Introduction to Electrodynamics 4th Edition Griffiths Solutions Manual

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Instructor’s Solution Manual Introduction to Electrodynamics Fourth Edition David J. Griffiths 2014

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2

Contents 1 Vector Analysis

4

2 Electrostatics

26

3 Potential

53

4 Electric Fields in Matter

92

5 Magnetostatics

110

6 Magnetic Fields in Matter

133

7 Electrodynamics

145

8 Conservation Laws

168

9 Electromagnetic Waves

185

10 Potentials and Fields

210

11 Radiation

231

12 Electrodynamics and Relativity

262

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3

Preface Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the figures, and I would like to thank him particularly for all his help. If you find errors, please let me know (griffith@reed.edu).

David Griffiths

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3

CHAPTER 1. VECTOR ANALYSIS

CHAPTER 1. VECTOR ANALYSIS 4

3 CHAPTER 1. VECTOR ANALYSIS 3

CHAPTER 1. VECTOR ANALYSIS

Chapter 1 Chapter 1 Chapter 1

1 VectorChapter Analysis

Vector Analysis Vector Analysis Vector Analysis Problem 1.1 ✒ ✣

Problem 1.1

}} }

✒ ✣

|C| sin θ2

C

+

C

C

✒ ✣

B

(a) From the diagram, |B1.1 + C| cos θ3 = |B| cos θ1 + |C| cos θ2 . Multiply by |A|.

Problem Problem 1.1 |A||B + C| cos θ(a) |A||B| θ1 + |A||C| cos θ2θ.3 = |B| cos θ1 + |C| cos θ2 . thecos diagram, |B + C| cos 3 =From From the diagram, |B + C| cos ✓= |B| cos ✓1θ ++|C| cos ✓2cos . Multiply by |A|.

3 = |A||B + C| cos θ |A||B| cos θ2θ.2 .

So:(a)A·(B + C) = A·B + A·C. (Dot product (a) From the diagram, |B |B| cos|A||C| θ1 + |C| cos 3 + C| cos θ3 = is 1 distributive)

|C| sin

B

B

C

+

C

}

C

+

|A||B + C| cos ✓3 + =A·(B |A||B| ✓=1 A·B + |A||C| cos ✓(Dot 2 . cosproduct |C| sin θ2 |A||B C| cos+ θ3cos = |A||B| cos+θ1A·C. + |A||C| θ2 .

So: C) is distributive) θ2 So: A·(B C) = A·B + A·C. (Dot product is Similarly: |B ++C| sin θ = |B| sin θ + |C| sin θ Mulitply by |A| n̂.

2 . distributive) So: A·(B3 + C) = A·B 1+ A·C. (Dot product is distributive) θ3 ✯ θ2 θ3 = |B|sin sinθθ1n̂. + |C| sin θ2 . Mulitply by |A| n̂.B |A||B + C| sin θ3 n̂Similarly: = |A||B||B sin+θ1C|n̂sin + |A||C| θ3 ✯ |B| sin θ1 2 θ2 Similarly: |B + C| sin θ = |B| sin θ + |C| sin θ .

Mulitply by |A| n̂.

Similarly: |B +|A||B C| sin+ ✓C| = |B| sin ✓ + |C| sin ✓ .

Mulitply by |A| n̂.

3 1 2 3 sin θ3 n̂ =1 |A||B| sin θ21 n̂ + |A||C| sin θ2 n̂. θ1 θ3 ✯ B |B| sin ✲ If n̂ is|A||B the unit vector pointing out of the it✓ follows that $|B| sin |A||B C|the sin θunit =vector |A||B| sin θpage, |A||C| n̂. θ1 3 n̂sin 1 n̂ + θ1 + C| sin ✓3 n̂ n̂+is = |A||B| ✓1 n̂ +pointing |A||C| sin n̂.sin 2of If out theθ2page, it follows that! "#θ1B $ !! "# A "# θ2 $ !✲ "# $ ✲ A×(B C)the = unit (A×B) (A×C).

(Cross product is distributive) |B| cos θ1 $ ! |C|"# cos If n̂ vector is the+unit vector out pointing out of theit page, it follows that A ! "# $ If + n̂ is pointing of the page, follows that A×(B + C) = (A×B) + (A×C). (Cross product is distributive) |B| cos θ1 |C| A cos θ2 A×(B + C) = (A×B) + (A×C). (Cross product is distributive) |B| cos θ1 |C| cos θ2 A⇥(B + C)case, = (A⇥B) + E. (A⇥C). product is distributive) (b) For the general see G. Hay’s (Cross Vector and Tensor Analysis, Chapter 1, Section 71,(dot product) and general E. Hay’s Vector and Analysis, Tensor Analysis, Section 7 (dot (b)(b) ForFor the the general case,case, see G.see E. G. Hay’s Vector and Tensor Chapter 1,Chapter Section 7 (dot product) and product) Section (cross product) Section 8 (cross (b) For8 the general case, see G. E. product) Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8 (cross product) Section 8 (cross product) Problem 1.2 Problem Problem 1.2 1.2 Problem 1.2 C CC ✻ TheThe triple cross-product is not isinassociative. general associative.

For example, triple cross-product not in general associative.

For example, The triple cross-product is not in general For example, ✻ ✻ The triple cross-product is not in general associative.

For example, suppose A = B and C is perpendicular to A, as in the diagram. suppose = B and C istoperpendicular to diagram.

A, as in the diagram. suppose A = B and C isAperpendicular A, as in the ✲ out-of-the-page, andasA×(B×C) points down, suppose A Then = Then B (B×C) and(B×C) C points is perpendicular to A, in theA×(B×C) diagram. ✲AA==BB ✲ A = B points out-of-the-page, and points down, Then (B×C) points A×(B×C) down, andpoints hasout-of-the-page, magnitude ABC. and But (A×B) = 0, sopoints (A×B)×C = 0 ̸= ❂ Then (B⇥C) out-of-the-page, and A⇥(B⇥C) points down, and has magnitude ABC. But (A×B) = 0, so=(A×B)×C =B×C 0 ̸= ❂ and has ABC. But (A×B) A×(B×C). ❂ ❄ A×(B×C) and magnitude has magnitude ABC. But (A⇥B)==0,0,so so (A×B)×C (A⇥B)⇥C = 00 6≠= B×C ❄ A×(B×C).

A×(B×C) B×C ❄ A×(B×C).

A×(B×C) A⇥(B⇥C).

