to cause a piece of paper to be attracted to the object. The test will also work for a negatively charged object because then the charges inside the molecules of paper will separate in the other direction.

25.11. Model: Model the charged masses as point charges. Visualize:

-

-

F2 on I

e 41 I

0

& % I

1 .o

x (m)

Chapter 25

25-4

Solve: (a)The charge q, exerts a force left. Using Coulomb’s law,

8

on

on q2 to the right, and the charge q2exerts a force

on q1to the

Klqlllq,l (9.OX1o9 ~m~ / c ~ ) ( ~ . o c ) ( I . o c ) eOn2 =FZon1 =-= 9.0 x lo9 N 5; (1.O ml2

(b) Newton’s second law on either q1or q2is

Assess: Coulomb is a pretty “big” unit. That is why

on2(orF,,

is such a large force.

25.12. Model: Model the plastic spheres as point charges.

-

Visualize:

+

d2

F2 0” 1

c-0

92

41 I

I

0

2.0

x (cm)

Solve: (a) The charge q1= -50.0 nC exerts a force

6

on

*on q2 = -50.0 nC to the right, and the charge q2 exerts a

+

force F, on I on q, to the left. Using Coulomb’s law, C)(50.0 x 109C)

(9.0 x lo9 N mz / C2)(50.0 x

K(q,I(q,I 4on2 = eon, =7 =

(2.0 x 10” m)‘

52

= 0.0563 N

(b) The ratio is

-F;on2 w

0.0563 N (2.0 x

kg)(9.8 m / sz)

= 2.87

25.13. Model: Model the glass bead and the ball bearing as point charges. Visualize:

Y (cm)

I

g1 = 20 nC 0

The ball bearing experiences a downward electric force Solve: Using Coulomb’s law,

E

on

*.By Newton’s third law, 4 w, = F; on 2 .

Because the force F;,,, is attractive and q1 is a positive charge, the charge q2 is a negative charge. Thus, q2 = -1.0 x IO-* c = -10 nC.

25.14. Model: Charges A, B, and C are point charges. Visualize: Please refer to Figure Ex25.14. Charge A experiences an electric force electric force left.

F,,,,

due to charge C. The force

F,,

F,,,

is directed to the right and the force

due to charge B and an

FconAis directed to the

Electric Charges and Forces

25-5

Solve: Coulomb’s law yields: ( 9 x 1 0 9 N m 2/C2)(1x104 C ) ( l ~ l 0 C) -~ FBonA

=

F.onA

=

(1 x

= 9.0 x 10-~N

m)’

( 9 x io9 N m2 / c’)(Ix 1 0 - ~~ ) ( x4 IO-^ C)

( 2 x IO-’ m)’

=9.0~10N -~

The net force on A is

F,,=

F

=(9.oX10-5 -~ E - O N ~ ~ ~~ ) i ^ + ( 9 . o x 1 0N)(-’)-

~

25.15. Model: Objects A and B are point charges. Visualize:

Y (cm)

Because there are only two charges A and B, the force on charge A is due to charge B only, and the force on B is due to charge A only. Solve: Coulomb’s law gives the magnitude of the forces between the charge. Thus,

(9 x io9 N mz / c’)(~o.o x 1 0 - ~C)(IO.Ox lo4 C ) FAonB

= FBonA =

= 4 . 5 0 ~ 1 0 ”N

(2.0 x IO-’ m)’

Because the charge on object A is positive and on object B is negative, FBOn A is upward and FAon is downward. Thus,

FBmA = G.50 x 10-3jN

FAonB = -4.50

x 10-’3 N

25.16. Model: Assume the glass bead, the proton, and the electron are point charges Visualize:

-

FIX,

0

-

7

0” p

x (cm)

/LO

0

Proton

-

- -

0

A

-

x (cm)

/LO

0

Electron

Solve: Coulomb’s law gives

(a) Newton’s second law is F = ma, so apokm

=

on pmmn

mpmen

2.88 x lo-‘’ N -

1.61 x lo-’’ kg

= 1.72 x 1014 m / s’

In vector form

apmcon = (1.72 x 10“ m / sz, away from bead)

.

25-6

Chapter 25

(b) Similarly,

= (3.16 x 10’’ m / s2, toward bead).

Thus

25.17. Model: The gravitation field is determined that creates the field. - by the mass of the- object is the gravitational force exerted on a Solve: The gravitational field 2 is defined as F, ,= m i , where

e,,

mass m placed at a point where the gravitational field is

2 . Thus, the strength of the gravitational field is

g=--6 0 . 0 N - 3 0 N / k g

2.0 kg

25.18. Model: A mass in a gravitational field experiences a force. Solve: The magnitude of the gravitational force on mass rn in a gravitational field

g is

F,,, = m g = 1 5 N In the same gravitational field and at the same point, a mass of 3m will experience a force Ensm = 3mg = 3F, ,= 3 x 15 N = 45 N

25.19. Model: The gravitational field of the earth extends through all of space. Solve: (a) The earth’s gravitational field is given by Equation 25.8:

Using meanh= 5.98 x loz4kg , re = 6.37 x lo6 m ,and G = 6.67 x lo-’’ N m2 / kg2,

id sudaEe = (9.83 N / kg, toward earth) (b) The gravitational field at the location of the moon’s orbit is

where r,,,is the radius of the moon’s orbit about the earth. With r, = 3.84 x 10’ m ,

,g

= (2.70 x

N / kg, toward earth)

Assess: The gravitational field at the radius of the moon’s orbit is small compared to the field at the earth’s surface because the gravitational field is inversely proportional to the square of the distance.

25.20. Model: The gravitational field of an object depends on its mass and extends through all of space. Solve: (a) The gravitational field strength due to a planet at the radius of its satellite’s orbit is 12 Nkg. That is, , toward planet = (12 N / kg, toward planet)

When the radius of the orbit is doubled,

gneu rxbit

- -G mplauet toward planet) (%b,l

l2 ’

=

(Lm, 4

toward planet = (3 N / kg, toward planet)

roL

(b) When the planet’s density is doubled. then mneu = p,,V=

2pV= 2mpbner. Thus, assuming that Vremains the same, = (24 N / kg, toward planet)

(c)

go,,,

does not depend on the satellite’s mass. Thus,

go,,, = (12 N / kg, toward planet).

Electric Charges and Forces

25-7

25.21. Model: The electric field is due to a charge and extends to all points in space. Solve: The magnitude of the electric field at a distance r from a charge 4 is

25.22. Model: The electric field is that of a positive charge on the glass bead. The charge is assumed to be a point charge. Solve: The electric field is

‘,

E=(9.Ox1O9 N m 2 / C 2 ) 6.0x10-9 i=1.35x105iN/C (2.0 x m)where F is the unit vector from the charge to the point at which we calculated the field. That is, the direction of the electric field is away from the bead.

