Instructors Solution ManualIntroduction to ElectrodynamicsThird EditionDavid J. GrithsRevised November, 20092Contents1 Vector Analysis 42 Electrostatics 253 Special Techniques 464 Electrostatic Fields in Matter 785 Magnetostatics 946 Magnetostatic Fields in Matter 1177 Electrodynamics 1288 Conservation Laws 1499 Electromagnetic Waves 16010 Potentials and Fields 18211 Radiation 19712 Electrodynamics and Relativity 221c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.3PrefaceAlthough I wrote these solutions, most of the typesetting was done by Jonah Gollub, Christopher Lee, andJames Terwilliger (any errors are, of course, entirely their fault). Chris also did most of the gures, and Iwould like to thank him particularly for all his help. All errors discovered before September 2009 have beencorrected in this version, but if you nd further mistakes, please let me know (grith@reed.edu).David Grithsc 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.4 CHAPTER 1. VECTOR ANALYSISChapter 1Vector AnalysisProblem 1.1CHAPTER 1. VECTOR ANALYSIS 3Chapter 1Vector AnalysisProblem 1.1-

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ABCB+C |B| cos 1 |C| cos 2}|B| sin 1}|C| sin 2123(a) From the diagram, |B+ C| cos 3 = |B| cos 1 + |C| cos 2.|A||B+C| cos 3 = |A||B| cos 1 + |A||C| cos 2.So: A(B+ C) = AB+AC. (Dot product is distributive)Similarly: |B+C| sin 3 = |B| sin 1 + |C| sin 2. Mulitply by |A| n.|A||B+C| sin 3 n = |A||B| sin 1 n + |A||C| sin 2 n.If n is the unit vector pointing out of the page, it follows thatA(B+C) = (AB) + (AC). (Cross product is distributive)(b) For the general case, see G. E. Hays Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)Problem 1.2-A = B6C=BC?A(BC)The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (BC) points out-of-the-page, and A(BC) points down,and has magnitude ABC. But (AB) = 0, so (AB)C = 0 =A(BC).Problem 1.3-y6z+x

BWAA = +1 x + 1 y 1z; A =3; B = 1 x + 1 y + 1z;AB = +1 + 1 1 = 1 = ABcos =33 cos cos . = cos1 13

70.5288Problem 1.4The 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):c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.(a) From the diagram, [B+C[ cos 3 = [B[ cos 1 +[C[ cos 2. Multiply by [A[.[A[[B+C[ cos 3 = [A[[B[ cos 1 +[A[[C[ cos 2.So: A(B+C) = AB+AC. (Dot product is distributive)Similarly: [B+C[ sin 3 = [B[ sin 1 +[C[ sin 2. Mulitply by [A[ n.[A[[B+C[ sin 3 n = [A[[B[ sin 1 n +[A[[C[ sin 2 n.If n is the unit vector pointing out of the page, it follows thatA(B+C) = (AB) + (AC). (Cross product is distributive)(b) For the general case, see G. E. Hays Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)Problem 1.2CHAPTER 1. VECTOR ANALYSIS 3Chapter 1Vector AnalysisProblem 1.1-

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ABCB+C |B| cos 1 |C| cos 2}|B| sin 1}|C| sin 2123(a) From the diagram, |B+ C| cos 3 = |B| cos 1 + |C| cos 2. Multiply by |A|.|A||B+C| cos 3 = |A||B| cos 1 + |A||C| cos 2.So: A(B+ C) = AB+AC. (Dot product is distributive)Similarly: |B+C| sin 3 = |B| sin 1 + |C| sin 2. Mulitply by |A| n.|A||B+C| sin 3 n = |A||B| sin 1 n + |A||C| sin 2 n.If n is the unit vector pointing out of the page, it follows thatA(B+C) = (AB) + (AC). (Cross product is distributive)(b) For the general case, see G. E. Hays Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)Problem 1.2-A = B6C=BC?A(BC)The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (BC) points out-of-the-page, and A(BC) points down,and has magnitude ABC. But (AB) = 0, so (AB)C = 0 =A(BC).Problem 1.3-y6z+x

BWAA = +1 x + 1 y 1z; A =3; B = 1 x + 1 y + 1z; B =3.AB = +1 + 1 1 = 1 = ABcos =33 cos cos = 13. = cos1 13

70.5288Problem 1.4The 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 + 0z; B = 1 x + 0 y + 3z.c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (BC) points out-of-the-page, and A(BC) points down,and has magnitude ABC. But (AB) = 0, so (AB)C = 0 ,=A(BC).Problem 1.3CHAPTER 1. VECTOR ANALYSIS 3Chapter 1Vector AnalysisProblem 1.1-

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ABCB+C |B| cos 1 |C| cos 2}|B| sin 1}|C| sin 2123(a) From the diagram, |B+ C| cos 3 = |B| cos 1 + |C| cos 2.|A||B+C| cos 3 = |A||B| cos 1 + |A||C| cos 2.So: A(B+ C) = AB+AC. (Dot product is distributive)Similarly: |B+C| sin 3 = |B| sin 1 + |C| sin 2. Mulitply by |A| n.|A||B+C| sin 3 n = |A||B| sin 1 n + |A||C| sin 2 n.If n is the unit vector pointing out of the page, it follows thatA(B+C) = (AB) + (AC). (Cross product is distributive)(b) For the general case, see G. E. Hays Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)Problem 1.2-A = B6C=BC?A(BC)The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (BC) points out-of-the-page, and A(BC) points down,and has magnitude ABC. But (AB) = 0, so (AB)C = 0 =A(BC).Problem 1.3-y6z+x

BWAA = +1 x + 1 y 1z; A =3; B = 1 x + 1 y + 1z;AB = +1 + 1 1 = 1 = ABcos =33 cos cos . = cos1 13

70.5288Problem 1.4The 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):c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.A = +1 x + 1 y 1z; A =3; B = 1 x + 1 y + 1z; B =3.AB = +1 + 1 1 = 1 = ABcos =33 cos cos = 13. = cos1_13_ 70.5288Problem 1.4The 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 + 0z; B = 1 x + 0 y + 3z.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 5AB = x y z1 2 01 0 3= 6 x + 3 y + 2z.This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by itslength:[AB[ =36 + 9 + 4 = 7. n = AB|AB| = 67 x + 37 y + 27z .Problem 1.5A(BC) = x y zAx Ay Az(ByCzBzCy) (BzCxBxCz) (BxCyByCx)= x[Ay(BxCyByCx) Az(BzCxBxCz)] + y() +z()(Ill just check the x-component; the others go the same way)= x(AyBxCyAyByCxAzBzCx +AzBxCz) + y() +z().B(AC) C(AB) = [Bx(AxCx +AyCy +AzCz) Cx(AxBx +AyBy +AzBz)] x + () y + () z= x(AyBxCy +AzBxCzAyByCxAzBzCx) + y() +z(). They agree.Problem 1.6A(BC)+B(CA)+C(AB) = B(AC)C(AB)+C(AB)A(CB)+A(BC)B(CA) = 0.So: A(BC) (AB)C = B(CA) = A(BC) C(AB).If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, orone is zero), or else BC = BA = 0, in which case B is perpendicular to A and C (including the case B = 0.)Conclusion: A(BC) = (AB)C either A is parallel to C, or B is perpendicular to A and C.Problem 1.7r = (4 x + 6 y + 8z) (2 x + 8 y + 7z) = 2 x 2 y + zr =4 + 4 + 1 = 3r = rr = 23 x 23 y + 13zProblem 1.8(a) Ay By + Az Bz = (cos Ay + sin Az)(cos By + sin Bz) + (sin Ay + cos Az)(sin By + cos Bz)= cos2AyBy + sin cos (AyBz + AzBy) + sin2AzBz + sin2AyBy sin cos (AyBz + AzBy) +cos2AzBz= (cos2 + sin2)AyBy + (sin2 + cos2)AzBz = AyBy +AzBz. (b) (Ax)2+ (Ay)2+ (Az)2= 3i=1AiAi = 3i=1_3j=1RijAj_ _3k=1RikAk_ = j,k (iRijRik) AjAk.This equals A2x +A2y +A2z provided 3i=1RijRik =_1 if j = k0 if j ,= k_Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sucient but alsonecessary. For suppose A = (1, 0, 0). Then j,k (iRijRik) AjAk = iRi1Ri1, and this must equal 1 (since wewant A2x+A2y+A2z = 1). Likewise, 3i=1Ri2Ri2 = 3i=1Ri3Ri3 = 1. To check the case j ,= k, choose A = (1, 1, 0).Then we want 2 = j,k (iRijRik) AjAk = iRi1Ri1 + iRi2Ri2 + iRi1Ri2 + iRi2Ri1. But we alreadyknow that the rst two sums are both 1; the third and fourth are equal, so iRi1Ri2 = iRi2Ri1 = 0, and soon for other unequal combinations of j, k.In matrix notation: RR = 1, where R is the transpose of R.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.6 CHAPTER 1. VECTOR ANALYSISProblem 1.9CHAPTER 1. VECTOR ANALYSIS 5ExTyzbXLooking down the axis:TysxCzTz

sy

Cx

cs

c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Looking down the axis:CHAPTER 1. VECTOR ANALYSIS 5ExTyzbXLooking down the axis:TysxCzTz

sy

Cx

cs

c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.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.R =__0 0 11 0 00 1 0__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) (AB) (A)(B) = (AB). That is, if C = AB, C C . No minus sign, in contrast tobehavior of an ordinary vector, as given by (b). If A and B are pseudovectors, then (AB) (A)(B) =(AB). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vectorand a pseudovector, one changes sign, the other doesnt, and therefore the cross-product is itself a vector.Angular momentum (L = rp) and torque (N = rF) are pseudovectors.(d) A(BC) (A)((B)(C)) = A(BC). So, if a = A(BC), then a a; a pseudoscalarchanges sign under inversion of coordinates.Problem 1.11(a)f = 2x x + 3y2 y + 4z3z(b)f = 2xy3z4 x + 3x2y2z4 y + 4x2y3z3z(c)f = exsin y ln z x +excos y ln z y +exsin y(1/z) zProblem 1.12(a) h = 10[(2y 6x 18) x + (2x 8y + 28) y]. h = 0 at summit, so2y 6x 18 = 02x 8y + 28 = 0 =6x 24y + 84 = 0_2y 18 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: h = 10[(2 6 18) x + (2 8 + 28) y] = 10(22 x + 22 y) = 220( x + y).[h[ = 2202 311 ft/mile ; direction: northwest.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 7Problem 1.13r = (x x