}

}

z

Problem 1.3 z✻✻ Problem 1.3 1.3 Problem 1.3 Problem z✻ √ p A = +1 x̂ + 1√ ŷ −p1 ẑ; A = 3; B√= 1 x̂ + 1 ŷ + 1 ẑ; √ A =x̂ +1 ẑ;=A =3;1 ŷB 3; B ŷ+√ + B = 1 ŷ3. ✣B =1 = 1 x̂1Ax̂ +=+1 1ŷ3; ẑ;ẑ;1B 3.+ 1 ẑ; A = +1 + 1x̂ŷ+−1 ŷ1Aẑ;=1A+1 ẑ; B131√ = x̂ = + 3 cos θ ⇒ cos θ.

A·B = +1 +x̂1+ − 1 =− 1= ABpcos θ = p B √ √ 11 √ √ ✣ θ ✣ ✲By & AB A·B + 11A·B 11−1 == 1%+1 = cos ✓=◦=1 3=3AB 3cos cos ✓θ⇒ ) cos ✓ 3= =cos ..θ ⇒ cos θ.

3 1 − 1 cos = A·B = +1=++1 1− = = AB θ = 3 θ cos θ 1 +cos 3 θ θ θ = cos 3 3 ≈ 70.5288 ✲y ❲ ✲y % & % & 1 θ= A ✰ cos◦−1 31 ≈ 70.5288◦ ✓ = 70.5288 −1cos1 1 3 ⇡ x ❲ θ = cos ❲ 3 ≈ 70.5288 A ✰ A ✰ x Problem 1.4 x Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, Problem 1.4 Problem 1.4 we might pick the base (A) and the left side (B): The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,

The cross-product The of any two vectorsofin the plane willingive a vector to the plane. example, cross-product any two vectors the plane willperpendicular give a vector perpendicular to For the plane.

For exam we might pick the base (A) and the left side (B): c base we Pearson might pick base (A) and the NJ. leftAllside (B): ⃝2005 Upper Saddle River, rights reserved. This material is we might pick the (A)Education, and the theInc., left side (B): A=

protected under all copyright laws as they currently exist. No portion of this material may be

1 x̂reproduced, + 2 ŷ +in0 any ẑ; B =or by 1 x̂any+means, 0 ŷ +without 3 ẑ. permission in writing from the publisher. form

A = −1 x̂ + 2 ŷ⃝2005 + 0 ẑ;Pearson B = Education, −1 x̂ + 0Inc., ŷ +Upper 3 ẑ. Saddle River, NJ. All rights reserved. c

This material is protected under all copyright laws as they currently exist. No portion of this material may be c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is reproduced, in any form or by any means, without permission in writing from the publisher. protected under copyright laws as they currently exist. c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ.all All rights reserved.

This material is No portion of this material may be in anyNo form or by any means, without permission in writing from the publisher. protected under all copyright laws as theyreproduced, currently exist. portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

5

CHAPTER 1. VECTOR ANALYSIS

x̂ ŷ ẑ 1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ.

1 0 3 This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its length: p A⇥B 6 3 2 |A⇥B| = 36 + 9 + 4 = 7. n̂ = |A ⇥B| = 7 x̂ + 7 ŷ + 7 ẑ .

A⇥B =

Problem 1.5

A⇥(B⇥C) =

(By Cz

x̂ Ax

Bz Cy ) (Bz Cx

ŷ Ay

Bx Cz ) (Bx Cy

ẑ Az

By Cx )

= x̂[Ay (Bx Cy By Cx ) Az (Bz Cx Bx Cz )] + ŷ() + ẑ() (I’ll just check the x-component; the others go the same way) = x̂(Ay Bx Cy Ay By Cx Az Bz Cx + Az Bx Cz ) + ŷ() + ẑ().

C(A·B) = [Bx (Ax Cx + Ay Cy + Az Cz ) Cx (Ax Bx + Ay By + Az Bz )] x̂ + () ŷ + () ẑ = x̂(Ay Bx Cy + Az Bx Cz Ay By Cx Az Bz Cx ) + ŷ() + ẑ(). They agree.

B(A·C) Problem 1.6

A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C) C(A·B)+C(A·B) A(C·B)+A(B·C) B(C·A) = 0.

So: A⇥(B⇥C) (A⇥B)⇥C = B⇥(C⇥A) = A(B·C) C(A·B).

If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or one is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.)

Conclusion: A⇥(B⇥C) = (A⇥B)⇥C () either A is parallel to C, or B is perpendicular to A and C. Problem 1.7

r

r

= (4 x̂ + 6 ŷ + 8 ẑ) p = 4+4+1= 3

r̂

=

r r

=

2 3 x̂

2 3 ŷ

(2 x̂ + 8 ŷ + 7 ẑ) = 2 x̂

2 ŷ + ẑ

+ 13 ẑ

Problem 1.8 (a) Āy B̄y + Āz B̄z = (cos Ay + sin Az )(cos By + sin Bz ) + ( sin Ay + cos Az )( sin By + cos Bz ) = cos2 Ay By + sin cos (Ay Bz + Az By ) + sin2 Az Bz + sin2 Ay By sin cos (Ay Bz + Az By ) + 2 cos Az Bz = (cos2 + sin2 )Ay By + (sin2 + cos2 )Az Bz = Ay By + Az Bz . X (b) (Ax )2 + (Ay )2 + (Az )2 = ⌃3i=1 Ai Ai = ⌃3i=1 ⌃3j=1 Rij Aj ⌃3k=1 Rik Ak = ⌃j,k (⌃i Rij Rik ) Aj Ak . ⇢ 1 if j = k This equals A2x + A2y + A2z provided ⌃3i=1 Rij Rik = 0 if j 6= k Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also necessary. For suppose A = (1, 0, 0). Then ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 , and this must equal 1 (since we 2 2 2 want Ax +Ay +Az = 1). Likewise, ⌃3i=1 Ri2 Ri2 = ⌃3i=1 Ri3 Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).

Then we want 2 = ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 + ⌃i Ri2 Ri2 + ⌃i Ri1 Ri2 + ⌃i Ri2 Ri1 . But we already know that the first two sums are both 1; the third and fourth are equal, so ⌃i Ri1 Ri2 = ⌃i Ri2 Ri1 = 0, and so on for other unequal combinations of j, k. X In matrix notation: R̃R = 1, where R̃ is the transpose of R. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6

CHAPTER 1. VECTOR ANALYSIS

CHAPTER 1. VECTOR ANALYSIS 5 CHAPTER 1. VECTOR ANALYSIS 5 Problem 1.9 y✻ y z′ ✻ ✻ y✻ y ❃ z′ ✻ ✻ ✿ ✿ ❃ down the axis: ✲ x LookingLooking down the axis: ✒ ❄ ✲x Looking down the axis: ✒ ❄ ′ ■ &y ′ ■z ✰ & ✠ y ✰ & z ′ ✰ z x ✠ x &x ✰′ z x

A 120 rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az , Ay = Ax , Az = Ay .

