El Llano En Llama Juan Rulfo

INSTRUCTOR'S SOLUTIONS MANUAL

INTRODUCTION to ELECTRODYNAMICS
Third Edition

David J. Griffiths

TABLE OF CONTENTS

Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12

Vector Analysis Electrostatics Special Techniques Electrostatic Fields in Matter Magnetostatics Magnetostatic Fields in Matter

1 22 42 73 89113 125 146 157 179 195 and Relativity 219

Electrod ynamics Conservation Electromagnetic Laws Waves

Potentials and Fields Radiation Electrodynamics

Chapter

1

Vector
Problem 1.1

Analysis

(a) From the diagram, IB + CI COSO3 IBI COSO1 ICI COSO2'Multiply by IAI. = + IAIIB + CI COSO3 IAIIBI COSO1 + IAIICI COSO2. = So: A.(B + C) = A.B + A.C. (Dot product is distributive.) IAIIB + CIsin 03 n = IAIIBI sin 01 n + IAIICI sin O2n. If n is the unit vector pointing out of the page, it follows that Ax(B + e) = (AxB) + (Axe). (Cross product is distributive.)

ICI sin 82

Similarly: IB + CI sin 03 = IBI sin 01 + ICI sin O2, Mulitply by IAI n.
IBlsin81

A

(b) For the general case, see G. E. Hay's Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8(cross product). Problem 1.2 The triple cross-product is not in general associative. For example, suppose A = ~ and C is perpendicular to A, as in the diagram. Then (B XC) points out-of-the-page, and A X(B XC) points down, and has magnitude ABC. But (AxB) = 0, so (Ax B) xC = 0 :f. Ax(BxC).

k-hB
BxC iAx(Bxe)
z

Problem 1.3
A

= + 1x + 1Y - H; A = /3;

B

= 1x + 1Y+

Hi B = /3. =>cosO=~. x y

A.B = +1 + 1-1 = 1 = ABcosO = /3/3coso
10 = COS-1(t) ~ 70.5288° Problem 1.4
I

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, we might pick the base (A) and the left side (B): A = -1 x + 2 y + 0 z; B = -1 x + 0 Y + 3 z.

1

2

CHAPTER

1. VECTOR ANALYSIS

x
AxB

y

Z

2 0 1= 6x + 3y + 2z. -1 0 3 Thishas the' right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its length: IAxBI=v36+9+4=7. '7X+ '7y + '7z . AXB 3 2 ft - IAX BI = 16
A

= I -1

I

Problem 1.5 Ax Ay Az (ByCz - BzCy) (BzCx - BxCz) (BxCy - ByCx) = x[Ay(BxCy - ByCx) - Az(BzCx - BxCz)] + yO + zO (I'll just check the x-component; the others go the same way.) = x(AyBxCy - AyByCx - AzBzCx +AzBxCz) + yO + zOo B(A.C) - C(A.B) = [Bx(AxCx + AyCy + AzCz) - Cx(AxBx + AyBy + AzBz)] x + 0 y + 0 z = x(AyBxCy + AzBxCz - AyByCx - AzBzCx) + yO + zOo They agree.
Problem
1.6

Ax(BxC)

=

x

y

Z

Ax(BXC)+Bx(CxA)+Cx(A-xB)
So: Ax(BxC) - (AxB)xC

= -Bx(CxA)

= B(A.C)-C(A.B)+C(A.B)-A(C.B)+A(B.C)-B(C.A)

= A(B.C) - C(A.B).

= o.

If this is zero, then either A is parallel to C(including the case in which they point in oppositedirections, 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:Ax(BxC) = (Ax B) xC

y
Z

= -y

= +y

sin

if>

+Z

+ z sinif>; multiply cos if>; multiply

by sinif>: ysinif> = +y by cos if>:z cos if> -y =

sin if>cos if>+ Z sin2

if>.

sin cos + Z
if>if>

COS2 if>.

= z(sin2if>+cos2if» z. Likewise,ycosif>-zsinif>=y. = So ~ = cosif>; ~ = - sinif>; ~v = sinif>; ~~ = cosif>. Therefore (VJ)y = U = ~~ + M~v = +cosif>(VJ)y +sinif>(VJ)z So V I transforms ,!,
Add: ysinif>+zcosif>

(V I) z

. ,!, = ~ = ~ ~ + !li..0:' = - sm'l' (V/) oz oy oz oz oz

y + coS'l'(V/) z }

as a vector.

qed

Problem

1.15

(a)V.va
(b)V.Vb

= tx (X2) +ty (3XZ2) tz (-2xz) = 2x + 0':'- 2x = O. +
= tx (xy) + ty (2yz) + tz (3xz) = y +
2x + 3x.

(c)V.vc Problem
V.v

= tx (y2) + ty(2xy + Z2)+ tz (2yz) = 0 + (2x) + (2y) = 2(x + y).

1.16

= tx(-?-)+ty(~)+tz(?-) 5 3
= 0-2

= tx3 [x(x2

+y2 +z2)-~]+ty
5

3

[y(X2 +y2 +Z2)-~]+tz
5

[z(x2 +y2 +z2)-~]

= 3r-3

+ x( -3/2)()-22x
- 3r-5(x2

+ y2 + Z2)

+ 0-2 + y( -3/2)0-22y +...