) x + (y y

) y + (z z

) z; r = _(x x

)2+ (y y

)2+ (z z

)2.(a) ( r 2) = x[(xx

)2+(yy

)2+(zz

)2] x+ y() y+ z() z = 2(xx

) x+2(yy

) y+2(zz

) z = 2 r .(b) ( 1r ) = x[(x x

)2+ (y y

)2+ (z z

)2]12 x + y()12 y + z()12 z= 12()322(x x

) x 12()322(y y

) y 12()322(z z

) z= ()32[(x x

) x + (y y

) y + (z z

) z] = (1/ r 3) r = (1/ r 2) r .(c) x( r n) = n r n1 rx = n r n1(121r 2 r x) = n r n1r x, so ( r n) = n r n1rProblem 1.14y = +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.So yy = cos ; yz = sin ; zy = sin ; zz = cos . Therefore(f)y = fy = fyyy + fzzy = +cos (f)y + sin (f)z(f)z = fz = fyyz + fzzz = sin (f)y + cos (f)z_ So f transforms as a vector. qedProblem 1.15(a)va = x(x2) + y(3xz2) + z(2xz) = 2x + 0 2x = 0.(b)vb = x(xy) + y(2yz) + z(3xz) = y + 2z + 3x.(c)vc = x(y2) + y(2xy +z2) + z(2yz) = 0 + (2x) + (2y) = 2(x +y)Problem 1.16v = x( xr3) + y( yr3) + z( zr3) = x_x(x2+y2+z2)32_+ y_y(x2+y2+z2)32_+ z_z(x2+y2+z2)32_= ()32 +x(3/2)()522x + ()32 +y(3/2)()522y + ()32+ z(3/2)()522z = 3r33r5(x2+y2+z2) = 3r33r3= 0.This conclusion is surprising, because, from the diagram, this vector eld is obviously diverging away from theorigin. How, then, can v = 0? The answer is that v = 0 everywhere except at the origin, but at theorigin our calculation is no good, since r = 0, and the expression for v blows up. In fact, v is innite atthat one point, and zero elsewhere, as we shall see in Sect. 1.5.Problem 1.17vy = cos vy + sin vz; vz = sin vy + cos vz.vyy = vyy cos + vzy sin =_vyyyy + vyzzy_ cos +_vzyyy + vzzzy_ sin . Use result in Prob. 1.14:=_vyy cos + vyz sin _ cos +_vzy cos + vzz sin _ sin .vzz = vyz sin + vzz cos = _vyyyz + vyzzz_ sin +_vzyyz + vzzzz_ cos = _vyy sin + vyz cos _ sin +_vzy sin + vzz cos _ cos . Soc 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.8 CHAPTER 1. VECTOR ANALYSISvyy + vzz = vyy cos2 + vyz sin cos + vzy sin cos + vzz sin2 + vyy sin2 vyz sin cos vzy sin cos + vzz cos2= vyy_cos2 + sin2_+ vzz_sin2 + cos2_ = vyy + vzz . Problem 1.18(a) va = x y zxyzx23xz22xz= x(0 6xz) + y(0 + 2z) +z(3z20) = 6xz x + 2z y + 3z2z.(b) vb = x y zxyzxy 2yz 3xz= x(0 2y) + y(0 3z) +z(0 x) = 2y x 3z y xz.(c) vc = x y zxyzy2(2xy +z2) 2yz= x(2z 2z) + y(0 0) +z(2y 2y) = 0.Problem 1.19v = y x +x y; or v = yz x +xz y +xy z; or v = (3x2z z3) x + 3 y + (x33xz2) z;or v = (sin x)(cosh y) x (cos x)(sinh y) y; etc.Problem 1.20(i) (fg) = (fg)x x + (fg)y y + (fg)z z =_f gx +gfx_ x +_f gy +gfy_ y +_fgz +gfz_ z= f_gx x + gy y + gz z_+g_fx x + fy y + fz z_ = f(g) +g(f). qed(iv) (AB) = x (AyBzAzBy) + y (AzBxAxBz) + z (AxByAyBx)= AyBzx +BzAyx AzByx ByAzx +AzBxy +BxAzy AxBzy BzAxy+AxByz +ByAxz AyBxz BxAyz= Bx_Azy Ayz_+By_Axz Azx_+Bz_Ayx Axy_Ax_Bzy Byz_Ay_Bxz Bzx_Az_Byx Bxy_ = B (A) A (B). qed(v) (fA) =_(fAz)y (fAy)z_ x +_(fAx)z (fAz)x_ y +_(fAy)x (fAx)y_z=_fAzy +Azfy fAyz Ayfz_ x +_fAxz +Axfz fAzx Azfx_ y+_fAyx +Ayfx fAxy Axfy_z= f__Azy Ayz_ x +_Axz Azx_ y +_Ayx Axy_z___Ayfz Azfy_ x +_Azfx Axfz_ y +_Axfy Ayfx_z_= f (A) A(f). qedProblem 1.21(a) (A) B =_AxBxx +AyBxy +AzBxz_ x +_AxByx +AyByy +AzByz_ y+_AxBzx +AyBzy +AzBzz_z.(b) r = rr = x x+y y+z zx2+y2+z2. Lets just do the x component.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 9[(r)r]x = 1_x x +y y +z z_ xx2+y2+z2= 1r_x_ 1 +x(12) 1( )32x_+yx_121( )32y_+zx_121( )32z__= 1r_xr 1r3_x3+xy2+xz2__ = 1r_xr xr3_x2+y2+z2__ = 1r_xr xr_ = 0.Same goes for the other components. Hence: (r) r = 0 .(c) (va) vb =_x2 x + 3xz2 y 2xz z_(xy x + 2yz y + 3xz z)= x2(y x + 0 y + 3z z) + 3xz2(x x + 2z y + 0z) 2xz (0 x + 2y y + 3xz)= _x2y + 3x2z2_ x +_6xz34xyz_ y +_3x2z 6x2z_z= x2_y + 3z2_ x + 2xz_3z22y_ y 3x2z zProblem 1.22(ii) [(AB)]x = x(AxBx +AyBy +AzBz) = Axx Bx +AxBxx + Ayx By +AyByx + Azx Bz +AzBzx[A(B)]x = Ay(B)zAz(B)y = Ay_Byx Bxy_Az_Bxz Bzx_[B(A)]x = By_Ayx Axy_Bz_Axz Azx_[(A)B]x = _Axx +Ayy +Azz_Bx = AxBxx +AyBxy +AzBxz[(B)A]x = BxAxx +ByAxy +BzAxzSo [A(B) +B(A) + (A)B+ (B)A]x= AyByx AyBxy AzBxz +AzBzx +ByAyx ByAxy BzAxz +BzAzx+AxBxx +AyBxy +AzBxz +BxAxx +ByAxy +BzAxz= BxAxx +AxBxx +By_Ayx Axy/ +Axy/ _+Ay_Byx Bxy/ +Bxy/ _+Bz_Axz/ +Azx + Axz/ _+Az_Bxz/ +Bzx + Bxz/ _= [(AB)]x (same for y and z)(vi) [(AB)]x = y(AB)z z(AB)y = y(AxByAyBx) z(AzBxAxBz)= Axy By +AxByy Ayy BxAyBxy Azz BxAzBxz + Axz Bz +AxBzz[(B)A(A)B+A(B) B(A)]x= BxAxx +ByAxy +BzAxz AxBxx AyBxy AzBxz +Ax_Bxx + Byy + Bzz_Bx_Axx + Ayy + Azz_= ByAxy +Ax_Bxx/ +Bxx/ +Byy + Bzz_+Bx_Axx/ Axx/ Ayy Azz_+Ay_Bxy_+Az_Bxz_+Bz_Axz_= [(AB)]x (same for y and z)Problem 1.23(f/g) = x(f/g) x + y(f/g) y + z(f/g) z= gfxf gxg2 x + gfyf gyg2 y + gfzf gzg2 z= 1g2_g_fx x + fy y + fzz_f_gx x + gy y + gzz__ = gffgg2 . qed(A/g) = x(Ax/g) + y(Ay/g) + z(Az/g)= gAxx Axgxg2 + gAyy Aygyg2 + gAzz Azgxg2= 1g2_g_Axx + Ayy + Azz__Axgx +Aygy +Azgz__ = gAAgg2 . qedc 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.10 CHAPTER 1. VECTOR ANALYSIS[(A/g)]x = y(Az/g) z(Ay/g)= gAzy Azgyg2 gAyz Aygzg2= 1g2_g_Azy Ayz__Azgy Aygz__= g(A)x+(Ag)xg2 (same for y and z). qedProblem 1.24(a) AB = x y zx 2y 3z3y 2x 0= x(6xz) + y(9zy) +z(2x26y2)(AB) = x(6xz) + y(9zy) + z(2x26y2) = 6z + 9z + 0 = 15zA = x_ y(3z) z(2y)_+ y_ z(x) x(3z)_+z_ x(2y) y(x)_ = 0; B(A) = 0B = x_ y(0) z(2x)_+ y_ z(3y) x(0)_+z_ x(2x) y(3y)_ = 5z; A(B) = 15z(AB) ?= B(A) A(B) = 0 (15z) = 15z. (b) AB = 3xy 4xy = xy ; (AB) = (xy) = x x(xy) + y y(xy) = y x x yA(B) = x y zx 2y 3z0 0 5= x(10y) + y(5x); B(A) = 0(A)B =_x x + 2y y + 3z z_(3y x 2x y) = x(6y) + y(2x)(B)A =_3y x 2x y_(x x + 2y y + 3z z) = x(3y) + y(4x)A(B) +B(A) + (A)B+ (B)A= 10y x + 5x y + 6y x 2x y + 3y x 4x y = y x x y = (AB). (c) (AB) = x_ y(2x26y2) z(9zy)_+ y_ z(6xz) x(2x26y2)_+z_ x(9zy) y(6xz)_= x(12y 9y) + y(6x + 4x) +z(0) = 21y x + 10x yA = x(x) + y(2y) + z(3z) = 1 + 2 + 3 = 6; B = x(3y) + y(2x) = 0(B)A(A)B+A(B) B(A) = 3y x 4x y 6y x + 2x y 18y x + 12x y = 21y x + 10x y= (AB). Problem 1.25(a) 2Tax2 = 2; 2Tay2 = 2Taz2 = 0 2Ta = 2.(b) 2Tbx2 = 2Tby2 = 2Tbz2 = Tb 2Tb = 3Tb = 3 sin xsin y sin z.(c) 2Tcx2 = 25Tc ; 2Tcy2 = 16Tc ; 2Tcz2 = 9Tc 2Tc = 0.(d) 2vxx2 = 2 ; 2vxy2 = 2vxz2 = 0 2vx = 22vyx2 = 2vyy2 = 0 ; 2vyz2 = 6x 2vy = 6x2vzx2 = 2vzy2 = 2vzz2 = 0 2vz = 0___2v = 2 x + 6x y.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 11Problem 1.26(v) = x_vzy vyz_+ y_vxz vzx_+ z_vyx vxy_=_2vzx y 2vzy x_+_2vxy z 2vxz y_+_2vyz x 2vyx z_ = 0, by equality of cross-derivatives.From Prob. 1.18: va = 6xz x + 2z y + 3z2z (va) = x(6xz) + y(2z) + z(3z2) =6z + 6z = 0. Problem 1.27(t) = x y zxyztxtytz= x_ 2ty z 2tz y_+ y_ 2tz x 2tx z_+z_ 2tx y 2ty x_= 0, by equality of cross-derivatives.In Prob. 1.11(b), f = 2xy3z4 x + 3x2y2z4 y + 4x2y3z3z, so(f) = x y zxyz2xy3z43x2y2z44x2y3z3= x(3 4x2y2z34 3x2y2z3) + y(4 2xy3z32 4xy3z3) +z(2 3xy2z43 2xy2z4) = 0. Problem 1.28(a) (0, 0, 0) (1, 0, 0). x : 0 1, y = z = 0; dl = dx x; v dl = x2dx;_ v dl = _10 x2dx = (x3/3)[10 = 1/3.(1, 0, 0) (1, 1, 0). x = 1, y : 0 1, z = 0; dl = dy y; v dl = 2yz dy = 0;_ v dl = 0.(1, 1, 0) (1, 1, 1). x = y = 1, z : 0 1; dl = dz z; v dl = y2dz = dz;_ v dl = _10 dz = z[10 = 1.Total: _ v dl = (1/3) + 0 + 1 = 4/3.(b) (0, 0, 0) (0, 0, 1). x = y = 0, z : 0 1; dl = dz z; v dl = y2dz = 0;_ v dl = 0.(0, 0, 1) (0, 1, 1). x = 0, y : 0 1, z = 1; dl = dy y; v dl = 2yz dy = 2y dy;_ v dl = _10 2y dy = y2[10 = 1.(0, 1, 1) (1, 1, 1). x : 0 1, y = z = 1; dl = dx x; v dl = x2dx;_ v dl = _10 x2dx = (x3/3)[10 = 1/3.Total: _ v dl = 0 + 1 + (1/3) = 4/3.(c) x = y = z : 0 1; dx = dy = dz; v dl = x2dx + 2yz dy +y2dz = x2dx + 2x2dx +x2dx = 4x2dx;_ v dl = _10 4x2dx = (4x3/3)[10 = 4/3.(d) _ v dl = (4/3) (4/3) = 0.Problem 1.29x, y : 0 1, z = 0; da = dxdy z; v da = y(z2 3) dxdy = 3y dxdy;_ v da = 3_20 dx_20 y dy =3(x[20)(y22 [20) = 3(2)(2) = -12. In Ex. 1.7 we got 20, for the same boundary line (the square in thexy-plane), so the answer is no: the surface integral does not depend only on the boundary line. The total uxfor the cube is 20 + 12 = 32.Problem 1.30_ T d = _ z2dxdy dz. You can do the integrals in any orderhere it is simplest to save z for last:_ z2__ __ dx_dy_dz.