0 1 0 01 R = @1 0 0A 0 10 Problem 1.10

(a) No change. (Ax = Ax , Ay = Ay , Az = Az ) (b) A !

A, in the sense (Ax =

Ax , Ay =

Ay , Az =

Az )

(c) (A⇥B) ! ( A)⇥( B) = (A⇥B). That is, if C = A⇥B, C ! C . No minus sign, in contrast to behavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A⇥B) ! (A)⇥(B) = (A⇥B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector and a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.

Angular momentum (L = r⇥p) and torque (N = r⇥F) are pseudovectors. (d) A·(B⇥C) ! ( A)·(( B)⇥( C)) = A·(B⇥C). So, if a = A·(B⇥C), then a ! changes sign under inversion of coordinates.

Problem 1.11

a; a pseudoscalar

(a)rf = 2x x̂ + 3y 2 ŷ + 4z 3 ẑ (b)rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ (c)rf = ex sin y ln z x̂ + ex cos y ln z ŷ + ex sin y(1/z) ẑ Problem 1.12 (a) rh = 10[(2y 6x 18) x̂ + (2x 8y + 28) ŷ]. rh = 0 at summit, so 2y 6x 18 = 0 2y 18 24y + 84 = 0.

2x 8y + 28 = 0 =) 6x 24y + 84 = 0 22y = 66 =) y = 3 =) 2x 24 + 28 = 0 =) x = 2.

Top is 3 miles north, 2 miles west, of South Hadley. (b) Putting in x = 2, y = 3: h = 10( 12 12 36 + 36 + 84 + 12) = 720 ft. (c) Putting in x = 1, y = 1: rh = 10[(2 6 18) x̂ + (2 p |rh| = 220 2 ⇡ 311 ft/mile; direction: northwest.

8 + 28) ŷ] = 10( 22 x̂ + 22 ŷ) = 220( x̂ + ŷ).

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be in any form or All by rights any means, without c ⃝2005 Pearson Education, Inc.,reproduced, Upper Saddle River, NJ. reserved.

Thispermission material isin writing from the publisher. c ⃝2005 Pearson Education, Inc., Upper Saddle NJ. All rights reserved.

This material is protected under all copyright laws River, as they currently exist.

No portion of this material may be protected under all copyright laws as they exist.

No portion of this may the be publisher. reproduced, in any form or bycurrently any means, without permission in material writing from reproduced, in any form or by any means, without permission in writing from the publisher.

7

CHAPTER 1. VECTOR ANALYSIS Problem 1.13

r

= (x

x0 ) x̂ + (y

y 0 ) ŷ + (z

r

z 0 ) ẑ;

=

p (x

x0 )2 + (y

y 0 )2 + (z

z 0 )2 .

(a) r( r 2 ) =

@ @x [(x

@ @ x0 )2 +(y y 0 )2 +(z z 0 )2 ] x̂+ @y () ŷ + @z () ẑ = 2(x x0 ) x̂+2(y y 0 ) ŷ +2(z z 0 ) ẑ = 2

(b) r( r1 ) =

@ @x [(x

x0 )2 + (y

(c)

1 2 ()

= =

()

@ @x (

r

n

3 2

3 2

2(x x ) x̂ [(x x0 ) x̂ + (y 0

)=n

r

r

n 1@ @x

1 2 ()

y 0 )2 + (z 3 2

z 0 )2 ] 1 2 ()

2(y y ) ŷ y ) ŷ + (z z 0 ) ẑ] = 0

0

=n

r

r 2r

n 1 1 1 (2

x)

1 2

x̂ +

3 2

@ @y () 0

1 2

ŷ +

2(z z ) ẑ (1/ r 3 ) r =

=n

r

n 1

r̂

Problem 1.14

x,

@ @z ()

1 2

r

.

ẑ

(1/ r 2 ) r̂ . so r( r n ) = n

r

n 1

r̂

y = +y cos + z sin ; multiply by sin : y sin = +y sin cos + z sin2 . z = y sin + z cos ; multiply by cos : z cos = y sin cos + z cos2 .

Add: y sin + z cos = z(sin2 + cos2 ) = z. Likewise, y cos z sin = y. @y @y @z @z So @y = cos ; @z = sin ; @y = sin ; @z = cos . Therefore ) @f @y @f @z (rf )y = @f @y = @y @y + @z @y = + cos (rf )y + sin (rf )z So rf transforms as a vector. @f @y @f @z (rf )z = @f sin (rf )y + cos (rf )z @z = @y @z + @z @z =

qed

Problem 1.15 (a)r·va =

@ 2 @x (x )

+

@ 2 @y (3xz )

(b)r·vb =

@ @x (xy)

+

@ @y (2yz)

(c)r·vc =

@ 2 @x (y )

+

@ @y (2xy

Problem 1.16

+

+

@ @z (

2xz) = 2x + 0

@ @z (3xz)

+ z2) +

2x = 0.

= y + 2z + 3x.

@ @z (2yz)

= 0 + (2x) + (2y) = 2(x + y)

h @ @ @ @ r·v = @x ( rx3 ) + @y ( ry3 ) + @z ( rz3 ) = @x x(x2 + y 2 + z 2 ) h i h i 3 3 @ @ + @y y(x2 + y 2 + z 2 ) 2 + @z z(x2 + y 2 + z 2 ) 2 3

5

= () 2 + x( 3/2)() 2 2x + () 5 + z( 3/2)() 2 2z = 3r 3 3r

3 2

5

3 2

i

3

+ y( 3/2)() 2 2y + () 2 5 2 (x + y 2 + z 2 ) = 3r 3 3r

3

= 0.

This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the origin. How, then, can r·v = 0? The answer is that r·v = 0 everywhere except at the origin, but at the origin our calculation is no good, since r = 0, and the expression for v blows up. In fact, r·v is infinite at that one point, and zero elsewhere, as we shall see in Sect. 1.5.