25.23. Model: The electric field is that of the charge on the object. Assume the charge on the object is a point charge. Solve: The electric field at a distance r from a point charge q is E=-- 1

4 ;

4Z&,

Because the electric field points toward the object,

i-’

E = (180,000 N / C)(-f).

-(180,000N/C)=(9.0x109 N mz /C’)

j4

(I

(2.0 x

Thus, = -8.0

x lo4 C = -8.0 nC

m)-

25.24. Model: Protons and electrons produce electric fields. Solve: (a) The electric field of the proton is

(b) The electric field of the electron is

E = (1.44 x 25.25. Model:

N / C)(-;) = (1.44 x lo-’ N / C, toward electron)

A field is the agent that exerts an electric force on a charge.

Visualize:

, ; ;O +

-

w

Solve: Newton’s second law on the plastic ball is

= Fmq - W .To balance the weight with the electric

X(&) P

force,

F D”9

=wJ-lqlE=mg*

mg E=-=

141

(1.0~10-’kg)(9.8N/kg) ~ . O X I O -c~

= 3.27 x lob N I C

Because Fonq must be upward and the charge is negative, the electric field at the location of the plastic ball must be pointing downward. Thus E = (3.27 x lo6 N / C, downward). Assess: F = qE means the sign of the charge q determines the direction of F or E . For positive q. pointing in the same direction. But E and F point in opposite directions when q is negative.

and

F are

25-8

Chapter 25

25.26. Model: A field is the agent that exerts an electric force on a charge.

-

+

wP

Solve:

We

(a) To balance the weight of a proton C(F,,et)v=

e,, = w

3 lqlE = mg

6”- w = 0 N. This means

mg = (1.67 X 10-Z7kg)(9.8N / kg) = 1.02 x io-’ N / c E= 141 1.60 x 10-l9c

Because F,,,, must be upward and the proton charge is positive, the electric field at the location of the proton must also be pointing upward. Thus = (1.02 x N / C, downward). (b) In the case of the electron,

E - mg

(9.11x10-3’ kg)(9.8N/kg) = 5.58 x lo-” N / C 1.60 x 10-l9c

141 Because F,,” e must be upward and the electron has a negative charge, the elecuic field at the location of the electron must be pointing downward. Thus I? = (5.58 x lo-’’ N / C, downward).

25.27.

Model: The electric field is that of a positive point charge located at the origin, Visualize: Y (cm)

I

The positions (5 cm, 0 cm), (-5 cm, 5 cm), and (-5 cm, -5 cm) are denoted by A, B, and C, respectively. Solve: (a) The electric field for a positive charge is

- (-!-T,

E=

~ X E ,r2

Using 1/4~q,= 9.0 x lo9 N mz / C’ and q = 10.0x

away from q

C,

1

I?=[90.0 N m 2 / C , away from q r’ The electric fields at points A, B, and C are

-

E, =

90.0 N m’ / C

(-5.0 x IO-’ m)’

+ (5.0 x lo-? m)’

-(-;+;)I

= (-1.27x104i+ 1 . 2 7 l0.l) ~ N/C

90.0 N m’ / C

E, =

(-5.OxlO-’ m)2+(-5.0x10-2 m)’ (b) The three vectors are shown in the diagram.

Assess:

The vectors

EA,ZBand E,

are pointing away from the positive charge.

N/c

Electric Charges and Forces

25-9

25.28. Model: The electric field is that of a negative charge located at the origin. Visualize:

Y (cm)

5.0 4) A + 7,

,

-10 nC

EA

-

x (cm)

I

0

-5.0

5.0

EY -5.0B

The positions (5 cm, 0 cm), (-5 cm, -5 cm), and (-5 cm, 5 cm) are denoted by A, B, and C, respectively. Solve: (a) The electric field for a positive charge is

Using 1/4m0 = 9.0 x lo9 N m2 / C 2 and q = -10.0 x

E=(

C,

90.0 N m2 / C , toward r2

The electric fields at points A, B, and C are - 9 0 . 0 N m 2 / C ( - j ) = -3.6 x 1 0 4 jN E, = (5.0 x m)

EB .. =

90.0 N m2 / C

(-5.0 x lo-’ m)’

E, =

+ (-5.0

[-)(~+j)]=(l.27x1O4L+l.27xlO4j) N / C x

m)

90.0 N m 2 / C (-5.0 x lo-’ rn)’

c

+ (5.0 x lo-’

m)

[ L ( L - ;)]

fi

= (1.27 x IO4

- 1.27 x 104.j)N / C

(b) The three vectors are shown in the diagram. and E, are pointing toward the negative charge. Assess: Note that the vectors

zA,&

25.29. Model: Use the charge model. Solve: The number of moles in the penny is n = -M = A

3.1 g = 0.04882 mol 63.5 g / mol

The number of copper atoms in the penny is

N = nN, = (0.04882 mo1)(6.02 x IOz3 mol-’) = 2.939 x 10” Since each copper atom has 29 electrons and 29 protons, the total positive charge in the copper penny is (29 x 2.939 x 1022)(1.60x

C) = 1.36 x lo5 C

Similarly, the total negative charge is -1.36 x IO5 C. Assess: Total positive and negative charges are equal in magnitude.

25.30. Model: Use the charge model and the model of a conductor as material through which electrons move. Solve: (a) The charge of a plastic rod decreases from -15 nC to -10 nC. That is, -5 nC charge has been removed from the plastic. Because it is the negatively charged electrons that are transferred, -5 nC has been added to the metal sphere.

