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Preview text SOLUTIONS MANUAL INTRODUCTION to ELECTRODYNAMICS Third Edition David J. Griffiths Errata Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Date: September 1, 2004 Page 4, Prob.

1.15 (b): last expression should read y 2z 3x. Page 4, Prob.1.16: at the beginning, insert the following Page 8, Prob. 1.26: last line should read From Prob.

1.18: va x 2z y 3z 2 va ) (2z) 2 (3z ) 6z 0. Page 8, Prob. 1.27, in the determinant for ), 3rd row, 2nd column: change y 3 to y 2.

Page 8, Prob. 1.29, line 2: the number in the box should be (insert minus sign).

Page 9, Prob. 1. Guard dog lift station control. 31, line 2: change 2x3 to 2z 3 line of part (c): insert comma between dx and dz. Page 12, Probl 1.39, line 5: remove comma after cos Page 13, Prob. 1.42(c), last line: insert z after ).

Page 14, Prob. 1.46(b): change to a. Page 14, Prob. 1.48, second line of J: change the upper limit on the r integral from to R. Fix the last line to read: Page 15, Prob. 1.49(a), line 3: in the box, change x2 to x3.

1 (4), r: 5 0. V dl r cos2 (dr) 45 r dr. 4 r2 4 5 4 r dr v dl 5 2 5 5 2 5 Total: v dl 0 2 2.

Page 21, Probl 1.61(e), line 2: change z z to z. Page 25, Prob. 2.12: last line should read Since Qtot 43 E 1 Q 0 R3 r (as in Prob. Page 26, Prob. 2.15: last expression in line of (ii) should be not d phi. Page 28, Prob.

2.21, at the end, insert the following V(r) 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0.5 1 1.5 2 2.5 3 r In the r is in units of R, and V (r) is in units of q 0 R. Page 30, Prob. 2.28: remove right angle sign in the Page 42, Prob. 3.5: subscript on V in last integral should be 3, not 2.

Page 45, Prob. 3.10: after the box, add: 1 1 1 x y x sin y, (2a)2 (2b)2 (2 a2 b2 )2 where cos a2 b2, sin a2 b2.

Q2 q2 a 1 2 a (a2 b2 3 b 1 x 2 b (a2 b2 y. W 2 1 q2 2 q2 1 1 1 1. 4 (2a) (2b) (2 a2 b2 ) a b a2 b2 Page 45, Prob. 3.10: in the second box, change to Page 46, Probl 3.13, at the end, insert the following: Technically, the series solution for is defective, since has produced a (naively) sum. More sophisticated of convergence permit one to work with series of this form, but it is better to sum the series and then (the second Page 51, Prob.

3.18, midpage: the reference to Eq. 3.71 should be 3.72. Page 53, Prob. 3.21(b), line 5: A2 should be 1 2R. 4 0 R next line, insert r2 after Page 55, Prob. 3.23, third displayed equation: remove the Page 58, Prob.

3.28(a), second line, integral: R3 should read R2. Page 59, Prob. 3.31(c): change V to W. Page 64, Prob. 3.41(a), lines 2 and 3: remove in the factor in the expressions for Eave in the second expression change to Page 69, Prob. 3.47, at the end add the following: Alternatively, start with the separable solution V (x, y) (C sin kx D cos kx) Aeky. Note that the is symmetric in x, so C 0, and V (x, 0) 0 B so (combining the constants) V (x, y) A cos kx sinh ky.

But V (b, y) 0, so cos kb 0, which means that kb, or k (2n, with n 1, 2, 3,. (negative k does not yield a sign can be absorbed into A). The general linear combination is V (x, y) An cos x sinh y, and it remains to the boundary condition: V (x, a) V0 4 An cos x sinh a. X Page 114, Prob. 6.4: last term in second expression for F should be z (plus, not minus). Page 119, Prob. 6.21(a): replace with the following: The magnetic force on the dipole is given Eq.

To move the dipole in from we must exert an opposite force, so the work done is r r F dl B) dl B(r) m U (I used the gradient theorem, Eq. As long as the magnetic goes to zero at then, U B. If the magnetic does not go to zero at one must stipulate that the dipole starts out oriented perpendicular to the Page 125, Prob. 7.2(b): in the box, c should be C. Page 129, Prob.

7.18: change two lines to read: Ia Ia I ln B B s s E Iloop R dQ dQ a dI ln(1. Dt dt dt aI a ln(1 dI Q ln(1 Page 131, Prob.

7.27: in the second integral, r should be s. Page 132, Prob. 7.32(c), last line: in the two equations, insert an I immediately after. Page 140, Prob.