Problem 1.17 v y = cos vy + sin vz ; v z = ⇣sin vy + cos v⌘z . ⇣ ⌘ @v y @vy @vy @y @vy @z @vz @vz @z z @y cos + @v sin . Use result in Prob. 1.14: @y = @y cos + @y sin = @y @y + @z @y @y @y + @z @y ⇣ ⌘ ⇣ ⌘ @vy @vy @vz @vz = @y cos + @z sin cos + @y cos + @z sin sin . ⇣ ⌘ ⇣ ⌘ @vy @vy @y @vy @z @v z @vz @vz @y @vz @z = sin + cos = + sin + + cos @z @z @y @z @z @z ⇣@z ⌘ @y @z ⇣ @z @z ⌘ @vy @vy @vz @vz = sin + cos . So @y sin + @z cos @y sin + @z cos c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8 @v y @y

+

@v z @z

=

@vy @y

cos2

@vy @z

+

CHAPTER 1. VECTOR ANALYSIS

sin cos +

sin cos +

@vz @y

=

@vy @y

cos

+ sin

+

2

2

@vz @y

sin cos + @vz @z

cos

sin2

@vz @z

@vz @z

sin2

+

@vy @y

@vy @z

sin2

sin cos

2

+ cos2

=

@vy @y

+

@vz @z .

X

Problem 1.18 (a) r⇥va =

x̂

ŷ

@ @x 2

@ @y

x̂

ŷ

ẑ

@ @x

@ @y

@ @z

x 3xz

(b) r⇥vb =

ẑ @ @z

= x̂(0

6xz) + ŷ(0 + 2z) + ẑ(3z 2

0) =

6xz x̂ + 2z ŷ + 3z 2 ẑ.

2xz

2

= x̂(0

2y) + ŷ(0

3z) + ẑ(0

x) =

2y x̂

3z ŷ

x ẑ.

xy 2yz 3xz

(c) r⇥vc =

x̂

ŷ

@ @x 2

@ @y

ẑ @ @z

= x̂(2z

2z) + ŷ(0

0) + ẑ(2y

2y) = 0.

y (2xy + z ) 2yz 2

Problem 1.19 As we go from point A to point B (9 o’clock to 10 o’clock), x increases, y increases, vx increases, and vy decreases, so @vx /@y > 0, while @vy /@y < 0. On the circle, vz = 0, and there is no dependence on z, so Eq. 1.41 says ✓ ◆ @vy @vx r ⇥ v = ẑ @x @y

y v v

v

B

A

x

points in the negative z direction (into the page), as the right hand rule would suggest. (Pick any other nearby points on the circle and you will come to the same conclusion.) [I’m sorry, but I cannot remember who suggested this cute illustration.]

z v Problem 1.20

v = y x̂ + x ŷ; or v = yz x̂ + xz ŷ + xy ẑ; or v = (3x2 z or v = (sin x)(cosh y) x̂ (cos x)(sinh y) ŷ; etc.

Problem 1.21 (i) r(f g) =

@(f g) @x

=f

⇣

x̂ +

@g @x

(iv) r·(A⇥B) =

= =

x̂ +

@(f g) @y @g @y

ŷ +

ŷ +

(v) r⇥ (f A) =

⇣

⇣ ⌘ ⇣ ⌘ ⇣ ⌘ @g @g @f @g @f ẑ = f @x + g @f x̂ + f + g ŷ + f + g ẑ @x @y @z @z ⌘ ⇣ ⌘ @y @f @f ẑ + g @f qed @x x̂ + @y ŷ + @z ẑ = f (rg) + g(rf ).

@g @z

Az By ) +

Ay

@(f Az ) @y

3xz 2 ) ẑ;

@(f g) @z

@ @ Ax Bz ) + @z (Ax By @y (Az Bx @A @B @Bz @Az @Bx y y z Ay @x + Bz @x Az @x By @x + Az @y + Bx @A @y @By @A x x +Ax @z + By @A Ay @B Bx @zy @z @z ⇣ ⌘ ⇣ ⌘ @A @A @Az @Ax y z x Bx @A + By @A + Bz @xy @y @z @z @x @y @ @x

(Ay Bz

z 3 ) x̂ + 3 ŷ + (x3

@Bx @z

@(f Ay ) @z

@Bz @x

⌘

x̂ +

⇣

Az

⇣

@(f Ax ) @z

@By @x

⌘

Ay Bx )

z Ax @B @y

Ax

⇣

@Bz @y

x Bz @A @y

@By @z

= B· (r⇥A) A· (r⇥B). ⌘ ⇣ ⌘ @(f Ay ) @(f Az ) @(f Ax ) ŷ + ẑ @x @x @y @Bx @y

⌘

qed

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9

CHAPTER 1. VECTOR ANALYSIS

⇣ ⌘ ⇣ ⌘ @A @f @f @Ax z z = f @A f @zy Ay @f f @A Az @f @y + Az @y @z x̂ + f @z + Ax @z @x @x ŷ ⇣ ⌘ @A x + f @xy + Ay @f f @A Ax @f @x @y @y ẑ h⇣ ⌘ ⇣ ⌘ i @Ay @Ay @Ax @Az @Ax z = f @A x̂ + ŷ + ẑ @y @z @x h⇣ @z ⌘ @x⇣ ⌘@y ⇣ ⌘ i @f @f @f @f @f Ay @f A x̂ + A A ŷ + A A z x z x y @z @y @x @z @y @x ẑ = f (r⇥A) A⇥ (rf ). qed Problem 1.22

⇣ ⌘ ⇣ ⌘ @By @By @By @Bx @Bx x (a) (A·r) B = Ax @B + A + A x̂ + A + A + A ŷ y z x y z @x @y @z @y @z ⇣ ⌘@x @Bz @Bz @Bz + Ax @x + Ay @y + Az @z ẑ. x x̂+y ŷ+z ẑ =p . Let’s just do the x component. x2 +y 2 +z 2 ⇣ ⌘ @ @ @ p 2 x2 2 [(r̂·r)r̂]x = p1 x @x + y @y + z @z x +y +z n h i h i h 1 1 p 3 2y + zx = 1r x p1 + x( 12 ) (p1 )3 2x + yx 2( )

(b) r̂ =

r r

=

1 r

x r

1 r3

x3 + xy 2 + xz 2

=

1 r

x r

x r3

1 1 p 3 2z 2( )

x2 + y 2 + z 2

=

1 r

x r

io

x r

= 0.