25- 10

Chapter 25

(b) Because each electron has a charge of 1.60 x C and a charge of 5 nC was transferred, the number of electrons transferred from the plastic rod to the metal sphere is 5x

C = 3.13 x 10" 1.60~10-l9c

25.31. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: (a) The charge of the glass rod decreases from +12 nC to +8 nC. Because it is the electrons that are transferred, -4 nC of electrons has been added to the glass rod. Thus, electrons are removed from the metal sphere and added to the glass rod. (b) Because each electron has a charge of 1.60 x lo-'' C and a charge of 4 nC was transferred, number of electrons transferred from the metal sphere to the glass rod is

25.32. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: Because the metal spheres are identical, the total charge is split equally between the two spheres. That is, C ) = -80 nC . qA= qB= 5 x 10" electrons. Thus, the charge on metal sphere A and B is (5 x lo")(-1.60 x 25.33. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: Plastic is an insulator and does not transfer charge from one sphere to the other. The charge of metal sphere A is (1.0 x 1OL2)(-1.6Ox C) = -160 nC and the charge of metal sphere B is 0 C. 25.34. Model: The protons are point charges. Solve: (a) The electric force between the protons is

FE= K

q,11q21 ( 9 . 0 ~ 1 0 'N m2 /C2)(1.60x10-19 C)(1.60x I= (2.0 x

r2

C)

= 57.6 N

m)'

(b) The gravitational force between the protons is Gm1m2

- (6.67 x lo-'' N m2 / kg2)(1.67x

kg)(1.67 x

kg) = 4.65 x

N

(2.0 x 10-l5m)2

F G = r ? -

(e)The ratio of the electric force to the gravitational force is

25.35. Model: The lZsXenucleus and the proton are point charges. That is, all the charge on the Xe nucleus is assumed to be at its center. Visualize: ' 2 5 ~nucleus e

Solve: (a) The magnitude of the force between the nucleus and the proton is given by Coulomb's law: Fnucleus on pmton

Klq,,,,,I/qP,,I r'

- ( 9 . 0 ~10' N mz / C2)(54x 1 . 6 0 lo-'' ~ C)(1.60 x (5.0 x

m)'

C)

= 498 N

Electric Charges and Forces

25-1 1

(b) Applying Newton’s second law to the proton,

25.36. Model: The two charged spheres are point charges. Solve: The electric force on one charged sphere due to the other charged sphere is equal to the sphere’s mass times its acceleration. Because the spheres are identical and equally charged, m, = = m and q1= q2= q . We have Fzonl

a q 2 = - =

mar? K

(1.0 X

Kq,q, - Kq2 - m a = < o n ? =-- r2 r2 kg)(225 m / s2)(2.0X lo-* m)2 = 1.0 x 10-l~c2 9 . 0 ~ 1 Nm’ 0 ~ /C2

3

= 1.0 x io-’

c

=

loo n c

25.37. Model: Objects A and B are point charges. Visualize:

-+

-r=lOcm-

mA = 0.10

kg

qA = *qB

mg = 0.10 kg qB

,

= 0.45 N. By Newton’s third law, FBonA = FAon= 0.45 N. Solve: (a) It is given that FAonB (b) Coulomb’s law is

(c) Newton’s second law is F , on A = mAa,. Hence,

25.38. Model: The charges are point charges. Visualize: Please refer to Figure F’25.38. Solve: The electric force on charge q, is the vector sum of the forces F, qz is the 2 nC charge, and q3is the other 2 nC charge. We have

(v, 1 =i

con, =

-

and F30n where q1is the 1 nC charge,

away from q2

(9.0 x lo9 N m’ I Cz)(lx 10” C ) ( 2 x (1 x IO-’ m)’

c), away from q2

=(1.8x10J N, awayfromq2)=(1.8x104 N)(cos60°~+sin600~)

1

away from q, = (1.8 x loJ N, away from q 3 )= (1.8 x IO4 N)(-cos6O0i^ + sin60j)

*Enl = ~ o n , + F ,=2(1.8x104 onl

N ) s i n 6 O 0 ~ = 3 . 1 2 x 1 O 4N~

So the force on the 1 nC charge is 3.12 x lo-‘ N directed upward.

25-12

Chapter 25

25.39. Model: The charges are point charges. Visualize: Please refer to Figure P25.39. Solve: The electric force on charge q, is the vector sum of the forces q2is the 2 nC charge, and q3is the -2 nC charges. We have

E,,~,=

6

on

and

6

on

where q1is the 1 nC charge,

(y 1 , away fiom q2

=i

(9.0 x IO9 N m 2/ C2)(1 x (1 x

C)(2x LO-'' C )

, away from q2

m)'

= (1.80 x lo4 N, away from q 2 )= ( 1 . 8 0 ~lo4 N)(cos60"f tsin60"j)

eon' =(y, cnl=Eml+Eonl

1

towardq, =(1.80x104 N, towardq,)=(1.80x104 N)(cos6O0f-sin60"j^). =(1.80x1O4 N ) ( 2 c o s 6 0 " ) ~ = 1 . 8 0 ~ 1 0 ~ ~ N

3

So, the force on the 1 nC charge is 1.80 x 10" N and it is directed to the right.

Solve: The electric force on charge ql is the vector sum of the forces

=i

E

On

(9.0 x IO9 N m2 / C2)(10x lo-' C)(5 x

= (4.5 x

and

6

on

We have

I

C) , away from q2

(1x10-2

N, away from q 2 )= 4.5 x 10-3jN

, toward q3

'[

( 9 . 0 lo9 ~ N m2 / Cz)(IO x

C)(15 x lO-'C)

( 3 . 0 ~ 1 0 m)' - ~ +(l.OxlO-'

, toward q,

m)'

= ( 1 . 3 5 ~ 1 0 -N, ~ towardq,) = (1.35~10-' N)(-cos&+sin\$) From the geometry of the figure,

1

Electric Charges and Forces

This means cos 8= 0.9487 and sin 8= 0.3163. Therefore, = ( - i . ~ X i o - ~ i ^ + 0.43X10-3j)

N

The magnitude and direction of the resultant force vector are

tan4 = 4‘07 lo” = 3.180 3 4 = tad(3.180) = 72.5“ below the --x axis i . 2 8 ~ 1 0 -N~

25.41. Model: The charges are point charges.

1.0 -

Solve: The electric force on charge q, is the vector sum of the forces

=(?,

Eonl

and

(9.0 x lo9 N m2 / C2)(10x lo4 C)(5 x

= (4.5 x

C)

(1.0 x IO-’ m)’

, toward q2

N, toward q 2 )= 4.5 X 10-3jN

(9.0 x lo9 N rn2 / C2)(10 x 10” C)(15 x

(3.0 x lo-’ rn)’ = (1.5 x

We have

1

towardq,

=i

con, fi,,,,

C) , toward q,

N, towardq,) = -1.5 x 10-3~’ N

* E,,= E

On I

+6

On

= (-1.5

x lo-,;

+ 4.5 x lo-”)

N

The magnitude and direction of the resultant force vector are = ,,/(-is x 10-3

tan@= 4S lo-’ 1.5 x lo-.’ N

N)’ + (4.5 x 10-3 N)‘

= 4.74 x 10-3 N

4 = tan-‘(3) = 71.6” above the --s axis

25-13

Chapter 25

25-14

042

4.0 I '

1 I 1

, I ,

3.0 I

I

,

I I I I

/

I

, I

2.0 -

I I

,

I

/

/

I I I

I I

,

I

/

1.0 -

,

/

I

I I I

Solve: The electric force on charge q1is the vector sum of the forces

=i

(9.0 x IO9 N m2 / C2)(5x

4

on,

and

C)(10 x lo+ C)

(4.0 x lo-' m)' +(3.0 x lo-' m)'

on 1.