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.12 CHAPTER 1. VECTOR ANALYSISThe sloping surface is x+y+z = 1, so the x integral is _(1yz)0 dx = 1yz. For a given z, y ranges from 0 to1z, so the y integral is _(1z)0 (1y z) dy = [(1z)y (y2/2)][(1z)0 = (1z)2[(1z)2/2] = (1z)2/2 =(1/2) z + (z2/2). Finally, the z integral is _10 z2(12 z + z22 ) dz = _10 (z22 z3+ z42 ) dz = (z36 z44 + z510)[10 =16 14 + 110 = 1/60.Problem 1.31T(b) = 1 + 4 + 2 = 7; T(a) = 0. T(b) T(a) = 7.T = (2x + 4y) x + (4x + 2z3) y + (6yz2)z; Tdl = (2x + 4y)dx + (4x + 2z3)dy + (6yz2)dz(a) Segment 1: x : 0 1, y = z = dy = dz = 0._Tdl = _10 (2x) dx = x210 = 1.Segment 2: y : 0 1, x = 1, z = 0, dx = dz = 0._Tdl = _10 (4) dy = 4y[10 = 4.Segment 3: z : 0 1, x = y = 1, dx = dy = 0._Tdl = _10 (6z2) dz = 2z310 = 2.____ba Tdl = 7. (b) Segment 1: z : 0 1, x = y = dx = dy = 0._Tdl = _10 (0) dz = 0.Segment 2: y : 0 1, x = 0, z = 1, dx = dz = 0._Tdl = _10 (2) dy = 2y[10 = 2.Segment 3: x : 0 1, y = z = 1, dy = dz = 0._Tdl = _10 (2x + 4) dx= (x2+ 4x)10 = 1 + 4 = 5.____ba Tdl = 7. (c) x : 0 1, y = x, z = x2, dy = dx, dz = 2xdx.Tdl = (2x + 4x)dx + (4x + 2x6)dx + (6xx4)2xdx = (10x + 14x6)dx._ba Tdl = _10 (10x + 14x6)dx = (5x2+ 2x7)10 = 5 + 2 = 7. Problem 1.32v = y + 2z + 3x_(v)d = _(y + 2z + 3x) dxdy dz = ____20 (y + 2z + 3x) dx_dy dz_(y + 2z)x + 32x220 = 2(y + 2z) + 6= ___20 (2y + 4z + 6)dy_dz_y2+ (4z + 6)y20 = 4 + 2(4z + 6) = 8z + 16= _20 (8z + 16)dz = (4z2+ 16z)20 = 16 + 32 = 48.Numbering the surfaces as in Fig. 1.29:(i) da = dy dz x, x = 2. vda = 2y dy dz._vda = __2y dy dz = 2y220 = 8.(ii) da = dy dz x, x = 0. vda = 0._vda = 0.(iii) da = dxdz y, y = 2. vda = 4z dxdz._vda = __4z dxdz = 16.(iv) da = dxdz y, y = 0. vda = 0._vda = 0.(v) da = dxdy z, z = 2. vda = 6xdxdy._vda = 24.(vi) da = dxdy z, z = 0. vda = 0._vda = 0._vda = 8 + 16 + 24 = 48 Problem 1.33v = x(0 2y) + y(0 3z) +z(0 x) = 2y x 3z y xz.da = dy dz x, if we agree that the path integral shall run counterclockwise. Soc 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 13(v)da = 2y dy dz._(v)da = ___2z0 (2y)dy_dzy22z0 = (2 z)2= _20 (4 4z +z2)dz = _4z 2z2+ z33_20= _8 8 + 83_ = 83ETzyddddddy=2zMeanwhile, vdl = (xy)dx + (2yz)dy + (3zx)dz. There are three segments.ETzyddddddE(1)dds(2)c(3)(1) x = z = 0; dx = dz = 0. y : 0 2._vdl = 0.(2) x = 0; z = 2 y; dx = 0, dz = dy, y : 2 0. vdl = 2yz dy._vdl = _02 2y(2 y)dy = _20 (4y 2y2)dy = _2y2 23y3_20 = _8 23 8_ = 83.(3) x = y = 0; dx = dy = 0; z : 2 0. vdl = 0._vdl = 0. So _ vdl = 83. Problem 1.34By Corollary 1, _(v)da should equal 43. v = (4z22x) x + 2z z.(i) da = dy dz x, x = 1; y, z : 0 1. (v)da = (4z22)dy dz; _(v)da = _10 (4z22)dz= (43z32z)10 = 43 2 = 23.(ii) da = dxdy z, z = 0; x, y : 0 1. (v)da = 0; _(v)da = 0.(iii) da = dxdz y, y = 1; x, z : 0 1. (v)da = 0; _(v)da = 0.(iv) da = dxdz y, y = 0; x, z : 0 1. (v)da = 0; _(v)da = 0.(v) da = dxdy z, z = 1; x, y : 0 1. (v)da = 2 dxdy; _(v)da = 2._(v)da = 23 + 2 = 43. Problem 1.35(a) Use the product rule (fA) = f(A) A(f) :_Sf(A) da =_S(fA) da +_S[A(f)] da =_PfA dl +_S[A(f)] da. qed(I used Stokes theorem in the last step.)(b) Use the product rule (AB) = B (A) A (B) :_VB (A)d =_V(AB) d +_VA (B) d =_S(AB) da +_VA (B) d. qed(I used the divergence theorem in the last step.)c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.14 CHAPTER 1. VECTOR ANALYSISProblem 1.36 r = _x2+y2+z2; = cos1_ zx2+y2+z2_; = tan1_yx_.Problem 1.37There are many ways to do this oneprobably the most illuminating way is to work it out by trigonometryfrom Fig. 1.36. The most systematic approach is to study the expression:r = x x +y y +z z = r sin cos x +r sin sin y +r cos z.If I only vary r slightly, then dr = r(r)dr is a short vector pointing in the direction of increase in r. To makeit a unit vector, I must divide by its length. Thus:r =rrrr; =rr; =rr.rr = sin cos x + sin sin y + cos z; rr2= sin2 cos2 + sin2 sin2 + cos2 = 1.r = r cos cos x +r cos sin y r sin z; r2= r2cos2 cos2 +r2cos2 sin2 +r2sin2 = r2.r = r sin sin x +r sin cos y; r2= r2sin2 sin2 +r2sin2 cos2 = r2sin2.r = sin cos x + sin sin y + cos z. = cos cos x+cos sin y sin z. = sin x + cos y.Check: rr = sin2(cos2 + sin2) + cos2 = sin2 + cos2 = 1, = cos sin cos + cos sin cos = 0, etc.sin r = sin2 cos x + sin2 sin y + sin cos z.cos = cos2 cos x + cos2 sin y sin cos z.Add these:(1) sin r + cos = +cos x + sin y;(2) = sin x + cos y.Multiply (1) by cos , (2) by sin , and subtract: x = sin cos r + cos cos sin .Multiply (1) by sin , (2) by cos , and add: y = sin sin r + cos sin + cos .cos r = sin cos cos x + sin cos sin y + cos2 z.sin = sin cos cos x + sin cos sin y sin2 z.Subtract these:z = cos r sin .c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 15Problem 1.38(a) v1 = 1r2r(r2r2) = 1r24r3= 4r_(v1)d = _(4r)(r2sin dr d d) = (4)_R0 r3dr_0 sin d_20 d = (4)_R44_(2)(2) = 4R4_v1da = _(r2r)(r2sin d dr) = r4_0 sin d_20 d = 4R4 (Note: at surface of sphere r = R.)(b) v2 = 1r2r_r2 1r2_ = 0 _(v2)d = 0_v2da = _ _ 1r2r_(r2sin d dr) = _sin d d = 4.They dont agree! The point is that this divergence is zero except at the origin, where it blows up, so ourcalculation of _(v2) is incorrect. The right answer is 4.Problem 1.39v = 1r2r(r2r cos ) + 1r sin (sin r sin ) + 1r sin (r sin cos )= 1r2 3r2cos + 1r sin r 2 sin cos + 1r sin r sin (sin )= 3 cos + 2 cos sin = 5 cos sin _(v)d = _(5 cos sin ) r2sin dr d d = _R0 r2dr_ 20__20 (5 cos sin ) d_ d sin 2(5 cos )=_R33_(10)_ 20 sin cos dsin2220= 12= 53 R3.Two surfacesone the hemisphere: da = R2sin d dr; r = R; : 0 2, : 0 2._vda = _(r cos )R2sin d d = R3_ 20 sin cos d_20 d = R3_12_(2) = R3.other the at bottom: da = (dr)(r sin d)(+) = r dr d (here = 2). r : 0 R, : 0 2._vda = _(r sin )(r dr d) = _R0 r2dr_20 d = 2R33 .Total: _vda = R3+ 23R3= 53R3. Problem 1.40 t = (cos + sin cos )r + (sin + cos cos ) + 1sin / (sin / sin )2t = (t)= 1r2r_r2(cos + sin cos )_+ 1r sin (sin (sin + cos cos )) + 1r sin (sin )= 1r2 2r(cos + sin cos ) + 1r sin (2 sin cos + cos2 cos sin2 cos ) 1r sin cos = 1r sin [2 sin cos / + 2 sin2 cos 2 sin cos / + cos2 cos sin2 cos cos ]= 1r sin _(sin2 + cos2) cos cos = 0. 2t = 0Check: r cos = z, r sin cos = x in Cartesian coordinates t = x +z. Obviously Laplacian is zero.Gradient Theorem: _ba tdl = t(b) t(a)Segment 1: = 2, = 0, r : 0 2. dl = dr r; tdl = (cos + sin cos )dr = (0 + 1)dr = dr._tdl = _20 dr = 2.Segment 2: = 2, r = 2, : 0 2. dl = r sin d = 2 d .tdl = (sin )(2 d) = 2 sin d._tdl = _ 20 2 sin d = 2 cos [20 = 2.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.16 CHAPTER 1. VECTOR ANALYSISSegment 3: r = 2, = 2; : 2 0.dl = r d = 2 d ; tdl = (sin + cos cos )(2 d) = 2 sin d._tdl = _022 sin d = 2 cos [02= 2.Total: _ba tdl = 2 2 + 2 = 2 . Meanwhile, t(b) t(a) = [2(1 + 0)] [0( )] = 2. Problem 1.41 From Fig. 1.42, s = cos x + sin y; = sin x + cos y; z = zMultiply rst by cos , second by sin , and subtract:s cos sin = cos2 x + cos sin y + sin2 x sin cos y = x(sin2 + cos2) = x.So x = cos s sin .Multiply rst by sin , second by cos , and add:s sin + cos = sin cos x + sin2 y sin cos x + cos2 y = y(sin2 + cos2) = y.So y = sin s + cos . z = z.Problem 1.42(a) v = 1ss_s s(2 + sin2)_+ 1s(s sin cos ) + z(3z)= 1s 2s(2 + sin2) + 1s s(cos2 sin2) + 3= 4 + 2 sin2 + cos2 sin2 + 3= 4 + sin2 + cos2 + 3 = 8.(b) _(v)d = _(8)s ds ddz = 8_20 s ds_ 20 d_50 dz = 8(2)_2_(5) = 40.Meanwhile, the surface integral has ve parts:top: z = 5, da = s ds dz; vda = 3z s ds d = 15s ds d._vda = 15_20 s ds_ 20 d = 15.bottom: z = 0, da = s ds dz; vda = 3z s ds d = 0._vda = 0.back: = 2, da = ds dz ; vda = s sin cos ds dz = 0._vda = 0.left: = 0, da = ds dz ; vda = s sin cos ds dz = 0._vda = 0.front: s = 2, da = s ddz s; vda = s(2 + sin2)s ddz = 4(2 + sin2)ddz._vda = 4_ 20 (2 + sin2)d_50 dz = (4)( + 4)(5) = 25.So _vda = 15 + 25 = 40. (c) v =_1s(3z) z(s sin cos )_s +_ z_s(2 + sin2)_ s(3z)_ +1s_ s(s2sin cos ) _s(2 + sin2)__z= 1s(2s sin cos s 2 sin cos ) z = 0.Problem 1.43(a) 3(32) 2(3) 1 = 27 6 1 = 20.(b) cos = -1.(c) zero.(d) ln(2 + 3) = ln 1 = zero.Problem 1.44(a) _22(2x + 3)13(x) dx = 13(0 + 3) = 1.(b) By Eq. 1.94, (1 x) = (x 1), so 1 + 3 + 2 = 6.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 17(c) _11 9x2 13 (x + 13) dx = 9_13_213 = 13.(d) 1 (if a > b), 0 (if a < b).Problem 1.45(a) _f(x)_x ddx(x)dx = xf(x)(x)[_ddx (xf(x)) (x) dx.The rst term is zero, since (x) = 0 at ; ddx (xf(x)) = xdfdx + dxdxf = xdfdx +f.So the integral is __xdfdx +f_(x) dx = 0 f(0) = f(0) = _f(x)(x) dx.So, x ddx(x) = (x). qed(b) _f(x)ddxdx = f(x)(x)[_dfdx(x)dx = f() _0dfdxdx = f() (f() f(0))= f(0) = _f(x)(x) dx. So ddx = (x). qedProblem 1.46(a) (r) = q3(r r