7.47: in the box, the top equation should have a minus sign in front, and in the bottom equation the plus sign should be minus. Page 141, Prob. 7.50, answer: R2 should read R2. Page 143, Prob. 7.55, penultimate displayed equation: tp should be Page 147, Prob. 8.2, top line, penultimate expression: change a2 to a4 in (c), in the box, change 16 to 8. Page 149, Prob.

8.5(c): there should be a minus sign in front of 2 in the box. Page 149, Prob. 8.7: almost all the here should be In line 1 change r to in the same line change dr to in the next line change dr to ds (twice), and change r to in the last line change r to s, dr to ds, and r to s (but leave r as is). 6 Page 153, Prob.

8.11, last line of equations: in the numerator of the expression for R change 2.01 to 2.10. Page 175, Prob. 9.34, penultimate line: n3 (not n3 ). Page 177, Prob.

9.38: down, remove minus sign in kx2 ky2 kz2. Page 181, Prob. 10.8: line: remove Page 184, Prob. 10.14: in the line, change (9.98) to (10.42). Page 203, Prob. 11.14: at beginning of second paragraph, remove Page Prob.

12.15, end of sentence: change comma to period. Page 225, Prob.

The contains two errors: the slopes are for (not and the intervals are incorrect. The correct solution is as follows: Page 227, Prob. 12.33: expression in third line, change c2 to c. 7 Chapter 1 Vector Analysis Problem 1.1 (a) From the diagram, IB CI IBI ICI IAI.

IAIIB CI IAIIBI COSO1 IAIICI COSO2. So: A.(B C) A.B A.C. (Dot product is distributive.) ICI sin 82 Similarly: IB CI sin 03 IBI sin 01 ICI sin O2, Mulitply IAI n. IAIIB CI sin 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.) IBlsin81 A (b) For the general case, see G. Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8 (cross product).

Problem 1.2 The triple is not in general associative. For example, suppose A and C is perpendicular to A, as in the diagram. Then (B XC) points and A X(B XC) points down, and has magnitude ABC. But (AxB) 0, so (Ax B) xC 0:f. 1x 1Y A B 1x A.B 1 ABcosO 10 BxC iAx(Bxe) z Problem 1.3 A Hi B y I x Problem 1.4 The 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 x 2 y 0 B x 0 Y 3 z.

1 2 CHAPTER x y 1. VECTOR ANALYSIS Z I 2 0 6x 3y 2z. 0 3 This has right direction, but the wrong magnitude. To make a unit vector out of it, simply divide its length:.

AXBBI 3 2 ft IAX AxB I A Problem 1.5 x Ax(BxC) y Z Ax Ay Az (Cz BzCy) (BzCx BxCz) (BxCy Cx) Cx) Az(BzCx yO zO just check the the others go the same way.) x(AyBxCy AyCx AzBzCx AzBxCz) yO zOo B(A.C) C(A.B) AyCy AzCz) Cx(AxBx Ay x 0 y 0 z x(AyBxCy AzBxCz AyCx AzBzCx) yO zOo They agree. Problem 1.6 So: Ax(BxC) (AxB)xC A(B.C) 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 A is parallel to C, or B is perpendicular to A and C. Problem 1.7 zl 4 1 yl4 3x 12A:rroblem 3Y 3z 2A lAI 1.8 (cos cpAy sin cpAz)(cos cp sin cpBz) (a) A.y A.zBz COS2 cpAy sin cos sin cos cpBz) sincpcoscp(AyBz Az) sin2 cpAzBz sin2 cpAy sin cpcoscp(AyBz Az) COS2cpAzBz sin2 cp)Ay (sin2 COS2cp)AzBz 2 2 2 3 (b) (Ax) (Ay) (Az).d d Th I A2 A2 A2 IS equa s x z provz e I 3 R R ij Ay 3 AzBz. I if j 0 if j 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 RijRik) AjAk RilRil, and this must equal 1 (since we. To check the caseJ choose A (1,1,0).

Then we want 2 RijRik) AjAk RilRil Ri2Ri2 RilRi2 Ri2Ril. But we already know that the first two sums are both the third and fourth are equal, so RilRi2 Ri2Ril 0, and so on for other unequal combinations of j, k. In matrix notation: RR 1, where R is the transpose of R. VECTOR ANALYSIS 4 Problem 1.13 (x x (y y (z Zl) Zj Vex (y (z ZI)2.

(a) tzOz (b) V(k) tx (y (z y tz 1, 1, 12 2( ) 2 ) 2 2( ) Z x (y y (z z (c) so i.1 I Problem 1.14 y sin cos cos z multiply cos Z sin Z sin2 sin cos Z COS2 So Therefore (VJ)y U So V I transforms,!, Add: (V I) z Problem. Oz oy oz oz oz y z tx (X2) ty tz 2x 2x O. (b)V.Vb tx (xy) ty (2yz) tz (3xz) y (c)V.vc tx (y2) ty(2xy tz (2yz) 0 (2x) (2y) 2(x y). V.v qed 1.15 (a)V.va Problem as a vector.