We have

, away from q2

=(1.8x104 N, awayfromq2)=(l.8x104 N)(-cos&-sin\$)

From the geometry of the figure, tan6 = 4.0 cm 3 6=53.13"*E,, 3.0 cm

- i

(9.0 x 10' N m2 / C2)(5x log C)(5 x lo4 C)

4 O " l

=

(3.0 x lo-? m)' 3

En = Eon + 6, I

I

I

= ( l . 8 x 1 0 4 N)(-0.6;-0.8j)

I

, toward q3 = (2.5 x lo4 N, toward q 3 )= 2.5 x IO";

= (1.42 x 10"; - 1.44 x 104f) N

The magnitude and direction of the resultant force vector are

F,, I = J(i.42 x io4 N)' q9 = tan-

(

,

+ (-1.4 x 104 N ) ~= 2.02 x 1 0 4 N

1.44~10" N

N ) = 45.4' below the +x axis

N

Electric Charges and Forces

25.43. Model: The charges are point charges. Visualize:

Y (cm)

I

,e

43

, I

I

I

,

I I

I I

I

I

have

Solve: The electric force on q1is the vector sum of the forces

(9.0~10' Nm' /C')(SX~O-~ C)(10~10-~C) , awayfromq, (4.0 x lo-' rn)'

= (2.81 x IO4 N, away frornq,) = -2.81 x 1 0 4 j N

(9.0 x lo9 N m' / C')(5

=[

x

C)(10 x

C)

, toward q3

(3.0x10-' m)'+(4.0~10-' m)'

= (1.8 x IO4 N, toward 4,) = (1.8 x lo4 N)(+cosk

+ sin\$)

From the geometry of the figure, tan0 =

con, = (1.08 x 10";

4.0 cm a cos0 = 0.6 and sin0 = 0.8 3.0 cm

-

+ 1.44 x 10~1) N

En = F,, + F30,,,= (1.08 x lo4;

The magnitude and direction of the resultant force vector are

F,,, =,/(l.08x104 N)' +(-1.37x104 N)' =1.74x104 N tan@=

lo-'

1 . 0 8 IO4 ~ N

jq?~

= 51.75" below the +x axis

- 1.37 x lO-"j) N

25-15

Chapter 25

25- 16

25.44. Model: The charges are point charges. Visualize: Please refer to Figure P25.44. Solve: Placing the 1 nC charge at the origin and calling it q l ,the q2 charge is in the first quadrant, the q, charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. The + electric force on q, is the vector sum of the forces 4 on 4 , The magnitude of these four I , 4on 1 , and 4 Dn forces is the same because all four charges are equal and equidistant from ql. So,

-

-

( 9 . 0 ~ 1 0 ' N m 2 / C ' ) ( ~ X ~ OC)(lxlO-' -~ C) F z m i =&on,

=

=<on1

=F40nl

(0.5 x lo-* m)* + (0.5 x IO-' m)'

-

Thus, F,, = (3.6 X lo" N, toward q2) + (3.6 x lo4 N, toward q,) toward qj). In component form,

En,= <",[(cos45"i'

+ sin45"j) + ( ~ 0 ~ 4 5 " ;sin45"j) + (-cos45"1-

+ (3.6 x lo4 sin45"j)

= 3.6 x io4 N

N, toward q4) + (3.6 x 10" N,

+ (-cos45"1+

sin45"j*)] = 6 N

25.45. Model: The charges are point charges. Visualize: Please refer to Figure p25.45. Solve: Placing the 1 nC charge at the origin and calling it ql, the q2charge is in the first quadrant, the q3 charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. The electric force on qi is the vector sum of the electric forces from the other four charges q2,q3,q4and q5. The magnitude of these four forces is the same because all four charges are equal in magnitude and are equidistant from q,. So, 4oni

( 9 . 0 ~ 1 0N~m 2 /C2)(2x10-9 C ) ( 1 ~ 1 0 -C) ~

=4ml = & a n i

=Fsm1

=

(0.5 x IO-? m)' + ( O S x

-

= 3.6 x lo4 N

m)*

Thus, = (3.6 x lo" N, away from q2)+ (3.6 x lo4 N, away from q3)+ (3.6 x lo" N, toward q4)+ (3.6 x lo" N, toward 4'). In component fonn,

Enl= F,,[(-cos45"i^

-sin45"j)

+ ( - c o s 4 5 " ~ + s i n 4 5 " ~+(-cos45"1 )

= (3.6 x lo4 N)(-4cos45"i') = -1.02

-sin45";)+

(-cos45"; +sin45';)]

low3;N

X

25.46. Model: The charges are point charges. Visualize: Please refer to Figure P25.46. Solve: Placing the 1 nC charge at the origin and calling it q l , the -6 nC is q3,the q,charge is in the first quadrant, and the q4charge is in the second quadrant. The net electric force on 4 ] is the vector sum of the electric forces from the other three charges q,, q3,and qw We have

=(y, 1 awayfiomq,

(9.0 x IO9 N m2 / Cz)(l X lo4 C)(2 x lo4 C) (5.0 x

I

, away from q2

m)'

= (0.72 x 10" N, away from q 2 )= (0.72 x lo-' N)(-cos45"~ -sin 45"j)

Go,,,

=(y, 1 towardq,

=i

C)(6 x lo-' C )

(9.0 x lo9 N m2 / C2)(l x (5.0 x lo-' m)'

N, away from q-()= 2.16 x 1O"j N

= (2.16 x

con,

3

(-

= Kt:!q41

E,,= E,, +

1

, away from q4 = (0.72 x

+ FAonl = [(2.16 x 10"

1

, toward q3

N)(cos45"3 - sin45'1)

N)-(0.72 x lo-' N)(2sin45")]~= 1.14 xlOdl N

Electric Charges and Forces

25-17

25.47. Model: The charges are point charges. Visualize: Please refer to Figure P25.47. Solve: Placing the 1 nC charge at the origin and calling it q l ,the -6 nC is q3,the q?charge is in the first quadrant, and the q4 charge is in the second quadrant. The net electric force on q, is the vector sum of the electric forces from the other three charges q2,q3,and qe We have

=(y, 1

EOn1

towardq,

(9.0 x lo9 N m2 / C')(1 x

C)(2 x

= (0.72 x

N, toward q,) = (0.72

=(y, 1 Eonl [y, 1 Eon]