). Check: _(r)d = q_3(r r

) d = q. (b) (r) = q3(r a) q3(r).(c) Evidently (r) = A(r R). To determine the constant A, we requireQ = _ d = _A(r R)4r2dr = A4R2. So A = Q4R2. (r) = Q4R2(r R).Problem 1.47(a) a2+aa +a2= 3a2.(b) _(r b)2 1533(r) d = 1125b2= 1125(42+ 32) = 15.(c) c2= 25 + 9 + 4 = 38 > 36 = 62, so c is outside 1, so the integral is zero.(d) (e (2 x + 2 y + 2z))2= (1 x + 0 y + (1) z)2= 1 + 1 = 2 < (1.5)2= 2.25, so e is inside 1,and hence the integral is e(d e) = (3, 2, 1)(2, 0, 2) = 6 + 0 + 2 = -4.Problem 1.48First method: use Eq. 1.99 to write J = _ er_43(r)_ d = 4e0= 4.Second method: integrating by parts (use Eq. 1.59).J = _Vrr2 (er) d +_Ser rr2 da. But _er_ =_ rer_r = err.=_ 1r2er4r2dr +_ er rr2 r2sin d dr = 4R_0erdr +eR_ sin d d= 4_er_R0 + 4eR= 4_eR+e0_+ 4eR= 4. _Here R = , so eR= 0._Problem 1.49 (a) F1 = x(0) + y(0) + z_x2_ = 0 ; F2 = xx + yy + zz = 1 + 1 + 1 = 3F1 = x y zxyz0 0 x2= y x_x2_ = 2x y ; F2 = x y zxyzx y z= 0c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.18 CHAPTER 1. VECTOR ANALYSISF2 is a gradient; F1 is a curl U2 = 12_x3+y2+z2_ would do (F2 = U2).For A1, we want_Ayz Azy_ = _Axz Azx_ = 0; Ayx Axy = x2. Ay = x33 , Ax = Az = 0 would do it.A1 = 13x2 y (F1 = A1) . (But these are not unique.)(b) F3 = x(yz) + y(xz) + z(xy) = 0; F3 = x y zxyzyz xz xy= x(x x) + y (y y) +z (z z) = 0.So F3 can be written as the gradient of a scalar (F3 = U3) and as the curl of a vector (F3 = A3). Infact, U3 = xyz does the job. For the vector potential, we have___Azy Ayz = yz, which suggests Az = 14y2z +f(x, z); Ay = 14yz2+g(x, y)Axz Azx = xz, suggesting Ax = 14z2x +h(x, y); Az = 14zx2+j(y, z)Ayx Axy = xy, so Ay = 14x2y +k(y, z); Ax = 14xy2+l(x, z)___Putting this all together: A3 = 14_x_z2y2_ x +y_x2z2_ y +z_y2x2_z_ (again, not unique).Problem 1.50(d) (a): F = (U) = 0 (Eq. 1.44 curl of gradient is always zero).(a) (c): _ F dl = _(F) da = 0 (Eq. 1.57Stokes theorem).(c) (b): _ba IF dl _ba IIF dl = _ba IF dl +_ab IIF dl = _ F dl = 0, so_ ba IF dl =_ ba IIF dl.(b) (c): same as (c) (b), only in reverse; (c) (a): same as (a) (c).Problem 1.51(d) (a): F = (W) = 0 (Eq 1.46divergence of curl is always zero).(a) (c): _ F da = _(F) d = 0 (Eq. 1.56divergence theorem).(c) (b): _I F da _II F da = _ F da = 0, so_IF da =_IIF da.(Note: sign change because for _ F da, da is outward, whereas for surface II it is inward.)(b) (c): same as (c) (b), in reverse; (c) (a): same as (a) (c) .Problem 1.52In Prob. 1.15 we found that va = 0; in Prob. 1.18 we found that vc = 0. Sovc can be written as the gradient of a scalar; va can be written as the curl of a vector.(a) To nd t:(1) tx = y2t = y2x +f(y, z)(2) ty = _2xy +z2_(3) tz = 2yzc 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 19From (1) & (3) we get fz = 2yz f = yz2+ g(y) t = y2x + yz2+ g(y), so ty = 2xy + z2+ gy =2xy +z2(from (2)) gy = 0. We may as well pick g = 0; then t = xy2+yz2.(b) To nd W: Wzy Wyz = x2; Wxz Wzx = 3z2x; Wyx Wxy = 2xz.Pick Wx = 0; thenWzx = 3xz2Wz = 32x2z2+f(y, z)Wyx = 2xz Wy = x2z +g(y, z).Wzy Wyz = fy +x2 gz = x2 fy gz = 0. May as well pick f = g = 0.W = x2z y 32x2z2z.Check: W = x y zxyz0 x2z 32x2z2= x _x2_+ y _3xz2_+z (2xz).You can add any gradient (t) to W without changing its curl, so this answer is far from unique. Someother solutions:W = xz3 x x2z y;W = _2xyz +xz3_ x +x2y z;W = xyz x 12x2z y + 12x2_y 3z2_ z.Probelm 1.53v = 1r2r_r2r2cos _+ 1r sin _sin r2cos _+ 1r sin _r2cos sin _= 1r24r3cos + 1r sin cos r2cos + 1r sin _r2cos cos _= r cos sin [4 sin + cos cos ] = 4r cos ._ (v) d =_ (4r cos )r2sin dr d d = 4R_0r3dr/2_0cos sin d/2_0d= _R4__12__2_ = R44 .Surface consists of four parts:(1) Curved: da = R2sin d dr; r = R. v da = _R2cos _ _R2sin d d_._ v da = R4/2_0cos sin d/2_0d = R4_12__2_ = R44 .c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.20 CHAPTER 1. VECTOR ANALYSIS(2) Left: da = r dr d ; = 0. v da = _r2cos sin _(r dr d) = 0._ v da = 0.(3) Back: da = r dr d ; = /2. v da = _r2cos sin _(r dr d) = r3cos dr d._ v da =R_0r3dr/2_0cos d = _14R4_(+1) = 14R4.(4) Bottom: da = r sin dr d ; = /2. v da = _r2cos _(r dr d) ._ v da =R_0r3dr/2_0cos d = 14R4.Total: _ v da = R4/4 + 0 14R4+ 14R4= R44 . Problem 1.54v = x y zxyzay bx 0= z (b a). So _(v) da = (b a)R2.v dl = (ay x +bx y) (dx x +dy y +dz z) = ay dx +bxdy; x2+y2= R22xdx + 2y dy = 0,so dy = (x/y) dx. So v dl = ay dx +bx(x/y) dx = 1y_ay2bx2_ dx.For the upper semicircle, y =R2x2, so v dl = a(R2x2)bx2R2x2 dx._ v dl =R_RaR2(a +b)x2R2x2dx =_aR2sin1_xR_(a +b)_x2_R2x2+ R22 sin1_xR___R+R= 12R2(a b) sin1(x/R)R+R= 12R2(a b)_sin1(1) sin1(+1)_ = 12R2(a b)_2 2_= 12R2(b a).And the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so_ v dl = R2(b a). Problem 1.55(1) x = z = 0; dx = dz = 0; y : 0 1. v dl = (yz2) dy = 0; _ v dl = 0.(2) x = 0; z = 22y; dz = 2 dy; y : 1 0. v dl = (yz2) dy +(3y +z) dz = y(22y)2dy (3y +22y)2 dy;_ v dl = 20_1(2y34y2+y 2) dy = 2_y42 4y33 + y22 2y_01= 143 .(3) x = y = 0; dx = dy = 0; z : 2 0. v dl = (3y +z) dz = z dz;_ v dl =0_2z dz = z2202= 2.Total: _ v dl = 0 + 143 2 = 83.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 21Meanwhile, Stokes thereom says _ v dl = _(v) da. Here da = dy dz x, so all we need is(v)x = y(3y +z) z(yz2) = 3 2yz. Therefore_ (v) da =_ _ (3 2yz) dy dz =_ 10__ 22y0(3 2yz) dz_ dy=_ 10_3(2 2y) 2y12(2 2y)2 dy =_ 10(4y3+ 8y210y + 6) dy= _y4+ 83y35y2+ 6y_10= 1 + 83 5 + 6 = 83. Problem 1.56Start at the origin.(1) = 2, = 0; r : 0 1. v dl = _r cos2_(dr) = 0._ v dl = 0.(2) r = 1, = 2; : 0 /2. v dl = (3r)(r sin d) = 3 d._ v dl = 3/2_0d = 32 .(3) = 2; r sin = y = 1, so r = 1sin , dr = 1sin2 cos d, : 2 0 tan1(1/2).v dl = _r cos2_(dr) (r cos sin )(r d) = cos2sin _cos sin2_ d cos sin sin2d= _cos3sin3+ cos sin _ d = cos sin _cos2 + sin2sin2_ d = cos sin3d.Therefore_ v dl = 0_/2cos sin3d = 12 sin20/2= 12 (1/5) 12 (1) = 52 12 = 2.(4) = 0, = 2; r :5 0. v dl = _r cos2_(dr) = 45r dr._ v dl = 450_5r dr = 45r2205= 45 52 = 2.Total:_ v dl = 0 + 32 + 2 2 = 32 .Stokes theorem says this should equal _(v) dav = 1r sin _ (sin 3r) (r sin cos )_ r + 1r_ 1sin _r cos2_ r(r3r)_ + 1r_ r(rr cos sin ) _r cos2__ = 1r sin [3r cos ] r + 1r[6r] + 1r[2r cos sin + 2r cos sin ] = 3 cot r 6 .c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.22 CHAPTER 1. VECTOR ANALYSIS(1) Back face: da = r dr d ; (v) da = 0._(v) da = 0.(2) Bottom: da = r sin dr d ; (v) da = 6r sin dr d. = 2, so (v) da = 6r dr d_ (v) da =1_06r dr/2_0d = 6 12 2 = 32 . Problem 1.57v dl = y dz.(1) Left side: z = a x; dz = dx; y = 0. Therefore _ v dl = 0.(2) Bottom: dz = 0. Therefore _ v dl = 0.(3) Back: z = a 12y; dz = 1/2 dy; y : 2a 0._ v dl =0_2ay_12 dy_ = 12y2202a= 4a24 = a2.Meanwhile, v = x, so _(v) da is the projection of this surface on the xy plane = 12 a 2a = a2. Problem 1.58v = 1r2r_r2r2sin _+ 1r sin _sin 4r2cos _+ 1r sin _r2tan _= 1r24r3sin + 1r sin 4r2_cos2 sin2_ = 4rsin _sin2 + cos2 sin2_= 4rcos2sin ._ (v) d =_ _4rcos2sin __r2sin dr d d_ =R_04r3dr/6_0cos2 d2_0d = _R4_(2)_2 + sin 24_/60= 2R4_ 12 + sin 604_ = R46_ + 332_ = R412_2 + 33_.Surface coinsists of two parts:(1) The ice cream: r = R; : 0 2; : 0 /6; da = R2sin d dr; vda = _R2sin _ _R2sin d d_ =R4sin2 d d._ vda = R4/6_0sin2 d2_0d = _R4_(2)_12 14 sin 2_/60= 2R4_ 12 14 sin 60_ = R46_ 332_(2) The cone: = 6; : 0 2; r : 0 R; da = r sin ddr =32 r dr d ; v da =3 r3dr d_ v da =3R_0r3dr2_0d =3 R44 2 =32 R4.Therefore _ v da = R42_3 32 +3_ = R412_2 + 33_. .c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 1. VECTOR ANALYSIS 23Problem 1.59(a) Corollary 2 says_(T)dl = 0. Stokes theorem says_(T)dl = _[(T)]da. So_[(T)]da = 0,and since this is true for any surface, the integrand must vanish: (T) = 0, conrming Eq. 1.44.(b) Corollary 2 says_(v)da = 0. Divergence theorem says_(v)da = _ (v) d. So_ (v) d= 0, and since this is true for any volume, the integrand must vanish: (v) = 0, conrming Eq. 1.46.Problem 1.60(a) Divergence theorem: _ v da = _(v) d. Let v = cT, where c is a constant vector. Using productrule #5 in front cover: v = (cT) = T(c) +c (T). But c is constant so c = 0. Therefore we have:_ c (T) d = _ Tc da. Since c is constant, take it outside the integrals: c _ T d = c _ T da. But cis any constant vectorin particular, it could be be x, or y, or zso each component of the integral on leftequals corresponding component on the right, and hence_ T d =_ T da. qed(b) Let v (v c) in divergence theorem. Then _ (v c)d = _(v c) da. Product rule #6 (v c) = c (v) v (c) = c (v). (Note: c = 0, since c is constant.) Meanwhile vectorindentity (1) says da (v c) = c (da v) = c (v da). Thus _ c (v) d = _ c (v da). Take coutside, and again let c be x, y, z then:_ (v) d = _ v da. qed(c) Let v = TU in divergence theorem: _ (TU) d = _ TU da. Product rule #(5) (TU) =T(U) + (U) (T) = T2U + (U) (T). Therefore_ _T2U + (U) (T)_ d =_ (TU) da. qed(d) Rewrite (c) with T U : _ _U2T + (T) (U)_ d = _(UT) da. Subtract this from (c), notingthat the (U) (T) terms cancel:_ _T2U U2T_ d =_ (TU UT) da. qed(e) Stokes theorem: _(v) da = _ v dl. Let v = cT. By Product Rule #(7): (cT) = T(c) c (T) = c (T) (since c is constant). Therefore, _(c (T)) da = _ Tc dl. Use vector indentity#1 to rewrite the rst term (c (T)) da = c (T da). So _ c (T da) = _ c T dl. Pull c outside,and let c x, y, and z to prove: _ T da = _ T dl. qedProblem 1.61(a) da = R2sin d dr. Let the surface be the northern hemisphere. The x and y components clearly integrateto zero, and the z component of r is cos , soa =_ R2sin cos d dz = 2R2z_ /20sin cos d = 2R2zsin22/20= R2z.(b) Let T = 1 in Prob. 1.60(a). Then T = 0, so _ da = 0. qedc 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.24 CHAPTER 1. VECTOR ANALYSIS(c) This follows from (b). For suppose a1 ,= a2; then if you put them together to make a closed surface,_ da = a1a2 ,= 0.(d) For one such triangle, da = 12(r dl) (since r dl is the area of the parallelogram, and the direction isperpendicular to the surface), so for the entire conical surface, a = 12_ r dl.(e) Let T = c r, and use product rule #4: T = (c r) = c (r) + (c )r. But r = 0, and(c )r = (cxx +cyy +czz)(x x +y y +z z) = cx x +cy y +czz = c. So Prob. 1.60(e) says_ T dl =_ (c r) dl = _ (T) da = _ c da = c _ da = c a = a c. qedProblem 1.62(1)v = 1r2r_r2