Same goes for the other components. Hence: (r̂·r) r̂ = 0 . ⇣ ⌘ @ @ @ (c) (va ·r) vb = x2 @x + 3xz 2 @y 2xz @z (xy x̂ + 2yz ŷ + 3xz ẑ)

= x2 (y x̂ + 0 ŷ + 3z ẑ) + 3xz 2 (x x̂ + 2z ŷ + 0 ẑ) 2xz (0 x̂ + 2y ŷ + 3x ẑ) = x2 y + 3x2 z 2 x̂ + 6xz 3 4xyz ŷ + 3x2 z 6x2 z ẑ = x2 y + 3z 2 x̂ + 2xz 3z 2

2y ŷ

3x2 z ẑ

Problem 1.23 (ii) [r(A·B)]x =

@ @x (Ax Bx

+ Ay By + Az Bz ) =

[A⇥(r⇥B)]x = Ay (r⇥B)z

@Ax @x Bx

Az (r⇥B)y = Ay

x + Ax @B @x +

@By @x

@Bx @y

@Ay @x By

+ Ay

@Bx @z

Az

@By @x

+

@Az @x Bz

z + Az @B @x

@Bz @x

@A

@Ax @Az x [B⇥(r⇥A)]x = By @xy Bz @A @y @z @x @Bx @Bx @ @ @ x [(A·r)B]x = Ax @x + Ay @y + Az @z Bx = Ax @B @x + Ay @y + Az @z @Ax @Ax @Ax [(B·r)A]x = Bx @x + By @y + Bz @z

So [A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A]x @B @Ay @Bz @Az x x x x = Ay @xy Ay @B Az @B By @A Bz @A @y @z + Az @x + By @x @y @z + Bz @x @Bx @Bx @Ax @Ax @Ax x +Ax @B @x + Ay @y + Az @z + Bx @x + By @y + Bz @z @Bx x = Bx @A @x + Ax @x + By

/

/

/

@Ax @y

+Bz + + Az = [r(A·B)]x (same for y and z) @Ax @z

z + @A @x

@Ay @x

(vi) [r⇥(A⇥B)]x = =

@Ax @z

/ + Ay @B@x @B / + @B@x + @B@z/ @z

x + @A @y x

y

z

/

@Bx @y

/

x + @B @y

x

@ @ @ Ay Bx ) @y (A⇥B)z @z (A⇥B)y = @y (Ax By @By @Ay @Ax @Bx @Az Ay @y @y By + Ax @y @y Bx @z Bx

[(B·r)A (A·r)B + A(r·B) B(r·A)]x @Ax @Ax x x x = Bx @A Ax @B Ay @B @x + By @y + Bz @z @x @y

x Az @B @z + Ax

@Bx @x

@ Ax Bz ) @z (Az Bx @Bx @Ax z Az @z + @z Bz + Ax @B @z

+

@By @y

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z + @B @z

Bx

@Ax @x

+

@Ay @y

z + @A @z

10 x = By @A @y + Ax

/

@Bx @x

/

x + @B @x +

CHAPTER 1. VECTOR ANALYSIS @By @y

+

@Bz @z x Bz @A @z

+ Bx

/

/

@Ax @x

@Ay @y

@Ax @x

@Az @z

x x + Ay @B + Az @B + @y @z = [r⇥(A⇥B)]x (same for y and z)

Problem 1.24 r(f /g) =

@ @ @ @x (f /g) x̂ + @y (f /g) ŷ + @z (f /g) ẑ @f @g @f @g g f g @x f @x g @f f @g x̂ + @y g2 @y ŷ + @z g2 @z ẑ 2 g h ⇣ ⌘ ⇣ @f @f @f @g @g 1 f @x x̂ + @y ŷ g 2 g @x x̂ + @y ŷ + @z ẑ

= =

r·(A/g) = = =

+

@ @ @ @x (Ax /g) + @y (Ay /g) + @z (Az /g) @A y @g @g x g Ay g @A Ax @x g @Az A @g @x + @y g2 @y + @z g2 z @x h g⇣2 ⌘ ⇣ @Ay @g @g @Ax @Az 1 Ax @x + Ay @y g2 g @x + @y + @z

[r⇥(A/g)]x = = = =

@ @ @y (Az /g) @z (Ay /g) @Az @A @g g @y Az @y g @zy Ay @g @z h g⇣2 ⌘g2 ⇣ @Ay @g @Az 1 Az @y g2 g @y @z g(r⇥A)x +(A⇥rg)x (same for g2

Ay @g @z

@g @z ẑ

⌘i

=

+ Az @g @z

grf f rg . g2

⌘i

=

qed

gr·A A·rg . g2

qed

⌘i

y and z). qed

Problem 1.25 (a) A⇥B =

x̂ ŷ ẑ x 2y 3z = x̂(6xz) + ŷ(9zy) + ẑ( 2x2 3y 2x 0

@ @ r·(A⇥B) = @x (6xz) + @y (9zy) + ⇣ ⌘ @ @ r⇥A = x̂ @y (3z) @z (2y) + ŷ ⇣ ⌘ @ @ r⇥B = x̂ @y (0) @z ( 2x) + ŷ ?

r·(A⇥B) = B·(r⇥A)

(b) A·B = 3xy

4xy =

6y 2 )

@ 2 6y 2 ) @z ( 2x @ @ @z (x) @x (3z) @ @z (3y)

A·(r⇥B) = 0

= 6z + 9z + 0 = 15z ⇣ ⌘ @ @ + ẑ @x (2y) @y (x) = 0; B·(r⇥A) = 0 ⇣ ⌘ @ @ @ (0) + ẑ ( 2x) (3y) = 5 ẑ; A·(r⇥B) = @x @x @y

15z

( 15z) = 15z. X

@ @ xy ; r(A·B) = r( xy) = x̂ @x ( xy) + ŷ @y ( xy) =

y x̂

x ŷ

x̂ ŷ ẑ A⇥(r⇥B) = x 2y 3z = x̂( 10y) + ŷ(5x); B⇥(r⇥A) = 0 0 0 5 ⇣ ⌘ @ @ @ (A·r)B = x @x + 2y @y + 3z @z (3y x̂ 2x ŷ) = x̂(6y) + ŷ( 2x) ⇣ ⌘ @ @ (B·r)A = 3y @x 2x @y (x x̂ + 2y ŷ + 3z ẑ) = x̂(3y) + ŷ( 4x)

A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A = 10y x̂ + 5x ŷ + 6y x̂ 2x ŷ + 3y x̂ 4x ŷ = y x̂ x ŷ = r·(A·B). X ⇣ ⌘ ⇣ @ @ @ @ @ (c) r⇥(A⇥B) = x̂ @y ( 2x2 6y 2 ) @z (9zy) + ŷ @z (6xz) @x ( 2x2 6y 2 ) + ẑ @x (9zy) = x̂( 12y 9y) + ŷ(6x + 4x) + ẑ(0) = 21y x̂ + 10x ŷ r·A =

@ @x (x)

+

@ @y (2y)

+

@ @z (3z)

= 1 + 2 + 3 = 6; r·B =

@ @x (3y)

+

@ @y (

⌘

@ @y (6xz)

2x) = 0

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11

CHAPTER 1. VECTOR ANALYSIS (B·r)A (A·r)B + A(r·B) = r⇥(A⇥B). X

B(r·A) = 3y x̂

4x ŷ

6y x̂ + 2x ŷ

18y x̂ + 12x ŷ =

21y x̂ + 10x ŷ

Problem 1.26 (a)

@ 2 Ta @x2

= 2;

(b)

@ 2 Tb @x2

=

(c)

@ 2 Tc @x2

= 25Tc ; @@yT2c =

(d)

@ 2 vx @x2 @ 2 vy @x2 @ 2 vz @x2

@ 2 Ta @y 2

@ 2 Tb @y 2

=

=

@ 2 Ta @z 2

@ 2 Tb @z 2

= 0 ) r2 Ta = 2.