F,

x lo-' N)(cos45"i^ + sin45"j)

towardq, = ( 2 . 1 6 ~ 1 0 " N, towardq3)=2.16x10-'jN

away fromq,

=

3

I

C) , toward q2

(5.0 x lo-' m)'

= (0.72

x lo-' N)(cos45"i^ - sin45"j)

, = Eon,+E, + Eon = (0.72 x 10" I

N)(2cos45");

+ (2.16 x

N)j

= ( 1 . 0 2 ~ l O ~ ~ i ^ + 2 . 1 6 ~N1 0 - ~ ~ )

25.48. Model: The charged particles are point charges. Visualize:

Y

41

-\ F2onp

q2

ennP E,,

Solve: (a) The mathematical problem is to find the position for which the forces and are equal in magnitude and opposite in direction. If the proton is at position x, it is a distance 1x1 from q1and Id - XI from q?, where d = 1 cm. The magnitudes of the forces are

Equating the two forces and using d = 1 cm,

The solutions to the equation are x = +0.414 cm and -2.41 cm. Both are points where the magnitudes of the two forces are equal, but N.414 cm is a point where the magnitudes are equal but the directions are the same. The solution we want is that the proton should be placed 2.41 cm from the 2 nC charge on the side opposite the 4 nC charge. (b)Yes. the net force on the electron located at x = -2.41 cm will be zero. This is because the solution in part (a) does not depend specifically on the type of the charge that experiences zero force from the other two charges. is zero for a proton, the electric field at that point must be zero. Thus, there will be no force on Furthermore, if any charged particle at that point.

cnp

25- 1 8

Chapter 25

25.49. Model:

The charged particles are point charges.

Visualize:

o q

2 cm Fqon 1

'\

3

Solve: The two 2 nC charges exert an upward force on the 1 nC charge. Since the net force on the 1 nC charge is zero, the unknown charge must exert a downward force of equal magnitude. This implies that q is a positive charge. The force of charges 2 on charge 1 is

- (9.0 x IO9 N m2 / C2)(1.0x

C)(2.0 x (0.020 m)2 + (0.030 m)'

-

C) (cos8

5>

+

From the figure, 8 = tan-'(2/3) = 33.69'. Thus

Eon,=(1.152xlO"

; + 0 . 7 6 8 ~ 1 0 -j^)N ~

6

From symmetry, on is the same except the x-component is reversed. When we add cancel and the y-components add to give

conmust I

6

on

and

6w),, the x-components

~ o , l = 1 . 5 3 6 ~ 1 0 -N 5f have the same magnitude, pointing in the

con, = 1.536 x IO"

N =-

-j^

direction, so

(1.536 x lo-' N)(0.020 m)* = o.68 nC (9.0 x lo9 N m2 / C')(1.0 x C)

A positive charge q = 0.68 nC will cause the net force on the 1 nC charge to be zero.

25.50. Model: The charged particles are point charges. Visualize: Please refer to Figure P25.50. Solve: The charge q1 is in static equilibrium, so the net electric field at the location of q2is zero. We have

and a negative We have used the f sign to indicate that a positive charge on q , leads to an electric field along +I charge on q1leads to an electric field along -I. Because the above equation can only be satisfied if we use +we I, infer that the charge q, is a positive charge. Thus,

- 2x10-9 C ( I ) = O N / C a q , = 8 n C *(+I) (0.-q m) (0.10 m)' 25.51. Model: The charged particles are point charges. Visualize:

V

Electric Charges and Forces

25-19

Solve: The force on q is the vector sum of the force from -Q and +Q. We have

From the figure we see that cos0 = a / J w . Thus

Assess: Note that (F,,,), = F,,, because the y-components of the two forces cancel each other out.

25.52. Model: The charged particles are point charges. Visualize: Lr I < 0 :

-Q

-

n

+Q @

I

+4 @--X

0

s < -u: +Y

8

+Q

-Q

-

@

I

X

0 k----u-u4 k-----X-----+

Solve:

(a) The force on q is the vector sum of the force from -Q and +Q. We have

To arrive at the final expression we used ( a - .)'(a (b) There are two cases when 1x1 > a . For .r > a ,

+ x)'

= [ ( a- .)(a

+ x)]'

= (a' - x2)?

)

, toward - Q =

( x -a)-

-(-:(x) + a ) KQ4

-4KQqar

..I....

,

.

.

.

.-

.

-

.

..-

25-20

Chapter 25

I I I I I I I

K'4el'q' away from 4Q

eOnq =(?,

L

45O1 +sin 45"j^)jFZonq = -L2

toward - Q

)=-(-y)y

(b) The net electric force on the charge +q is the vector sum of the electric forces from the other three charges. The net force is

Electric Charges and Forces

25-21

25.54. Model: The charges are point charges. Visualize: Y I

I I

I

-

-

q,

F20n3

I

Fl o n 3

I

~

\

2

L I I I

Outside the charges

-., F

I I

I

I

93 2

w

o

Inside the charges

n

3

j I

I

Outsidethe charges

We must first identify the region of space where the third charge q3 is located. You can see from the figure that the forces canâ&#x20AC;&#x2122;t possibly add to zero if q3 is above or below the axis or outside the charges. However, at some point on the x-axis between the two charges the forces from the two charges will be oppositely directed. Solve: The mathematical problem is to find the position for which the forces F;o,3 and are equal in magnitude. If q3is the distance x from ql, it is the distance L - x from q2.The magnitudes of the forces are

Equating the two forces,

The solution x = -L is not allowed as you can see from the figure. To find the magnitude of the charge q3,we apply the equilibrium condition to charge ql:

We are now able to check the static equilibrium condition for the charge 4q (or q2):

The sign of the third charge q3 must be negative. A positive sign on q3will not have a net force of zero either on the charge q or the charge 4q. In summary, a charge of -44 placed x = S L from the charge q will cause the 3-charge system to be in static equilibrium.

25.55. Model: Use the charge model and assume the copper spheres are point objects with point charges. Solve: (a) The mass of the copper sphere is M

(

= pV = F r 3 ) = (8920 kg / m3)

= 3.736 x

kg = 0.03736 g

The number of moles in the sphere is

The number of copper atoms in the sphere is

N = n N , = (5.884 x lo4 mo1)(6.02 x lo2â&#x20AC;&#x2122; mol-â&#x20AC;&#x2122;) = 5.542 X lozo

25-22

Chapter 25

The number of electrons in the copper sphere is thus 29 x 3.542 x lo2’ = 1.027 x lo**. The total positive or negative charge in the sphere is (1.027 X 10”)(1.60 x C) = 1643 C. Hence, the spheres will have a net charge of low9X 1643 C = 1.643 X 10“ C . The force between two such spheres is

F=

Kq,q* = (9.0 x IO9 N m2 I Cz)(1.643x lo6 (1.0 x IO-* m)’ ( 1 . 0 ~IO-* m)’

C)’

= 243 N

(b) This is a force that is easily detectable. Since we don’t observe such forces, any difference between the proton charge and the electron charge must be smaller than 1 part in lo9.