1r_ = 1r2r(r) = 1r2.For a sphere of radius R:_ v da = _ _1Rr_

_R2sin d dr_ = R_ sin d d = 4R._(v) d = _ _ 1r2_ _r2sin dr d d_ =_R_0dr___ sin d d_ = 4R.___So divergencetheorem checks.Evidently there is no delta function at the origin.(rnr) = 1r2r_r2rn_ = 1r2r_rn+2_ = 1r2(n + 2)rn+1= (n + 2)rn1(except for n = 2, for which we already know (Eq. 1.99) that the divergence is 43(r)).(2) Geometrically, it should be zero. Likewise, the curl in the spherical coordinates obviously gives zero.To be certain there is no lurking delta function here, we integrate over a sphere of radius R, usingProb. 1.60(b): If (rnr) = 0, then _(v) d = 0 ?= _ v da. But v = rnr and da =R2sin d dr are both in the r directions, so v da = 0. c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 2. ELECTROSTATICS 25Chapter 2ElectrostaticsProblem 2.1(a) Zero.(b) F = 140qQr2 , where r is the distance from center to each numeral. F points toward the missing q.Explanation: by superposition, this is equivalent to (a), with an extra q at 6 oclocksince the force of alltwelve is zero, the net force is that of q only.(c) Zero.(d) 140qQr2 , pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) asa cancellation in pairs of opposite charges (1 oclock against 7 oclock; 2 against 8, etc.), with one unpaired qdoing the job, then youll need a dierent explanation for (d).Problem 2.2(a) Horizontal components cancel. Net vertical eld is: Ez = 1402 qr 2 cos .Exzsqsqd2d2eeeeeeTE!eeurPHere r 2= z2+_d2_2; cos = zr , so E = 1402qz_z2+_d2_2_3/2 z. rWhen z d youre so far away it just looks like a single charge 2q; the eldshould reduce to E = 1402qz2 z. And it does (just set d 0 in the formula).(b) This time the vertical components cancel, leavingE = 1402 qr 2 sin x, or rE = 140qd_z2+_d2_2_3/2 x.ExzsqsqeeeeeeE E!eeFrom far away, (z d), the eld goes like E 140qdz3 z, which, as we shall see, is the eld of a dipole. (If weset d 0, we get E = 0, as is appropriate; to the extent that this conguration looks like a single point chargefrom far away, the net charge is zero, so E 0.)c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.26 CHAPTER 2. ELECTROSTATICSProblem 2.3r1Problem 2.3-x6zzL. .x/dq = dxKEz = 140_ L0dx2 cos ; (2= z2+ x2; cos = z)= 140z_ L01(z2+x2)3/2dx= 140z_ 1z2xz2+x2_L0= 140zLz2+L2.Ex = 140_ L0dx2 sin = 140_ x dx(x2+z2)3/2= 140_ 1x2+z2_L0= 140_1z 1z2+L2_.E = 140z__1 + zz2+ L2_ x +_ Lz2+ L2_z_.For z L you expect it to look like a point charge q = L: E 140Lz2 z. It checks, for with z L the xterm 0, and the z term 140zLzz.Problem 2.4From Ex. 2.1, with L a2 and z _z2+_a2_2(distance from center of edge to P), eld of one edge is:E1 = 140a_z2+ a24_z2+ a24 + a24.There are 4 sides, and we want vertical components only, so multiply by 4 cos = 4 zqz2+a24:E = 1404az_z2+ a24__z2+ a22z.Problem 2.5rzKHorizontal components cancel, leaving: E = 140__ dl2 cos _z.Here, 2= r2+ z2, cos = z (both constants), while_dl = 2r. SoE = 140(2r)z(r2+ z2)3/2 z.c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Ez = 140_L0dxr 2 cos ; ( r 2= z2+x2; cos = zr )= 140z_L01(z2+x2)3/2dx= 140z_ 1z2xz2+x2_L0= 140zLz2+L2.Ex = 140_L0dxr 2 sin = 140_ x dx(x2+z2)3/2= 140_ 1x2+z2_L0= 140_1z 1z2+L2_.E = 140z__1 + zz2+L2_ x +_ Lz2+L2_z_.For z L you expect it to look like a point charge q = L: E 140Lz2 z. It checks, for with z L the xterm 0, and the z term 140zLzz.Problem 2.4From Ex. 2.1, with L a2 and z _z2+_a2_2(distance from center of edge to P), eld of one edge is:E1 = 140a_z2+ a24_z2+ a24 + a24.There are 4 sides, and we want vertical components only, so multiply by 4 cos = 4 zqz2+a24:E = 1404az_z2+ a24__z2+ a22z.Problem 2.51Problem 2.3-x6zzL. . x/dq = dx K Ez = 140_ L0dx2 cos ; (2= z2+ x2; cos = z)= 140z_ L01(z2+x2)3/2dx= 140z_ 1z2xz2+x2_L0= 140zLz2+L2.Ex = 140_ L0dx2 sin = 140_ x dx(x2+z2)3/2= 140_ 1x2+z2_L0= 140_1z 1z2+L2_.E = 140z__1 + zz2+ L2_ x +_ Lz2+ L2_z_.For z L you expect it to look like a point charge q = L: E 140Lz2 z. It checks, for with z L the xterm 0, and the z term 140zLzz.Problem 2.4From Ex. 2.1, with L a2 and z _z2+_a2_2(distance from center of edge to P), eld of one edge is:E1 = 140a_z2+ a24_z2+ a24 + a24.There are 4 sides, and we want vertical components only, so multiply by 4 cos = 4 zqz2+a24:E = 1404az_z2+ a24__z2+ a22z.Problem 2.5rzKHorizontal components cancel, leaving: E = 140__ dl2 cos _z.Here, 2= r2+ z2, cos = z (both constants), while_dl = 2r. SoE = 140(2r)z(r2+ z2)3/2 z.c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Horizontal components cancel, leaving: E = 140__ dlr 2 cos _z.Here, r 2= r2+z2, cos = zr (both constants), while _dl = 2r. SorE = 140(2r)z(r2+z2)3/2 z.Problem 2.6Break it into rings of radius r, and thickness dr, and use Prob. 2.5 to express the eld of each ring. Totalcharge of a ring is 2r dr = 2r, so = dr is the line charge of each ring.Ering = 140(dr)2rz(r2+z2)3/2; Edisk = 1402z_ R0r(r2+z2)3/2dr.Edisk = 1402z_1z 1R2+z2_z.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 2. ELECTROSTATICS 27For R z the second term 0, so Eplane = 1402z = 20z.For z R, 1R2+z2 = 1z_1 + R2z2_1/2 1z_1 12R2z2_, so [ ] 1z 1z + 12R2z3 = R22z3,and E = 1402R22z2 = 140Qz2, where Q = R2. Problem 2.7E is clearly in the z direction. From the diagram,dq = da = R2sin d d,r 2= R2+z22Rz cos ,cos = zRcos r . rSo1Contents=x-y6zRzEz = 140