=

Tb ) r2 Tb =

2

3Tb =

3 sin x sin y sin z.

2

9Tc ) r2 Tc = 0.

9 2 2 = 2 ; @@yv2x = @@zv2x = 0 ) r2 vx = 2 > = @2v @2v r2 v = 2 x̂ + 6x ŷ. = @y2y = 0 ; @z2y = 6x ) r2 vy = 6x > ; @ 2 vz @ 2 vz 2 = @y2 = @z2 = 0 ) r vz = 0

Problem 1.27

r·(r⇥v) = ⇣ 2 @ vz = @x @y

@ @x

⇣

16Tc ; @@zT2c =

@vy @z

@vz @y

@ 2 vz @y @x

⌘

+

From Prob. 1.18: r⇥va = Problem 1.28

⇣

⌘

@ 2 vx @y @z

@ @ @ @x @y @z @t @t @t @x @y @z

@vx @z

@ @y

@ 2 vx @z @y

⌘

⇣

@vy @vz @ @x + @z @x ⇣ 2 ⌘ @ vy @ 2 vy + @z @x @x @z

@vx @y

= x̂

@2t @y @z

@2t @z @y

+ ŷ

@2t @z @x

⌘

= 0, by equality of cross-derivatives.

6xz x̂+2z ŷ+3z 2 ẑ ) r·(r⇥va ) =

x̂ ŷ ẑ r⇥(rt) =

+

@2t @x @z

@ @x (

@ @ 6xz)+ @y (2z)+ @z (3z 2 ) =

+ ẑ

@2t @x @y

6z+6z = 0.

@2t @y @x

= 0, by equality of cross-derivatives.

In Prob. 1.11(b), rf = 2xy 3 z 4 x̂ + 3x2 y 2 z 4 ŷ + 4x2 y 3 z 3 ẑ, so x̂ ŷ ẑ @ @ @ r⇥(rf ) = @x @y @z 3 4 2 2 4 2 3 3 2xy z 3x y z 4x y z = x̂(3 · 4x2 y 2 z 3 4 · 3x2 y 2 z 3 ) + ŷ(4 · 2xy 3 z 3 2 · 4xy 3 z 3 ) + ẑ(2 · 3xy 2 z 4

Problem 1.29 (a) (0, 0, 0) (1, 0, 0) (1, 1, 0) R Total: (b) (0, 0, 0) (0, 0, 1) (0, 1, 1) R Total:

3 · 2xy 2 z 4 ) = 0. X

R R1 ! (1, 0, 0). x : 0 ! 1, y = z = 0; dl = dx x̂; v · dl = x2 dx; v · dl R= 0 x2 dx = (x3 /3)|10 = 1/3. ! (1, 1, 0). x = 1, y : 0 ! 1, z = 0; dl = dy ŷ; v · dl = 2yz dy = 0; v · dl = 0.

R R1 ! (1, 1, 1). x = y = 1, z : 0 ! 1; dl = dz ẑ; v · dl = y 2 dz = dz; v · dl = 0 dz = z|10 = 1.

v · dl = (1/3) + 0 + 1 = 4/3.

R ! (0, 0, 1). x = y = 0, z : 0 ! 1; dl = dz ẑ; v · dl = y 2 dz = 0; v · dl = 0.

R R1 ! (0, 1, 1). x = 0, y : 0 ! 1, z = 1; dl = dy ŷ; v · dl = 2yz dy = 2y dy; v · dl = 0 2y dy = y 2 |10 = 1.

R R1 ! (1, 1, 1). x : 0 ! 1, y = z = 1; dl = dx x̂; v · dl = x2 dx; v · dl = 0 x2 dx = (x3 /3)|10 = 1/3.

v · dl = 0 + 1 + (1/3) = 4/3.

(c) x = y = z : 0 ! 1; dx = dy = dz; v · dl = x2 dx + 2yz dy + y 2 dz = x2 dx + 2x2 dx + x2 dx = 4x2 dx; R R1 v · dl = 0 4x2 dx = (4x3 /3)|10 = 4/3.

H (d) v · dl = (4/3) (4/3) = 0. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12

CHAPTER 1. VECTOR ANALYSIS

Problem 1.30 x, y : 0 ! 1, z = 0; da = dx dy ẑ; v · da = y(z 2

3) dx dy =

2

R 3y dx dy; v · da =

3

R2 0

dx

R2 0

y dy =

3(x|20 )( y2 |20 ) = 3(2)(2) = 12. In Ex. 1.7 we got 20, for the same boundary line (the square in the xy-plane), so the answer is no: the surface integral does not depend only on the boundary line. The total flux for the cube is 20 + 12 = 32.

Problem 1.31 R R T d⌧ = z 2 dx dy dz. You can do the integrals in any order—here it is simplest to save z for last: Z ✓Z ◆ Z z2 dx dy dz.

R (1 y z) The sloping surface is x+y +z = 1, so the x integral is 0 dx = 1 y z. For a given z, y ranges from 0 to R (1 z) (1 z) 1 z, so the y integral is 0 (1 y z) dy = [(1 z)y (y 2 /2)]|0 = (1 z)2 [(1 z)2 /2] = (1 z)2 /2 = R R 1 z2 2 4 3 1 2 1 z z4 z5 1 2 (1/2) z + (z /2). Finally, the z integral is 0 z ( 2 z + 2 ) dz = 0 ( 2 z 3 + z2 ) dz = ( z6 4 + 10 )|0 = 1 6

1 4

+

1 10

= 1/60.

Problem 1.32 T (b) = 1 + 4 + 2 = 7; T (a) = 0. ) T (b)

T (a) = 7.

rT = (2x + 4y)x̂ + (4x + 2z 3 )ŷ + (6yz 2 )ẑ; rT ·dl = (2x + 4y)dx + (4x + 2z 3 )dy + (6yz 2 )dz 9 R R1 1 > (a) Segment 1: x : 0 ! 1, y = z = dy = dz = 0. rT ·dl = 0 (2x) dx = x2 0 = 1. =R R R1 b 1 rT ·dl = 7. X Segment 2: y : 0 ! 1, x = 1, z = 0, dx = dz = 0. rT ·dl = 0 (4) dy = 4y|0 = 4. a > R R1 2 1 ; 3 Segment 3: z : 0 ! 1, x = y = 1, dx = dy = 0. rT ·dl = 0 (6z ) dz = 2z 0 = 2.