25.56. Model: The electron and the proton are point charges. Solve: The electric Coulomb force between the electron and the proton provides the centripetal acceleration for the electron’s circular motion. Thus, -=-K(e)(e) myz r2 r jj-= -=

mr’

9.1 1 x

kg)(5.3 x lo-”)

-mr02

1 rev = 4.12 x 10l6radls x -= 6.56 x lOI5 revls 27r rad

25.57. Model: Model the metal plate and yourself as point charges. Solve: At the beginning, both you and the metal plates are neutral. As the electrons are pumped from the metal plate into you, the plate becomes as much positively charged as you become negatively charged. When enough charge difference builds up between you and the metal plate, your weight will be counterbalanced by the upward electrical force. You will slowly begin to hang suspended in the air when

Dividing this charge by the charge on an electron yields 3.2 x lOI5 electrons.

25.58. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’s law. Solve: Let q be the charge on each plastic bead. The repulsive force between the beads pushes the beads apart. The spring is stretched until the restoring spring force on either bead is equal to the repulsive Coulomb force. That is,

The spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.O cm. Thus m g = k(l.O x lo-’ m)

k=

(1.0 x lo-’ kg)(9.8 N / kg)

1.0 x IO-’ m

= 0.98 N / m

(0.98 N I m)(4.5 x IO-’ m - 4.0 x lo-’ m)(4.5 x lo-’ m)’ 9.0 x l o 9 N m’ I C’

25.59. Model: The charged spheres are point charges. Visualize:

Pictorial representation

Free-body diagram

x

= 33.2 nC

Electric Charges and Forces

25-23

Each sphere is in static equilibrium and the string makes an angle 0 with the vertical. The three forces acting on each sphere are the electric force, the weight of the sphere, and the tension force. Solve: In static equilibrium, Newton's first law is = ? 6 F, = 0. In component form,

- Fna + + (eet)* = T, +w, + ( F , ) x = 0 N (eJy= T, + w y 3

-Tsine+O N+- Kq2 = 0 N

3

Kq' TsinO = -=

TcosO-mg+O N = O N

d2

d2

=0 N

Kq2

TcosO = +mg

(2LsinO)'

Dividing the two equations, Kq2 = (9.0 x lo9 N m2 / c2)(100X 10" c)*= 4.59 x lo4 sin2e t a n e = 4L2mg 4(1.0 m)'(5.0 x 10" kg)(9.8 N / kg) For small-angles, tan0 = sine. With this approximation we obtain sin0 = 0.07714 radand 0 = 4.42".

25.60. Model: The charged spheres are point charges. Visualize:

Pictorial representation

Free-body diagram Y

--x

Charged sphere W

Each sphere is in static equilibrium when the string makes an angle of 20" with the vertical. The three forces acting on each sphere are the electric force, the weight of the sphere, and the tension force. - Solve: In the static equilibrium, Newton's first law is \$,et = ? + G + F, = 0. In component form,

(e&= T Y + W y + ( 4 ) y = O N

(F,JX= T , + w , + ( F , ) x=ON

= -TsinO+O j

N+- Kq2 = 0 N d2

Kq2 TsinO=-=

d2

+Kq2 (2Lsir10)~

TcosO- mg+O N = 0 N TcosO = +mg

Dividing the two equations and solving for q. 4(sin' 2O0tan20")(1.O m)*(3.0 x

kg)(9.S N / kg)

9.0 x lo9 N m2 / C2

= 746nC

25.61. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P25.61. Place the 10 nC charge at the origin. Solve: The electric field is. (9.0 x IO9 N m' / C2)(10x 10-9C)

[

= 90.0 ",m'

r' IC

away from q

1

, away from q

1

25-24

Chapter 25

At each of the three points,

90.0 N m2 / C

(5.0 x IO-' m) = (3.6 x io4 N

I I

, away from q = (3.6 x IO4 N / C)(cos&

+ sin 6 )

c)(+F+ ~ j= (2.88 ) x 10째F + 2.16 x 1 0 4 j )N c

90.0Nm' / C

, away from q = 5.63 x lo"; N 1 C

(4.0 x IO-' m)

25.62. Model: The electric field is that of a positive point charge. Visualize: Please refer to Figure p25.62. Place point 1 at the origin. Solve: The electric field is (9.0 x IO9 N m2 / C')(2 x r2 = (18.0

C) , toward q

:y2 / C , toward q

The electric fields at the two points are 18.0Nm2/C

E, = (

( ~ . o x I o -m) ~

= (1.8 x

E, =(

2 ,

IO5 N/C, 60" counter clockwise from the +x-axis or 60" north of east)

18.0Nm2 / C (1.0~

= (1.8 x

I

toward q

, toward q

m)

IO5 N/C, 60" clockwise from the -x-axis or 60" north of west)

25.63. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P25.63. Place the 5 nC charge at the origin. Solve: The electric field is (9 x 10' N m2 / c2)(5 x 1 0 - ~C) r' = (45.0 !;mi

, away from q

1 C , away from q

At each of the three points,

E, =[

45.0 N m' / C

(2.0 x IO-' m)? + (1.0 x IO-' m)-

1

, away from q = (9.0 x lo4 N / C)(cos& + sin\$)

JJ' +JTJ~ ) = [ 4 . 0 2 x l O 4 ~ + 8 . 0 5 x i O 4 ~ ) N / C

=(9.Ox1O4 N/C)(-!-'

Electric Charges and Forces

45.0

m2

(1.0 x

'C,

1

, away from q = 4.5 x 105; N / c

m)

45.0 N m2 / C

m)'

(2.0 x

25-25

+ (1.0 x IO-'

I

, away from q =(4.02 x lo4; - 8.05 x 1041)N / C

m)

25.64. Model: The electric field is that of a negative charge at (x, y) = (2.0 cm, 1.0 cm). Visualize:

Y tcm)

w .6. b '

2.0 -

-2.0 I

ga

,

.ec

c

@ ,a -

'cy'

2.0

4.0

,/'-

04

- -2.0

Solve: (a) The electric field of a negative charge points toward the charge, so we can roughly locate where the field has a particular value by inspecting the signs of E, and E,. At point a, the electric field has no y-component and the x-component points to the left, so its location must be to the right of the charge along a horizontal line. Using the equation for the field of a point charge,