R2sin d d(z Rcos )(R2+ z22Rz cos )3/2 .

d = 2.= 140(2R2)

0(z Rcos ) sin (R2+ z22Rz cos )3/2 d. Let u = cos ; du = sin d;

= 0 u = +1 = u = 1

.= 140(2R2)

11z Ru(R2+ z22Rzu)3/2du. Integral can be done by partial fractionsor look it up.= 140(2R2)

1z2zu RR2+ z22Rzu

11= 1402R2z2

(z R)|z R| (z R)|z + R|

.For z > R (outside the sphere), Ez = 1404R2z2 = 140qz2, so E = 140qz2 z.For z < R (inside), Ez = 0, so E = 0.Problem 2.8According to Prob. 2.7, all shells interior to the point (i.e. at smaller r) contribute as though their chargewere concentrated at the center, while all exterior shells contribute nothing. Therefore:E(r) = 140Qintr2 r,where Qint is the total charge interior to the point. Outside the sphere, all the charge is interior, soE = 140Qr2 r.Inside the sphere, only that fraction of the total which is interior to the point counts:c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Ez = 140_ R2sin d d(z Rcos )(R2+z22Rz cos )3/2 ._d = 2.= 140(2R2)_ 0(z Rcos ) sin (R2+z22Rz cos )3/2 d. Let u = cos ; du = sin d;_ = 0 u = +1 = u = 1_.= 140(2R2)_ 11z Ru(R2+z22Rzu)3/2du. Integral can be done by partial fractionsor look it up.= 140(2R2)_ 1z2zu RR2+z22Rzu_11= 1402R2z2_(z R)[z R[ (z R)[z +R[_.For z > R (outside the sphere), Ez = 1404R2z2 = 140qz2, so E = 140qz2 z.For z < R (inside), Ez = 0, so E = 0.Problem 2.8According to Prob. 2.7, all shells interior to the point (i.e. at smaller r) contribute as though their chargewere concentrated at the center, while all exterior shells contribute nothing. Therefore:E(r) = 140Qintr2 r,where Qint is the total charge interior to the point. Outside the sphere, all the charge is interior, soE = 140Qr2 r.Inside the sphere, only that fraction of the total which is interior to the point counts:Qint =43r343R3Q = r3R3Q, so E = 140r3R3Q 1r2 r = 140QR3r.Problem 2.9(a) = 0E = 01r2r_r2 kr3_ = 01r2k(5r4) = 50kr2.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.28 CHAPTER 2. ELECTROSTATICS(b) By Gausss law: Qenc = 0_ Eda = 0(kR3)(4R2) = 40kR5.By direct integration: Qenc = _ d = _R0 (50kr2)(4r2dr) = 200k_R0 r4dr = 40kR5.Problem 2.10Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surfaceof this larger cube gets the same ux as every other one, so:2Qint =43r343R3Q = r3R3Q, so E = 140r3R3Q 1r2 r = 140QR3r.Problem 2.9(a) = 0E = 01r2r_r2 kr3_ = 01r2k(5r4) = 50kr2.(b) By Gausss law: Qenc = 0_ Eda = 0(kR3)(4R2) = 40kR5.By direct integration: Qenc =_ d =_R0 (50kr2)(4r2dr) = 200k_R0 r4dr = 40kR5.Problem 2.10Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surfaceof this larger cube gets the same ux as every other one, so:_onefaceEda = 124_wholelargecubeEda.The latter is 10q, by Gausss law. Therefore_onefaceEda = q240.Problem 2.11Er% Gaussian surface: Inside:_ Eda = E(4r2) = 10Qenc = 0 E = 0.EGaussian surface: Outside: E(4r2) = 10(4R2) E = R2

0r2 r.}(As in Prob. 2.7.)Problem 2.12

rcR%Gaussian surface_ Eda = E 4r2= 10Qenc = 1043r3. SoE = 130rr.Since Qtot = 43R2, E = 140QR3r (as in Prob. 2.8).Problem 2.13Ts. .lGaussian surface _ Eda = E 2s l = 10Qenc = 10l. SoE = 20ss (same as Ex. 2.1).Problem 2.14c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._onefaceEda = 124_wholelargecubeEda.The latter is 1

0q, by Gausss law. Therefore_onefaceEda = q240.Problem 2.112Qint =43r343R3Q = r3R3Q, so E = 140r3R3Q 1r2 r = 140QR3r.Problem 2.9(a) = 0E = 01r2r_r2 kr3_ = 01r2k(5r4) = 50kr2.(b) By Gausss law: Qenc = 0_ Eda = 0(kR3)(4R2) = 40kR5.By direct integration: Qenc =_ d =_R0 (50kr2)(4r2dr) = 200k_R0 r4dr = 40kR5.Problem 2.10Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surfaceof this larger cube gets the same ux as every other one, so:_onefaceEda = 124_wholelargecubeEda.The latter is 10q, by Gausss law. Therefore_onefaceEda = q240.Problem 2.11Er% Gaussian surface: Inside:_ Eda = E(4r2) = 10Qenc = 0 E = 0.EGaussian surface: Outside: E(4r2) = 10(4R2) E = R2

0r2 r.}(As in Prob. 2.7.)Problem 2.12

rcR%Gaussian surface_ Eda = E 4r2= 10Qenc = 1043r3. SoE = 130rr.Since Qtot = 43R2, E = 140QR3r (as in Prob. 2.8).Problem 2.13Ts. .lGaussian surface _ Eda = E 2s l = 10Qenc = 10l. SoE = 20ss (same as Ex. 2.1).Problem 2.14c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Problem 2.122Qint =43r343R3Q = r3R3Q, so E = 140r3R3Q 1r2 r = 140QR3r.Problem 2.9(a) = 0E = 01r2r_r2 kr3_ = 01r2k(5r4) = 50kr2.(b) By Gausss law: Qenc = 0_ Eda = 0(kR3)(4R2) = 40kR5.By direct integration: Qenc =_ d =_R0 (50kr2)(4r2dr) = 200k_R0 r4dr = 40kR5.Problem 2.10Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surfaceof this larger cube gets the same ux as every other one, so:_onefaceEda = 124_wholelargecubeEda.The latter is 10q, by Gausss law. Therefore_onefaceEda = q240.Problem 2.11Er% Gaussian surface: Inside:_ Eda = E(4r2) = 10Qenc = 0 E = 0.EGaussian surface: Outside: E(4r2) = 10(4R2) E = R2

0r2 r.}(As in Prob. 2.7.)Problem 2.12

rcR%Gaussian surface_ Eda = E 4r2= 10Qenc = 1043r3. SoE = 130rr.Since Qtot = 43R2, E = 140QR3r (as in Prob. 2.8).Problem 2.13Ts. .lGaussian surface _ Eda = E 2s l = 10Qenc = 10l. SoE = 20ss (same as Ex. 2.1).Problem 2.14c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._ Eda = E 4r2= 1

0Qenc = 1

043r3. SoE = 130rr.Since Qtot = 43R3, E = 140QR3 r (as in Prob. 2.8).Problem 2.131ContentsProblem 2.13Ts lGaussian surface Eda = E 2s l = 10Qenc = 10l. SoE = 20ss (same as Ex. 2.1).Problem 2.14

r %Gaussian surfaceProblem 2.15(i) Qenc = 0, so E = 0.(ii)

Eda = E(4r2) = 10Qenc = 10

d = 10

k r2 r2sin d r d d= 4k0

ra d r = 4k0(r a) E = k

0

r ar2

r.(iii) E(4r2) = 4k0

ba d r = 4k0(b a), soE = k

0

b ar2

r.ErT|E|a bProblem 2.16c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._ Eda = E 2s l = 1

0Qenc = 1

0l. SoE = 20ss (same as Ex. 2.1).Problem 2.141ContentsProblem 2.13Ts lGaussian surface Eda = E 2s l = 10Qenc = 10l. SoE = 20ss (same as Ex. 2.1).Problem 2.14

r %Gaussian surfaceProblem 2.15(i) Qenc = 0, so E = 0.(ii)

Eda = E(4r2) = 10Qenc = 10

d = 10

k r2 r2sin d r d d= 4k0

ra d r = 4k0(r a) E = k

0

r ar2

r.(iii) E(4r2) = 4k0

ba d r = 4k0(b a), soE = k

0

b ar2

r.ErT|E|a bProblem 2.16c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._ Eda = E 4r2= 1

0Qenc = 1

0_ d = 1

0_(k r)( r2sin d r d d)= 1

0 k 4_r0 r3d r = 4k

0r44 = k

0 r4. E = 140kr2r.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 2. ELECTROSTATICS 29Problem 2.15(i) Qenc = 0, so E = 0.(ii) _ Eda = E(4r2) = 1

0Qenc = 1

0_ d = 1

0_ k r2 r2sin d r d d= 4k

0_ra d r = 4k

0 (r a) E = k

0_r ar2_r.(iii) E(4r2) = 4k

0_ba d r = 4k

0 (b a), soE = k

0_b ar2_r.1ContentsProblem 2.13Ts lGaussian surface Eda = E 2s l = 10Qenc = 10l. SoE = 20ss (same as Ex. 2.1).Problem 2.14

r %Gaussian surfaceProblem 2.15(i) Qenc = 0, so E = 0.(ii)

Eda = E(4r2) = 10Qenc = 10

d = 10

k r2 r2sin d r d d= 4k0

ra d r = 4k0(r a) E = k

0

r ar2

r.(iii) E(4r2) = 4k0

ba d r = 4k0(b a), soE = k

0

b ar2

r.ErT|E|a bProblem 2.16c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Problem 2.16(i)1ContentsProblem 2.13Problem 2.16(i)l