9 R R1 > (b) Segment 1: z : 0 ! 1, x = y = dx = dy = 0. rT ·dl = 0 (0) dz = 0. > R R1 > 1 Segment 2: y : 0 ! 1, x = 0, z = 1, dx = dz = 0. rT ·dl = 0 (2) dy = 2y|0 = 2. = R b R R1 rT ·dl = 7. X a > Segment 3: x : 0 ! 1, y = z = 1, dy = dz = 0. rT ·dl = 0 (2x + 4) dx > > 1 ; = (x2 + 4x) 0 = 1 + 4 = 5. (c) x : 0 ! 1, y = x, z = x2 , dy = dx, dz = 2x dx.

rT ·dl = (2x + 4x)dx + (4x + 2x6 )dx + (6xx4 )2x dx = (10x + 14x6 )dx.

Rb R1 1 rT ·dl = 0 (10x + 14x6 )dx = (5x2 + 2x7 ) 0 = 5 + 2 = 7. X a

Problem 1.33

r·v = y + 2z + 3x o R R RR nR 2 (r·v)d⌧ = (y + 2z + 3x) dx dy dz = (y + 2z + 3x) dx dy dz 0 ⇥ ⇤2 (y + 2z)x + 32 x2 0 = 2(y + 2z) + 6 n o R R2 = (2y + 4z + 6)dy dz 0 ⇥ 2 ⇤2 y + (4z + 6)y 0 = 4 + 2(4z + 6) = 8z + 16

,!

,!

=

R2 0

(8z + 16)dz = (4z 2 + 16z)

2 0

= 16 + 32 = 48.

Numbering the surfaces as in Fig. 1.29: c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

13

CHAPTER 1. VECTOR ANALYSIS R RR 2 (i) da = dy dz x̂, x = 2. v·da = 2y dyRdz. v·da = 2y dy dz = 2y 2 0 = 8. (ii) da = dy dz x̂, x = 0. v·da = 0. v·daR= 0.

RR (iii) da = dx dz ŷ, y = 2. v·da = 4z dx R dz. v·da = 4z dx dz = 16. (iv) da = dx dz ŷ, y = 0. v·da = 0. v·da R = 0. (v) da = dx dy ẑ, z = 2. v·da = 6x dxRdy. v·da = 24. (vi)R da = dx dy ẑ, z = 0. v·da = 0. v·da = 0. ) v·da = 8 + 16 + 24 = 48 X

Problem 1.34

r⇥v = x̂(0 2y) + ŷ(0 3z) + ẑ(0 x) = 2y x̂ 3z ŷ x ẑ. da = dy dz x̂, if we agree that the path integral shall run counterclockwise. So (r⇥v)·da = 2y dy dz. o R R nR 2 z (r⇥v)·da = ( 2y)dy dz 0 z6 2 2 z 2 @ y 0 = (2 z) ⇣ ⌘ @ 2 R2 3 @ = (4 4z + z 2 )dz = 4z 2z 2 + z3 0 0 @ 8 = 8 8 + 83 = @ 3 @ y Meanwhile, v·dl = (xy)dx + (2yz)dy + (3zx)dz. There are three segments. y

,!

=

2

z

z6 @ @@ I (2) (3) @@ @ ? @ @ - y (1) R (1) x = z = 0; dx = dz = 0. y : 0 ! 2. v·dl = 0.

(2) x = 0; z = 2 y; dx = 0, dz = dy, y : 2 ! 0. v·dl = 2yz dy.

R R0 R2 2 v·dl = 2 2y(2 y)dy = (4y 2y 2 )dy = 2y 2 32 y 3 0 = 8 23 · 8 = 0 R H (3) x = y = 0; dx = dy = 0; z : 2 ! 0. v·dl = 0. v·dl = 0. So v·dl = 83 . X

8 3.

Problem 1.35

R

By Corollary 1, (r⇥v)·da should equal 43 . r⇥v = (4z 2 (i) da = dy dz x̂, x = 1; y, z : 0 ! 1. (r⇥v)·da = (4z 2 ( 43 z 3

1 0

2x)x̂ + 2z ẑ.

R R1 2)dy dz; (r⇥v)·da = 0 (4z 2

= 2z) = 2= R (ii) da = dx dy ẑ, z = 0; x, y : 0 ! 1. (r⇥v)·da = 0; R (r⇥v)·da = 0. (iii) da = dx dz ŷ, y = 1; x, z : 0 ! 1. (r⇥v)·da = 0; (r⇥v)·da = 0.

R (iv) da = dx dz ŷ, y = 0; x, z : 0 ! 1. (r⇥v)·da = 0; (r⇥v)·da = 0.

R (v) da = dx dy ẑ, z = 1; x, y : 0 ! 1. (r⇥v)·da = 2 dx dy; (r⇥v)·da = 2.

R ) (r⇥v)·da = 23 + 2 = 43 . X 4 3

2 3.

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2)dz

14

CHAPTER 1. VECTOR ANALYSIS

Problem 1.36 (a) Use the product rule r⇥(f A) = f (r⇥A) A ⇥ (rf ) : Z Z Z I Z f (r⇥A) · da = r⇥(f A) · da + [A ⇥ (rf )] · da = f A · dl + [A ⇥ (rf )] · da. qed S

S

S

P

S

(I used Stokes’ theorem in the last step.) (b) Use the product rule r·(A ⇥ B) = B · (r⇥A) A · (r⇥B) : Z Z Z I Z B · (r⇥A)d⌧ = r·(A ⇥ B) d⌧ + A · (r⇥B) d⌧ = (A ⇥ B) · da + A · (r⇥B) d⌧. qed V

V

V

S

V

(I used the divergence theorem in the last step.)

Problem 1.37 r = Problem 1.38

p x2 + y 2 + z 2 ;

✓ = cos

1

✓

p

z x2 +y 2 +z 2

◆

;

= tan

1

y x

.

There are many ways to do this one—probably the most illuminating way is to work it out by trigonometry from Fig. 1.36. The most systematic approach is to study the expression: r = x x̂ + y ŷ + z ẑ = r sin ✓ cos x̂ + r sin ✓ sin ŷ + r cos ✓ ẑ. @ If I only vary r slightly, then dr = @r (r)dr is a short vector pointing in the direction of increase in r. To make it a unit vector, I must divide by its length. Thus:

r̂ = @r @r @r @✓ @r @

= sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ; = r cos ✓ cos x̂ + r cos ✓ sin ŷ =

@r @r @r @r

; ✓ˆ =

@r 2 @r

r sin ✓ ẑ;

r sin ✓ sin x̂ + r sin ✓ cos ŷ;

@r 2 @

@r @✓ @r @✓

; ˆ=

= sin2 ✓ cos2 @r 2 = @✓ 2 2

@r @ @r @

.