Thus, point a is at the position (xa,y,) = (4 cm, 1 cm). (b) Point b is above and to the left of the charge. The magnitude of the field at this point is

E = J ~ = J ( 1 6 1 , 0 0 0 N / C ) ' +(80,500N/C)2 = 180,00ON/C Using the equation for the field of a point charge,

This gives the total distance but not the horizontal and vertical components. However, we can determine the angle 8because 2,points straight toward the negative charge. Thus,

The horizontal and vertical distances are then d , = rbcos0 = 2.00 cm and d, = rbsine = 1.00 cm . Thus, point b is at the position (xb.yb) = (0 cm, 2 cm). (c) Point c, which is below and to the left of the charge, is calculated by following a similar procedure. We first find that E = 36,000 N / C . From this we find that the total distance r, = 5.00 cm. The angle @ is

which gives the distances d, = rc cos@= 4.00cm and d, = rc sin4 = 3.00 cm . Thus point c is at position (xc, ye) = (-2 cm, -2 cm).

25-26

Chapter 25

25.65. Model: The electric field is that of a positive charge at (x,y ) = (1 .O cm, 2.0 cm). Visualize: Y (cm)

,

1.0 -

\ \ \

I

I

0

\

I

'

I

2.0 '3.0

1.0

4.0

n(cm)

\ \ \ \ \ \

Y (9.0 x IO9 N m2 / C2)(10.0x

C)

225,000 N / C

.,2

Thus, ( x ~y,p )= (-1 cm, 2 cm). (b) Point b is above and to the right of the charge. The magnitude of the field at this point is

-/,

= J(161,oOa N / C)' +(80,500 N / C)' = 180,000N / C

E=

Using the field of a point charge, (9.0 x lo9 N m2 / Cz)(lO.Ox

180,000 N / C

C) = 2.236 cm

This gives the total distance but not the horizontal and vertical components. However, we can determine the angle e because E, points straight away from the positive charge. Thus,

6 = tan-'(

I]

)

= tan-'( 80,500 N / C = tan-'(+) 161,000 N / C

= 26.57'

The horizontal and vertical distances are then d, = r, cos0 = (2.236 cm)cos26.57' = 2.00 cm and d, = rbsinf3= 1.00 cm . Thus, point b is at position (xb,yb)= (3 cm, 3 cm). (c)To calculate point c, which is below and to the right of the charge, a similar procedure is followed. We first find E = 36,000 N / C from which we find the total distance r, = 5.00 cm. The angle @ is

which gives distances d, = 7 cos@= 3.00 cm and d,, = rc sin\$ = 4.00 cm. Thus, point c is at position (xc,yc) = (4 cm, -2 cm).

25.66. Model: The electric field is that of three point charges. Visualize: q, = 1 nC 0.

I

,

Electric Charges and Forces

25-27

Solve: (a) In the figure, the distances are '1 = '3 = d(1 cm)2 +(3 cm)* = 3.162 cm and the angle is 8 = tan-](l/ 3) = 18.43". Using the equation for the field of a point charge, ~(q,( E = E =-= 1

3

'12

(9.0 x IO9 N m2 / C2)(1.0 x lo4 C)' = 9000N/C (0.03162 m)'

We now use the angle 8 to find the components of the field vectors:

-

E, = E, cos& - E, sin(\$

,\$ = E3cos& + E3sin6 = (85401 + 28403 ) N / C

)

= (85401 - 28403 N / C

E2 is easier since it has only an x-component. Its magnitude is ~lq,l

E2=-r,'

-

(9.0 x lo9 N m2 / C2)(1.0 x

C)

(0.0300 m)'

= 10,000N / C

E, = E,;

(b) The electric field is defined in terms of an electric force acting on charge q:

= lO,OOOi^N / C

E = F/q. Since forces obey a

fi

principle of superposition ( = + F; + ...) it follows that the electric field due to several charges also obeys a principle of superposition. - - . . . (c) The net electric field at a point 3 cm to the right of q2is = E, + E2 + E3 = 27,100; N / C. The y-components of

E, and E, cancel, giving a net field pointing along the x-axis.

25.67. Model: Model the proton and the electron as point charges. Solve: (a) The force that an electric field E exerts on a charge q is = qE . A proton has q = e. Thus,

Fpromn =e(200i^ + 4ooj^) N / C where we used e = 1.60 x C. (b) The charge on an electron is q = -e. Thus,

Fe,-,,

=

= (3.201

+ 6 . 4 O j ^ ) ~ l O " ~N

-Fpown= (-3.201 - 6.403) x lo-''

N

(c) From Newton's second law,

(d) The electron experiences a force of the same magnitude but it has a different mass. Thus,

The forces may be the same, but the electron has a much larger acceleration due to its much smaller mass. Assess: The two forces in parts (a) and (b) are equal in magnitude but opposite in direction.

25.68. Model: The charged ball attached to the string is a point charge. Visualize:

-

w

The ball is in static equilibrium in the external electric field when the string makes an angle 8 = 20" with the vertical. The three forces acting on the charged ball are the electric force due to the field, the weight of the ball, and the tension force.

.

~

-

.-

...

.-

25-28

Chapter 25

Solve: In static equilibrium, Newton’s second law for the ball is

(F,), = T, + O N + q E = 0 N

Fm= f + G +

(%a)y

= 6. h component form,

= T, -mg+O N = 0 N

The above two equations simplify to

‘a

Tsin0 = qE

Tcos0 = mg

Dividing both equations, we get

m g m 0 ( 5 . 0 ~ 1 0 -kg)(9.8N/kg)tan20° ~ q = -= 1.78 x E 100,000 N / C

tan8 =-9E

w

C = 178 nC

25.69. Model: The charged ball attached to the string is the point charge. visualize:

I

I

I - +

--++

E=

\$e

-

W

The charged ball is in static equilibrium in the external electric field when the string makes an angle 8 with the vertical. The three forces acting on the charge are the electric force due to the electric field, the weight of the ball, and the tension force. - Solve: In static equilibrium, Newton’s second law for the charged ball is = T + G + 4 = 0. In component form,

Ea

(eet), = r, + O N + q E = 0 N

(F,t)y = T, - mg+O N = 0 N

These two equations become Tsin0 = qE and Tcos0 = mg . Dividing the equations gives a n 0 = - q=E mg

(25 X 10-9 c)(200,000 N / C) (2.0 x 10” kg)(9.8 N / kg)

= 0.255

0 = 14.3”

Solve: (a) How many excess electrons on a dust particle produce an electric field of magnitude 1.0 x lo6N/C a distance of 1.O pm from the dust particle? (a) The number of electrons is

25.70.