Gaussian surface

Eda = E 2s l = 10Qenc = 10s2l;E = s20s.(ii)l

Gaussian surface6s Eda = E 2s l = 10Qenc = 10a2l;E = a220ss.(iii)l

Gaussian surface6s

Eda = E 2s l = 10Qenc = 0;E = 0.-s6|E|a bc 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._ Eda = E 2s l = 1

0Qenc = 1

0s2l;E = s20s.(ii)1ContentsProblem 2.13Problem 2.16(i)l

Gaussian surface

Eda = E 2s l = 10Qenc = 10s2l;E = s20s.(ii)l

Gaussian surface6s Eda = E 2s l = 10Qenc = 10a2l;E = a220ss.(iii)l

Gaussian surface6s

Eda = E 2s l = 10Qenc = 0;E = 0.-s6|E|a bc 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._ Eda = E 2s l = 1

0Qenc = 1

0a2l;E = a220ss.(iii)1ContentsProblem 2.13Problem 2.16(i)l

Gaussian surface

Eda = E 2s l = 10Qenc = 10s2l;E = s20s.(ii)l

Gaussian surface6s Eda = E 2s l = 10Qenc = 10a2l;E = a220ss.(iii)l

Gaussian surface6s

Eda = E 2s l = 10Qenc = 0;E = 0.-s6|E|a bc 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._ Eda = E 2s l = 1

0Qenc = 0;E = 0.1ContentsProblem 2.13Problem 2.16(i)l

Gaussian surface

Eda = E 2s l = 10Qenc = 10s2l;E = s20s.(ii)l

Gaussian surface6s Eda = E 2s l = 10Qenc = 10a2l;E = a220ss.(iii)l

Gaussian surface6s

Eda = E 2s l = 10Qenc = 0;E = 0.-s6|E|a bc 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Problem 2.17On the xz plane E = 0 by symmetry. Set up a Gaussian pillbox with one face in this plane and theother at y.1ContentsProblem 2.17On the xz plane E = 0 by symmetry. Set up a Gaussian pillbox with one face in this plane and theother at y.y- =Gaussian pillboxQenc = 10Ad E =

0d y (for y > d).-6Eyddd

0Problem 2.18From Prob. 2.12, the eld inside the positive sphere is E+ = 30r+, where r+ is the vector from the positivecenter to the point in question. Likewise, the eld of the negative sphere is 30r. So the total eld isE = 30(r+r)But (see diagram) r+r = d. So E = 30d.

*

+r+rdc 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher._ Eda = E A = 1

0Qenc = 1

0Ay;E =

0y y (for [y[ < d).c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.30 CHAPTER 2. ELECTROSTATICSQenc = 1

0Ad E =

0d y (for y > d).2Problem 2.17On the xz plane E = 0 by symmetry. Set up a Gaussian pillbox with one face in this plane and theother at y.y- =Gaussian pillbox

Eda = E A = 10Qenc = 10Ay;E =

0y y (for |y| < d).Qenc = 10Ad E =

0d y (for y > d).-6Eyddd

0Problem 2.18From Prob. 2.12, the eld inside the positive sphere is E+ = 30r+, where r+ is the vector from the positivecenter to the point in question. Likewise, the eld of the negative sphere is 30r. So the total eld isE = 30(r+r)But (see diagram) r+r = d. So E = 30d.

*

+r+rdProblem 2.19E = 140

2 d = 140

2

d (since depends on r

, not r)= 0 (since

2

= 0, from Prob. 1.62).Problem 2.20(1) E1 = k

x y zxyzxy 2yz 3zx

= k [ x(0 2y) + y(0 3z) +z(0 x)] = 0,so E1 is an impossible electrostatic eld.(2) E2 = k

x y zxyzy22xy + z22yz

= k [ x(2z 2z) + y(0 0) +z(2y 2y)] = 0,so E2 is a possible electrostatic eld.c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Problem 2.18From Prob. 2.12, the eld inside the positive sphere is E+ = 30r+, where r+ is the vector from the positivecenter to the point in question. Likewise, the eld of the negative sphere is 30r. So the total eld isE = 30(r+r)But (see diagram) r+r = d. So E = 30d. rr!B'+r+rdProblem 2.19E = 140_ rr 2 d = 140_ __ rr 2__ d (since depends on r

, not r)= 0 (since _ rr 2_ = 0, from Prob. 1.62).Problem 2.20(1) E1 = k x y zxyzxy 2yz 3zx= k [ x(0 2y) + y(0 3z) +z(0 x)] ,= 0,so E1 is an impossible electrostatic eld.(2) E2 = k x y zxyzy22xy +z22yz= k [ x(2z 2z) + y(0 0) +z(2y 2y)] = 0,so E2 is a possible electrostatic eld.Lets go by the indicated path:&&&&&axEyTz&&&&aIEIITIIIr(x0, y0, z0) Edl = (y2dx + (2xy +z2)dy + 2yz dz)kStep I: y = z = 0; dy = dz = 0. Edl = ky2dx = 0.Step II: x = x0, y : 0 y0, z = 0. dx = dz = 0.Edl = k(2xy +z2)dy = 2kx0y dy._II Edl = 2kx0_y00 y dy = kx0y20.Step III : x = x0, y = y0, z : 0 z0; dx = dy = 0.Edl = 2kyz dz = 2ky0z dz.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 2. ELECTROSTATICS 31_III Edl = 2y0k_z00 z dz = ky0z20.V (x0, y0, z0) = (x0,y0,z0)_0Edl = k(x0y20 +y0z20), or V (x, y, z) = k(xy2+yz2).Check: V =k[ x(xy2+yz2) x+ y(xy2+yz2) y+ z(xy2+yz2) z]=k[y2 x+(2xy+z2) y+2yz z]=E. Problem 2.21V (r) = _rEdl.___Outside the sphere (r > R) : E = 140qr2r.Inside the sphere (r < R) : E = 140qR3rr.So for r > R: V (r) = _r_ 140q r2_d r = 140q_1 r_r= q401r,and for r < R: V (r) = _R_ 140q r2_d r _rR_ 140qR3 r_d r = q40_1R 1R3_r2R22__= q4012R_3 r2R2_.When r > R, V = q40r_1r_r = q401r2r, so E = V = q401r2r. When r < R, V = q4012Rr_3 r2R2_r = q4012R_2rR2_r = q40rR3r; so E = V = 140qR3rr.0.5 1 1.5 2 2.5 30.20.40.60.811.21.41.6rV(r)(In the gure, r is in units of R, and V (r) is in units of q/40R.)Problem 2.22E = 1402s s (Prob. 2.13). In this case we cannot set the reference point at , since the charge itselfextends to . Lets set it at s = a. ThenV (s) = _sa_ 1402 s_d s = 1402ln_sa_.(In this form it is clear why a = would be no goodlikewise the other natural point, a = 0.)V = 1402 s_ln_sa__s = 14021ss = E. Problem 2.23V (0) = _0Edl = _b_k

0(ba)r2_dr _ab_k

0(ra)r2_dr _0a(0)dr = k

0(ba)b k

0_ln_ab_+a_1a 1b__= k

0_1 ab ln_ab_1 + ab_ = k

0ln_ba_.Problem 2.24Using Eq. 2.22 and the elds from Prob. 2.16:V (b) V (0) = _b0 Edl = _a0 Edl _ba Edl = 20_a0 s ds a220_ba1sdsc 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.32 CHAPTER 2. ELECTROSTATICS= _ 20_ s22a0+ a220 ln s[ba = a240_1 + 2 ln_ba__.Problem 2.25(a) V = 1402q_z2+_d2_2.(b) V = 140_LLdxz2+x2 = 40 ln(x +z2+x2)LL= 40ln_ L +z2+L2L +z2+L2_ = 20ln_L +z2+L2z_. r1ContentsProblem 2.25(a) V = 1402q_z2+_d2_2.(b) V = 140_LLdxz2+x2 = 40ln(x +z2+x2)LL= 40ln_ L +z2+L2L +z2+L2_ = 20ln_L+z2+L2z_.xz(c) V = 140_R0 2r drr2+z2 = 1402 (r2+z2)R0 = 20__R2+ z2z_.In each case, by symmetry Vy = Vx = 0. E = Vz z.(a) E = 1402q_12_ 2zz2+(d2)23/2 z = 1402qz_z2+_d2_2_3/2 z (agrees with Prob. 2.2a).(b) E = 40_ 1(L+z2+L2)121z2+L22z 1(L+z2+L2)121z2+L22z_z= 40zz2+L2_L+z2+L2Lz2+L2(z2+L2)L2_z = 2L401zz2+L2z (agrees with Ex. 2.1).(c) E = 20_121R2+z22z 1_z = 20_1 zR2+z2_z (agrees with Prob. 2.6).If the right-hand charge in (a) is q, then V = 0 , which, naively, suggests E = V = 0, in contradictionwith the answer to Prob. 2.2b. The point is that we only know V on the z axis, and from this we cannothope to compute Ex = Vx or Ey = Vy . That was OK in part (a), because we knew from symmetry thatEx = Ey = 0. But now E points in the x direction, so knowing V on the z axis is insucient to determine E.c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.(c) V = 140_ R0 2r drr2+z2= 1402 (_r2+z2)R0= 20__R2+z2z_.In each case, by symmetry Vy = Vx = 0. E = Vzz.(a) E = 1402q_12_ 2z_z2+_d2_2_3/2 z = 1402qz_z2+_d2_2_3/2 z (agrees with Prob. 2.2a).(b) E = 40_ 1(L +z2+L2)121z2+L22z 1(L +z2+L2)121z2+L22z_z= 40zz2+L2_L +z2+L2L z2+L2(z2+L2) L2_z = 2L401zz2+L2z (agrees with Ex. 2.1).(c) E = 20_121R2+z22z 1_z = 20_1 zR2+z2_z (agrees with Prob. 2.6).If the right-hand charge in (a) is q, then V = 0 , which, naively, suggests E = V = 0, in contradictionwith the answer to Prob. 2.2b. The point is that we only know V on the z axis, and from this we cannothope to compute Ex = Vx or Ey = Vy . That was OK in part (a), because we knew from symmetry thatEx = Ey = 0. But now E points in the x direction, so knowing V on the z axis is insucient to determine E.Problem 2.262-6ab hhc 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.rrV (a) = 140_ 2h0_2rr_ d r = 24012(2h) = h20(where r = r /2)c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 2. ELECTROSTATICS 33V (b) = 140_ 2h0_2rr_ d r (where r =_h2+ r 22hr )= 24012_ 2h0r_h2+ r 22hr d r= 220__h2+ r 22hr + h2 ln(2_h2+ r 22hr + 2 r 2h)_2h0= 220_h + h2 ln(2h + 22h 2h) h h2 ln(2h 2h)_= 220h2_ln(2h +2h) ln(2h 2h)_ = h40ln_2 +22 2_ = h40ln_(2 +2)22_= h20ln(1 +2). V (a) V (b) = h20_1 ln(1 +2)_.Problem 2.273Problem 2.27 .. zL2. .x .. Ldx- Cut the cylinder into slabs, as shown in the gure, anduse result of Prob. 2.25c, with z x and dx:V = 20z+L/2_zL/2_R2+ x2x_dx= 2012_xR2+ x2+ R2ln(x +R2+ x2) x2z+L/2zL/2= 408 z._c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Cut the cylinder into slabs, as shown in the gure, anduse result of Prob. 2.25c, with z x and dx:V = 20z+L/2_zL/2_R2+x2x_dx= 2012_xR2+x2+R2ln(x +R2+x2) x2z+L/2zL/2= 408 z._ V = 40 2 2_z_01zr2dr +R_z1rr2dr_ =