+ sin2 ✓ sin2

r2 cos2 ✓ cos2

= r sin ✓ sin

2

+ cos2 ✓ = 1.

+ r2 cos2 ✓ sin2

+ r2 sin ✓ cos2 2

+ r2 sin2 ✓ = r2 .

= r2 sin ✓.

2

r̂ = sin ✓ cos x̂ + sin ✓ sin ŷ + cos ✓ ẑ. ) ✓ˆ = cos ✓ cos x̂ + cos ✓ sin ŷ sin ✓ ẑ. ˆ = sin x̂ + cos ŷ.

Check: r̂·r̂ = sin2 ✓(cos2 + sin2 ) + cos2 ✓ = sin2 ✓ + cos2 ✓ = 1, X ˆ ˆ = cos ✓ sin cos + cos ✓ sin cos = 0, X etc. ✓· sin ✓ r̂ = sin2 ✓ cos x̂ + sin2 ✓ sin ŷ + sin ✓ cos ✓ ẑ. cos ✓ ✓ˆ = cos2 ✓ cos x̂ + cos2 ✓ sin ŷ sin ✓ cos ✓ ẑ.

Add these: (1) sin ✓ r̂ + cos ✓ ✓ˆ = + cos x̂ + sin ŷ; ˆ = sin x̂ + cos ŷ.

(2) Multiply (1) by cos , (2) by sin , and subtract: x̂ = sin ✓ cos r̂ + cos ✓ cos ✓ˆ

sin

ˆ.

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Introduction to Electrodynamics 4th Edition Griffiths Solutions Manual Full Download: https://alibabadownload.com/product/introduction-to-electrodynamics-4th-edition-griffiths-solutions-manual/

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CHAPTER 1. VECTOR ANALYSIS Multiply (1) by sin , (2) by cos , and add: ŷ = sin ✓ sin r̂ + cos ✓ sin ✓ˆ + cos

ˆ.

cos ✓ r̂ = sin ✓ cos ✓ cos x̂ + sin ✓ cos ✓ sin ŷ + cos2 ✓ ẑ. sin ✓ ✓ˆ = sin ✓ cos ✓ cos x̂ + sin ✓ cos ✓ sin ŷ sin2 ✓ ẑ.

Subtract these: ˆ sin ✓ ✓.

ẑ = cos ✓ r̂ Problem 1.39

@ (a) r·v1 = r12 @r (r2 r2 ) = r12 4r3 = 4r ⇣ 4⌘ R R R RR R⇡ R (r·v1 )d⌧ = (4r)(r2 sin ✓ dr d✓ d ) = (4) 0 r3 dr 0 sin ✓ d✓ 2⇡ (2)(2⇡) = 4⇡R4 0 d = (4) 4 R R 2 R R ⇡ 2⇡ v1 ·da = (r r̂)·(r2 sin ✓ d✓ d r̂) = r4 0 sin ✓ d✓ 0 d = 4⇡R4 X (Note: at surface of sphere r = R.) R @ (b) r·v2 = r12 @r r2 r12 = 0 ) (r·v2 )d⌧ = 0 R R 1 R 2 v2 ·da = r 2 r̂ (r sin ✓ d✓ d r̂) = sin ✓ d✓ d = 4⇡.

They don’t agree!

The point is that this divergence is zero except at the origin, where it blows up, so our R calculation of (r·v2 ) is incorrect. The right answer is 4⇡.

Problem 1.40 @ 1 @ 1 @ r·v = r12 @r (r2 r cos ✓) + r sin ✓ @✓ (sin ✓ r sin ✓) + r sin ✓ @ (r sin ✓ cos ) 1 1 1 = r2 3r2 cos ✓ + r sin ✓ r 2 sin ✓ cos ✓ + r sin ✓ r sin ✓( sin ) = 3 cos ✓ + 2 cos ✓ sin = 5 cos ✓ sin R R RR R ✓ hR 2⇡ (r·v)d⌧ = (5 cos ✓ sin ) r2 sin ✓ dr d✓ d = 0 r2 dr 02 0 (5 cos ✓ =

=

⇣

R3 3

⌘

(10⇡)

R

⇡ 2

0

sin ) d

i

,!2⇡(5 cos ✓)

d✓ sin ✓

sin ✓ cos ✓ d✓

⇡ 2

,! sin2 ✓ 0 2

=

1 2

5⇡ 3 3 R .

Two surfaces—one the hemisphere: da = R2 sin ✓ d✓ d r̂; r = R; : 0 ! 2⇡, ✓ : 0 ! ⇡2 .

R R R⇡ R 2⇡ v·da = (r cos ✓)R2 sin ✓ d✓ d = R3 02 sin ✓ cos ✓ d✓ 0 d = R3 12 (2⇡) = ⇡R3 . ˆ = r dr d ✓ˆ (here ✓ = ⇡ ). r : 0 ! R, : 0 ! 2⇡. other the flat bottom: da = (dr)(r sin ✓ d )(+✓) 2 R R R R 2 R 2⇡ R3 v·da = (r sin ✓)(r dr d ) = r dr d = 2⇡ .

3 0 0 R Total: v·da = ⇡R3 + 23 ⇡R3 = 53 ⇡R3 . X

Problem 1.41 rt = (cos ✓ + sin ✓ cos )r̂ + ( sin ✓ + cos ✓ cos )✓ˆ +

1 sin /✓

( sin / ✓ sin ) ˆ

r2 t = r·(rt) @ 1 @ 1 @ = r12 @r r2 (cos ✓ + sin ✓ cos ) + r sin ✓ @✓ (sin ✓( sin ✓ + cos ✓ cos )) + r sin ✓ @ ( sin ) 2 1 1 1 2 = r2 2r(cos ✓ + sin ✓ cos ) + r sin ✓ ( 2 sin ✓ cos ✓ + cos ✓ cos sin ✓ cos ) r sin ✓ cos 2 2 1 2 = r sin ✓ [2 sin ✓ cos ✓ + 2 sin ✓ cos 2 sin ✓ cos ✓ + cos ✓ cos sin ✓ cos cos ] ⇥ ⇤ 2 1 2 = r sin cos = 0. ✓ (sin ✓ + cos ✓) cos

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