1.16 3 5 x( tx3 5 3 5 y( z( O. Y2 Z2) This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the 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, V.v is infinite at that one point, and zero elsewhere, as we shall see in Sect. Problem 1.17 origin. How, then, can V.v O?

The answer is that V.v Vy Vz,!,. P b 114.!lJL.!bl.!lJL.8Z.,!, U 8y 8y oy 8y 8z 8y 8y 8y 8z oy se resu t m ra. COg sin!lJL.!lJL.sin sin 8y cos OZ ( ( sin ) ( 8y sin i: 8z ( ) OZ 8z sin ) sin sin!lJL.8. 8y 8z oz 8z )!lJL.sin sin sin 8y!lJL.sin2 8z 5 oy (cOS2 sin2 (sin2 COS2 A. ) Problem 8z.( 1.18 x0 (a) Vxva (b) VXVb 8x (c) Vxvc z0 3xz2 oy 2xz x y Z 0 8y 0 oz xy 2yz 3xz I!Ix x(O 6xz) 2z) z(3z2 0) oz 8 ox y2 Problem y0 x2 x v or v y 0 oy (2xy z2) x(O 2y) y(O 3z) x) x 2z y 3Z2z.1 3zy xz.! Z 0 oz 2yz x(2z 2z) y(O 0) z(2y 2y) 1.19 yx or v yzx xzy xy Zjor v (3x2z (sin x) (cosh Z3) 3y (x3 y) x (cos x) (sinh y) yj etc. Problem 1.20 (i) V(fg) x y Z gU) x gU) y J g(VI).

Qed!Ix (AyBz (iv)V.(AxB) Az)!ly (AzBx AxBz)!lz (Ax AyBx) B y ML A 8Bz B ML Ax 8x z 8y x 8y Ay Bz Az 8Bv 8x Bx x B y ML 8z A y 8B. 8z B x ML ML Bz B y ( 8z 8y OZ 8z ) ( ( ) ML OA. A 8y ) B z oA. Oy x 8z Az B. (v) Vx (fA) y A J Mox J 8y z 8y 8z Y8z ) OZ z oz z 8x )y z oAz 8y 8z y 8x x 8y ) ML ML Z J M.

8Az ( fu fu )Y ( ) A ) A AzU) x (AzU (AxU AyU) J(VxA) Problem Ax (VI). Qed 1.21 (a ) (A.

B (A A A Zfu8B.) X A x 8x ( (b) f!r y oy (A z. Z ) just do the x component. (x z tz) AZfu8Bv )Y A ) qed 7 u 9 A 2 g2 (same for y and z). G2 Problem 1.24 x y Z (a) AxB x 2y 3z 0 3y I x(6xz) Qed y(9zy) z(2X2 6y2) t.,(6xz) ty(9zy) 6y2) 6z 9z 0 15z V.(AxB) VxA x (tz(x) ty(x)) B.(VxA) 0 VxB x (ty(O) y (tz(3y) t., (0)) z ty(3Y)) A.(VxB)? V.(AxB) (b)A.B 3xy 4xy x y A.(VxB) 0 15z. Xt.,( xy V(A.B) z x 2y 3z Bx(VxA) 0 I I (A.V)B 0 0 I (xt., 2yty 3ztz) (3yx 2xy) x(6y) x(3y) y( (B.V)A (3y to: 2xty) (xx 2yy 3zz) Ax(VxB) Bx(VxA) (A.V)B (B.V)A 5xy 6yx 2xy 3yx 4xy xy V.(A.B). (c) V X (AxB) x (ty tz y (tz (6xz) t., z (t., (9zy) ty (6xz)) 9y) y(6x 4x) z(O) lOxy V.A t.,(x) ty(2y) tz(3z) V.B t.,(3y) 0.

(B.V)A (A.V)B A(V.B) B(V.A) Vx(AxB). 3yx 4xy 6yx 2xy 12xy Problem 1.25 (a) (b) 0 I 2.1 W 8271 W 8271 8271 n. I I (c) 25Tc I (d) 82v 2 82v 0.1 0 2 82v W W 0 F 6x 2 Vy 6x 0 0 lOxy 8 CHAPTER Problem 1.26 V. VECTOR ANALYSIS ( a2vz ax 8y az a2vz ) (ayoz 8y 8x From Prob.

1.18: VXVb 82vz oz oy ). Al t f d 0 b ax 8z ), y equ 1 y 0 enva Ives. ) JL ax az ax ay ( az ) JL ) (oz ax 3zy xz O.