N=

( 1 . 5 ~ 1 0N/C)(l.OxlOd ~ m)’ (9.0~ l o 9 N m2 / C2)(1.60x

= 1042 C)

25.71. Solve: (a) Two equal charges separated by 1.50 cm exert repulsive forces of 0.020 N on each other. What is the magnitude of the charge? (b) The charge is (0.020 N)(0.0150 m)* = 22.4 nC 9.0 x lo9 N m2 / C2 The problem does not give the direction of the force. So, the charges could be both positive or both negative.

25.72. Solve: (a) At what distance from a 15 nC charge is the electric field strength 54,000 N/C? (b) The distance is (9.0 x lo9 N m2 / C2)(15x 10” C) 54,000 N I C

= 0.050 m = 5.0 cm

Electric Charges and Forces

25-29

25.73. Solve: (a) A 1.0 nC charge is placed at (0 cm, 2 cm) and another 1.0 nC charge is placed at (0 cm, -2 cm). A third charge +q is placed along a line halfway between q1 and q2 such that the angle between the forces on q due to each of the other two charges is 60". The resultant force on q is (5.0 x 10" N, 2). What is the magnitude of the charge q? (b) q, = 1 nC

q2= 1 nC

We have (9.0 x IO9 N m2 / C2)(1.0x 40"3=

(0.020 r n / s i n 3 0 ~ ) ~

I

C)q , away from q3 = (5625 N / C)q(cos 30"; - sin 30"j)

eon3(5625 cn32 =

=

N / C)q(cos3O0l+sin30j) x (5625 N I C)qcos3O0i

2(5625 N I C)qcos30째 = 5.0 x

N

3

q=

(~.oxIo-~N)

= 5.13 nC

2(5625 N / C)cos30"

25.74. Model: The charged balls are point charges. Visualize: Pictorial representation In-plane physical representation

Fm-body d i m Y

F0"3

-

F20n3\$

Flon3

Side view

Top view

Because of symmetry and the fact that the three balls have the same charge, the magnitude of the electric force on each ball is the same. The other forces acting on each ball are the weight of the ball and the tension force. Solve: The force on ball 3 is the sum of the force from ball 1 and ball 2. We have

- 2Kq' * F,, =rz 3

F, =

cos3O"j 3

=

2Kp2 ~ 0 ~ 3=0 F,,,? " = F,,, r-

2(9.0 x lo9 N m' I C ? ) q 2cos30째 (0.20 m)'

= F,

= (3.897 x 10" q 2 )N / C'

25-30

Chapter 25

The distance I, between one of the balls and the center of the equilateral triangle, is 0.10 m l~Os30"= = 0.10 m j 1 = = 0.1155 m 2 cos30" Thus, the angle made by the string with the plane containing the three balls is

-

cos0 = - = 0*1155 L 0.80m From the free-body diagram, we have TsinO-mg=ON

0 = 81.700

j

-Tcos0+F,=ON

ms

*tan0=-=mg

5

(3.897 x 10"q2)N / C2

(3.0 x IO" kg)(9.8 N / kg) (3.897 x 10" N / C2)tan81.700 = 1.05 ~ 1 0 C~ =' 105 nC

25.75.

Model: The charged spheres are point charges. Visualize: Free-body diagram

Negative charge

The figure shows the free-body diagram of the forces on the sphere with the negative charge that is shown in Figure CP25.75. The force FEis due to the external electric field. The force F, is the attractive force between the positive and the negative spheres. The tension in the string and the weight are the remaining two forces on the spheres. Solve: The two electrical forces are calculated as follows:

F, = )q)E= ( 1 0 0 io+ ~ c)(105 N / c)= 1.0 x io-2 N (9.0 x IO9 N m 2 / C2)(100X

K)q,IlqZl c=-= r'

C)' - 9.0 x lo-' N m2 I C r2

r'

From the geometry of Figure CP25.75,

r = 2(Lsin10") = 2(0.50 m)sinlO" = 0.1736 m a F, =

9.0 x 10" N m' / C = 2.986 (0.1736 m)'

From the free-body diagram, TcoslO" = mg

TsinlO"+ F, = F,

Rearranging and dividing the two equations,

-sin 10" K-F, - tanlo" = cos 10"

m=-- F, - F, gtanI0"

25.76.

-

mg

1.OX lo-' N - 0.2986X IO-' N = 0.406 x lo-' kg = 4.06 g ( 9 . 8 N / kg)tan10"

Solve: (a) Kinetic energy is K = +mv', so the velocity squared is v2 = 2Wm. From kinematics, a particle J of moving through distance Ax with acceleration a, starting from rest, finishes with 3 = 2aAx. To gain K = 2 x kinetic energy in Ax = 2.0 pm requires an acceleration 2.0 x lo-'' J v2 2 K I m ---K a=-=-= 1.10x 10'' m / s2 2Ax 2Ax m h ( 9 . 1 1 ~ 1 0 - ~kg)(2.0x1O4 ' m)

Electric Charges and Forces

25-3 1

(b) The force that produces this acceleration is

F = ma = (9.11 x

kg)(l.lO x 10" m / s2)= 1.0 x lo-'' N

(c) The electric field is

(d) The force on an electron due to charge q is F = Klqle/r2 . To have a breakdown, the force on the electron must be at least 1.O x lo-'? N. The minimum charge that could cause a breakdown will be the charge that causes exactly a force of 1.O x IO-" N:

Iq

=

*

= 1.0 x lo-" N

r'

=\$

(0.01 m)2(1.0 x 10-l~N) r2F 14)= = Ke (9.0 x lo9 N m2 / C2)(1.6x

C)

= 6.9 x lo-* C = 68 nC

25.77. Model: Charge Q and the dipole charges (q and -4) are point charges. Visualize: Please refer to Figure CP25.77. Solve: (a) The force on the dipole is the vector s u m of the force on q and -q. We have =

(

KIQllq' , away from Q ( r+ s / 2 )

(b) The net force F,, is toward the charge Q, because the attractive force due to Q on the negative charge of the dipole is more than the repulsive force of Q on the positive charge. (c) Using the binomial approximation, we get

(d) Coulomb's law applies only to the force between two point charges. A dipole is not a point charge, so there's no reason that the force between a dipole and a point charge should be an inverse-square force. Assess: Note that when s + 0 m, F,,, -+ 0 N. In this limit the dipole is a point with zero charge.

Chapter 25

20~10-~ c 1.60~10-l9 c 25.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one m...