0_1zz33 + R2z22_ = 20_R2 z23_.But = q43R3, so V (z) = 1203q4R3_R2 z23_ = q80R_3 z2R2_; V (r) = q80R_3 r2R2_. Problem 2.292V = 1402__ r_d = 140_(r

)_2 1r _d (since is a function of r

, not r)= 140_(r

)[43(r r

)] d = 1

0(r). Problem 2.30.(a) Ex. 2.4: Eabove = 20 n; Ebelow = 20 n ( n always pointing up); EaboveEbelow =

0 n. Ex. 2.5: At each surface, E = 0 one side and E =

0 other side, so E =

0. Prob. 2.11: Eout = R2

0r2r =

0r ; Ein = 0 ; so E =

0r. (b) Outside: _Eda = E(2s)l = 1

0Qenc =

0(2R)l E =

0Rss =

0s (at surface).Inside: Qenc = 0, so E = 0. E =

0s. 1ContentsProblem 2.30.(a) Ex. 2.4: Eabove = 20 n; Ebelow = 20 n ( n always pointing up); EaboveEbelow = 0 n. Ex. 2.5: At each surface, E = 0 one side and E = 0other side, so E = 0. Prob. 2.11: Eout = R20r2r = 0r ; Ein = 0 ; so E = 0r. (b) Outside:_Eda = E(2s)l = 10Qenc = 10(2R)l E = 0Rss = 0s (at surface).Inside: Qenc = 0, so E = 0. E = 0s. 6s 6R. .l(c) Vout = R20r = R0(at surface); Vin = R0; so Vout = Vin. Voutr = R20r2 = 0(at surface); Vinr = 0 ; so Voutr Vinr = 0. Problem 2.31(a) V = 140

qirij= 140_qa + q2a + qa_ = q40a_2 + 12_. W4 = qV = q240a_2 + 12_.(2)+(3)(4)+(1)(b) W1 = 0, W2 = 140_q2a_; W3 = 140_ q22 a q2a_; W4 = (see (a)).Wtot = 140q2a_1 + 12 1 2 + 12_ = 1402q2a_2 + 12_.c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.(c) Vout = R2

0r = R

0 (at surface); Vin = R

0 ; so Vout = Vin. Voutr = R2

0r2 =

0 (at surface); Vinr = 0 ; so Voutr Vinr =

0. c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 2. ELECTROSTATICS 35Problem 2.31(a) V = 140

qirij= 140_qa + q2a + qa_ = q40a_2 + 12_. W4 = qV = q240a_2 + 12_.r(2)+ r(3)r(4)+r(1)(b) W1 = 0, W2 = 140_q2a_; W3 = 140_ q22 a q2a_; W4 = (see (a)).Wtot = 140q2a_1 + 12 1 2 + 12_ = 1402q2a_2 + 12_.Problem 2.32(a) W = 12_V d. From Prob. 2.21 (or Prob. 2.28): V = 20_R2 r23_ = 140q2R_3 r2R2_W = 12 140q2R_ R0_3 r2R2_4r2dr = q40R_3r33 1R2r55_R0= q40R_R3 R35_= q50R2= qR250q43R3 = 140_35q2R_.(b) W = 02_E2d. Outside (r > R) E = 140qr2r ; Inside (r < R) E = 140qR3rr. W = 021(40)2q2__ R1r4(r24 dr) +_ R0_ rR3_2(4r2dr)_= 140q22__1r_R+ 1R6_r55_R0_ = 140q22_1R + 15R_ = 14035q2R.(c) W = 02_ _S V Eda +_V E2d_, where 1 is large enough to enclose all the charge, but otherwisearbitrary. Lets use a sphere of radius a > R. Here V = 140qr.W = 02__r=a_ 140qr__ 140qr2_r2sin d d +_ R0E2d +_ aR_ 140qr2_2(4r2dr)_= 02_ q2(40)21a4 + q2(40)245R + 1(40)24q2_1r_aR_= 140q22_1a + 15R 1a + 1R_ = 14035q2R.As a , the contribution from the surface integral_ 140q22a_ goes to zero, while the volume integral_ 140q22a(6a5R 1)_ picks up the slack.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.36 CHAPTER 2. ELECTROSTATICSProblem 2.332Problem 2.32(a) W = 12_V d. From Prob. 2.21 (or Prob. 2.28): V = 20_R2r23_ = 140q2R_3 r2R2_W = 12 140q2R_ R0_3 r2R2_4r2dr = q40R_3r33 1R2r55_R0= q40R_R3R35_= q50R2= qR250q43R3 = 140_35q2R_.(b) W = 02_E2d. Outside (r > R) E = 140qr2r ; Inside (r < R) E = 140qR3rr. W = 021(40)2q2__ R1r4(r24 dr) +_ R0_ rR3_2(4r2dr)_= 140q22__1r_R+ 1R6_r55_R0_ = 140q22_ 1R + 15R_ = 14035q2R.(c) W = 02_ _S V Eda +_V E2d_, where V is large enough to enclose all the charge, but otherwisearbitrary. Lets use a sphere of radius a > R. Here V = 140qr.W = 02__r=a_ 140qr__ 140qr2_r2sin d d +_ R0E2d +_ aR_ 140qr2_2(4r2dr)_= 02_ q2(40)21a4 + q2(40)245R + 1(40)24q2_1r_aR_= 140q22_1a + 15R1a + 1R_ = 14035q2R.As a , the contribution from the surface integral_ 140q22a_ goes to zero, while the volume integral_ 140q22a( 6a5R1)_ picks up the slack.Problem 2.33 q*r)d qc 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.dW = d q V = d q_ 140_ qr, ( q = charge on sphere of radius r). q = 43r3 = q r3R3 (q = total charge on sphere).d q = 4r2dr = 4r243R3q dr = 3qR3r2dr.dW = 140_qr3R3_ 1r_3qR3r2dr_ = 1403q2R6 r4drW = 1403q2R6_ R0r4dr = 1403q2R6R55 = 140_35q2R_.Problem 2.34(a) W = 02_ E2d. E = 140qr2 (a < r < b), zero elsewhere.W = 02_ q40_2 _ba_ 1r2_24r2dr = q280_ba1r2 = q280_1a 1b_.(b) W1 = 180q2a , W2 = 180q2b , E1 = 140qr2 r (r > a), E2 = 140qr2 r (r > b). SoE1 E2 =_ 140_2q2r4 , (r > b), and hence _ E1 E2d = _ 140_2q2_b1r44r2dr = q240b.Wtot = W1 +W2 +0_ E1 E2d = 180q2_1a + 1b 2b_ = q280_1a 1b_.Problem 2.35(a) R = q4R2 ; a = q4a2 ; b = q4b2.(b) V (0) = _0Edl = _b_ 140qr2_dr _ab (0)dr _Ra_ 140qr2_dr _0R(0)dr = 140_qb + qR qa_.(c) b 0 (the charge drains o); V (0) = _a(0)dr _Ra_ 140qr2_dr _0R(0)dr = 140_qR qa_.Problem 2.36(a) a = qa4a2; b = qb4b2; R = qa +qb4R2 .(b) Eout = 140qa +qbr2 r, where r = vector from center of large sphere.(c) Ea = 140qar2ara, Eb = 140qbr2brb, where ra (rb) is the vector from center of cavity a (b).(d) Zero.(e) R changes (but not a or b); Eoutside changes (but not Ea or Eb); force on qa and qb still zero.c 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.CHAPTER 2. ELECTROSTATICS 37Problem 2.37Between the plates, E = 0; outside the plates E = /0 = Q/0A. SoP = 02 E2= 02Q2

20A2 = Q220A2.Problem 2.38Inside, E = 0; outside, E = 140Qr2r; so3Problem 2.34(a) W = 02

E2d. E = 140qr2 (a < r < b), zero elsewhere.W = 02

q40

2 ba

1r2

24r2dr = q280

ba1r2 = q280

1a 1b

.(b) W1 = 180q2a , W2 = 180q2b , E1 = 140qr2 r (r > a), E2 = 140qr2 r (r > b). SoE1 E2 =

140

2q2r4 , (r > b), and hence

E1 E2 d =

140

2q2 b1r44r2dr = q240b.Wtot = W1 + W2 + 0

E1 E2 d = 180q2 1a + 1b 2b

= q280

1a 1b

.Problem 2.35(a) R = q4R2 ; a = q4a2 ; b = q4b2.(b) V (0) =

0Edl =

b

140qr2

dr

ab (0)dr

Ra

140qr2

dr

0R(0)dr = 140

qb + qR qa

.(c) b 0 (the charge drains o); V (0) =

a(0)dr

Ra

140qr2

dr

0R(0)dr = 140

qR qa

.Problem 2.36(a) a = qa4a2; b = qb4b2; R = qa + qb4R2 .(b) Eout = 140qa + qbr2 r, where r = vector from center of large sphere.(c) Ea = 140qar2ara, Eb = 140qbr2brb, where ra (rb) is the vector from center of cavity a (b).(d) Zero.(e) R changes (but not a or b); Eoutside changes (but not Ea or Eb); force on qa and qb still zero.Problem 2.37Between the plates, E = 0; outside the plates E = /0 = Q/0A. SoP = 02 E2= 02Q2

20A2 = Q220A2.Problem 2.38Inside, E = 0; outside, E = 140Qr2r; so6z

EEave = 12140QR2 r; fz = (Eave)z; = Q4R2.Fz =

fzda =

Q4R2

12

140QR2

cos R2sin d d= 120

Q4R

22

/20 sin cos d = 10

Q4R

2 12 sin2

/20 = 120

Q4R

2= Q232R2

0.c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.Eave = 12140QR2 r; fz = (Eave)z; = Q4R2.Fz = _fzda = __ Q4R2_12_ 140QR2_cos R2sin d d= 120_ Q4R_22_/20 sin cos d = 10_ Q4R_2 _12 sin2_/20 = 120_ Q4R_2= Q232R2

0.Problem 2.39Say the charge on the inner cylinder is Q, for a length L. The eld is given by Gausss law:_Eda = E 2s L = 1

0Qenc = 1

0Q E = Q20L1ss. Potential dierence between the cylinders isV (b) V (a) = _ baEdl = Q20L_ ba1sds = Q20L ln_ba_.As set up here, a is at the higher potential, so V = V (a) V (b) = Q20L ln_ba_.C = QV = 20Lln(ba), so capacitance per unit length is 20ln_ba_.Problem 2.40(a) W = (force)(distance) = (pressure)(area)(distance) = 02 E2A.(b) W = (energy per unit volume)(decrease in volume) =_

0E22_(A). Same as (a), conrming that theenergy lost is equal to the work done.Problem 2.414Problem 2.39Say the charge on the inner cylinder is Q, for a length L. The eld is given by Gausss law:_Eda = E 2s L = 10Qenc = 10Q E = Q20L1s s. Potential dierence between the cylinders isV (b) V (a) = _ baEdl = Q20L_ ba1sds = Q20L ln_ba_.As set up here, a is at the higher potential, so V = V (a) V (b) = Q20L ln_ba_.C = QV = 20Lln(ba), so capacitance per unit length is 20ln_ba_.Problem 2.40(a) W = (force)(distance) = (pressure)(area)(distance) = 02 E2A.(b) W = (energy per unit volume)(decrease in volume) =_

0E22_(A). Same as (a), conrming that theenergy lost is equal to the work done.