Problem 1.27 V X (Vt) x ya z8 8x fJy fJz I fJta fJt fJy fJx fJ2t x ( fJy fJz fJt az fJ2t fJz fJy ) fJ2t Y( fJz fJx fJ2t fJx fJz a2t ) z ( ax fJy a2t fJyax ) 0, equality of In Prob. 1.l1(b), V f 2xy3z4 X 3x2y2z4 y 4x2y3z3z, so X a VX(Vf) fJx 2xy3z4 y a fJy 3x2y3z4 z a az 4x2y3z3 x(3. 3x2y2z3) y(4. Problem 1.28 (a) (0,0, 0) (1,0, 0) (1,1, 0) (1,0,0). X: 0 1, y z OJdl dx Xjv. Dl Total: J v. Dl (b) (0,0, 0) (0,0, 1) (0,1, 1) x2 dxj J v.

Dl X2 dx (1,1,0). X 1,Y: 0 1,z dy v.

Dl 2yz dy OJJ v. X Y 1,z: 0 Ij dl dz Zjv. Dl y2dz dzj J v.

Dl Jo1dz zlA 1. X y 0, z: 0 dz v. Dl y2 dz J v. X O,y: 0 l,z Ijdl 2yzdy 2ydy 1. X: 0 1,y z Ij dl v. Dl x2 dx Total: J v. Dl 0 1 (c) x y z: 0 Ij dx dy dzj v.

Dl X2 dx 2yz dy y2 dz x2 dx 2X2 dx x2 dx 4X2 J v. Dl 4x2 dx (d) f v. Dl Problem 1.29 x,y: 0 l,z Ojda da y(z2 3)dxdy da In Ex. 1.7 we got 20, for the same boundaryline (the square in the plane), so the answer is Ino: Ithe surface integral does not depend only on the boundary line. The total flux for the cube is 20 12 Problem 1.30 J T dr Z2dx dy dz. You can do the integrals in any Z2 it is simplest to save z for last: The sloping surface is 1, so the x integral is dx For a y ranges from 0 to z, so the y integral is y z) dy z)y z)2 CHAPTER 1.

VECTOR ANALYSIS 10 J(Vxv).da y z Z)2 Meanwhile, v.dl (xy)dx (2yz)dy (3zx)dz. There are three segments. Y z (3)1 y z dx dz O. Z 2 dx 0, dz y: 2 O.v.dl 2yzdy. (1) x (2) x Jv.dl y)dy J:(4y (S.S) (3) x y dx dy z: 2 O.v.dl O.Jv.dl O.

So Problem1.34 Corollary 1, J(Vxv).da (i) da dydz x, should equal VXv (4z2 2x)x 2zz. (Vxv).da J(Vxv).da 2)dz 2 (ii) da z OJx,y: 0 1.(Vxv).da J(Vxv).da O. Da dxdzy, y x,z: 0 1. (Vxv).da f(Vxv).da O. (iv) da y x, z: 0 1.

(Vxv).da J(Vxv).da O. (v) da dxdyz, z x,y: 0 1. (Vxv).da J(Vxv).da 2. 2 Problem 1.35 (a) Use the product rule V X(fA) f(V XA) L A x (V f): qed. (I used theorem in the last step.) (b) Use the product rule V.(A x B) B. (VxB): (I used the divergence theorem in the last step.) IvA,(VXB)dr. 11 Problem 1.361 r,jx2 y2 Problem 1.37 0 cos ( ) q, m.

There are many ways to do this the most illuminating way is to work it out trigonometry from Fig. The most systematic approach is to study the expression: r x x y Y z z r sin 0 cos q, x r sin 0 sin q, y r cos 0 z.

If I only vary r slightly, then dr a short vector pointing in the direction of increase in r. To make it a unit vector, I must divide its length. Thus: 8r!k 8r sinOcosq,x sinOsinq,y sinO cosq,x sinOsin q,y cosO Z.

F r2 sin2 Osin2 q, r2 sin2 OCOS2 q, cosOcos q, x cosOsinq,y sinOz. Sinq,x cosq,y. Check: f.f sin2 0(COS2q, COS2 0 sin20 COS2 0 1, etc.

SinOf sin2 0 cosq,x sin2 Osin q,y sin OcosOz. CosOIi COS2Boos q,x COS20 sin q,y sinOcosOz. Add these: (1) (2) Multiply (1) (2) and subtract: I x sinOcosq,f cosO cos q,Ii Multiply (1) (2) and add: Iy cosOf sinO Ii sinO cosOcos q,x sinOsinq,f 1. CosOsinq,1i sin OcosO sin q,y COS2 0 z. SinO cosO cosq,x sinO cosO sinq,y sin2 0 z. Subtract these: I r2 sin2 0.