Instucto s Solution Mnul Intoduction to Electodynmics Fouth Edition Dvid J Giffiths
Contents Vecto Anlysis 4 Electosttics 6 3 Potentil 53 4 Electic Fields in Mtte 9 5 Mgnetosttics 6 Mgnetic Fields in Mtte 33 7 Electodynmics 45 8 Consevtion Lws 68 9 Electomgnetic Wves 85 Potentils nd Fields Rdition 3 Electodynmics nd Reltivity 6 c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
3 Pefce Although I wote these solutions, much of the typesetting ws done by Jonh Gollub, Chistophe Lee, nd Jmes Tewillige ny mistkes e, of couse, entiely thei fult Chis lso did mny of the figues, nd I would like to thnk him pticully fo ll his help If you find eos, plese let me know giffith@eededu Dvid Giffiths c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
4 CHAPTER VECTOR ANALYSIS Chpte Vecto Anlysis Poblem Fom the digm, B + C cos θ 3 = B cos θ + C cos θ Multiply by A A B + C cos θ 3 = A B cos θ + A C cos θ So: A B + C = A B + A C Dot poduct is distibutive Similly: B + C sin θ 3 = B sin θ + C sin θ Mulitply by A ˆn A B + C sin θ 3 ˆn = A B sin θ ˆn + A C sin θ ˆn If ˆn is the unit vecto pointing out of the pge, it follows tht A B + C = A B + A C Coss poduct is distibutive θ 3 }{{}}{{} θ B B + C C θ }{{} B cos θ }{{} C cos θ } C sin θ } B sin θ b Fo the genel cse, see G E Hy s Vecto nd Tenso Anlysis, Chpte, Section 7 dot poduct nd Section 8 coss poduct Poblem The tiple coss-poduct is not in genel ssocitive Fo exmple, suppose A = B nd C is pependicul to A, s in the digm Then B C points out-of-the-pge, nd A B C points down, nd hs mgnitude ABC But A B =, so A B C = A B C B C C A B C A A = B Poblem 3 z A = + ˆx + ŷ ẑ; A = 3; B = ˆx + ŷ + ẑ; B = 3 A B = + + = = AB cos θ = 3 3 cos θ cos θ = 3 θ = cos 3 7588 Poblem 4 x B θ A y The coss-poduct of ny two vectos in the plne will give vecto pependicul to the plne Fo exmple, we might pick the bse A nd the left side B: A = ˆx + ŷ + ẑ; B = ˆx + ŷ + 3 ẑ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 5 ˆx ŷ ẑ A B = = 6 ˆx + 3 ŷ + ẑ 3 This hs the ight diection, but the wong mgnitude To mke unit vecto out of it, simply divide by its length: A B = 36 + 9 + 4 = 7 ˆn = A B = 6 A B 7 ˆx + 7ŷ 3 + 7ẑ Poblem 5 ˆx ŷ ẑ A B C = A x A y A z B y C z B z C y B z C x B x C z B x C y B y C x = ˆxA y B x C y B y C x A z B z C x B x C z ] + ŷ + ẑ I ll just check the x-component; the othes go the sme wy = ˆxA y B x C y A y B y C x A z B z C x + A z B x C z + ŷ + ẑ BA C CA B = B x A x C x + A y C y + A z C z C x A x B x + A y B y + A z B z ] ˆx + ŷ + ẑ = ˆxA y B x C y + A z B x C z A y B y C x A z B z C x + ŷ + ẑ They gee Poblem 6 A B C+B C A+C A B = BA C CA B+CA B AC B+AB C BC A = So: A B C A B C = B C A = AB C CA B If this is zeo, then eithe A is pllel to C including the cse in which they point in opposite diections, o one is zeo, o else B C = B A =, in which cse B is pependicul to A nd C including the cse B = Conclusion: A B C = A B C eithe A is pllel to C, o B is pependicul to A nd C Poblem 7 = 4 ˆx + 6 ŷ + 8 ẑ ˆx + 8 ŷ + 7 ẑ = ˆx ŷ + ẑ = 4 + 4 + = 3 ˆ = = 3 ˆx 3ŷ + 3ẑ Poblem 8 Āy B y + Āz B z = cos φa y + sin φa z cos φb y + sin φb z + sin φa y + cos φa z sin φb y + cos φb z = cos φa y B y + sin φ cos φa y B z + A z B y + sin φa z B z + sin φa y B y sin φ cos φa y B z + A z B y + cos φa z B z = cos φ + sin φa y B y + sin φ + cos φa z B z = A y B y + A z B z b A x + A y + A z = Σ 3 i= A ia i = Σ 3 i= Σ 3 j= R ij A j Σ 3 k= R ik A k = Σj,k Σ i R ij R ik A j A k { } if j = k This euls A x + A y + A z povided Σ 3 i= R ijr ik = if j k Moeove, if R is to peseve lengths fo ll vectos A, then this condition is not only sufficient but lso necessy Fo suppose A =,, Then Σ j,k Σ i R ij R ik A j A k = Σ i R i R i, nd this must eul since we wnt A x+a y+a z = Likewise, Σ 3 i= R ir i = Σ 3 i= R i3r i3 = To check the cse j k, choose A =,, Then we wnt = Σ j,k Σ i R ij R ik A j A k = Σ i R i R i + Σ i R i R i + Σ i R i R i + Σ i R i R i But we ledy know tht the fist two sums e both ; the thid nd fouth e eul, so Σ i R i R i = Σ i R i R i =, nd so on fo othe uneul combintions of j, k In mtix nottion: RR =, whee R is the tnspose of R c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
6 CHAPTER VECTOR ANALYSIS Poblem 9 z y x Looking down the xis: z x z y y x A ottion cies the z xis into the y = z xis, y into x = y, nd x into z = x So A x = A z, A y = A x, A z = A y R = Poblem No chnge A x = A x, A y = A y, A z = A z b A A, in the sense A x = A x, A y = A y, A z = A z c A B A B = A B Tht is, if C = A B, C C No minus sign, in contst to behvio of n odiny vecto, s given by b If A nd B e pseudovectos, then A B A B = A B So the coss-poduct of two pseudovectos is gin pseudovecto In the coss-poduct of vecto nd pseudovecto, one chnges sign, the othe doesn t, nd theefoe the coss-poduct is itself vecto Angul momentum L = p nd toue N = F e pseudovectos d A B C A B C = A B C So, if = A B C, then ; pseudoscl chnges sign unde invesion of coodintes Poblem f = x ˆx + 3y ŷ + 4z 3 ẑ b f = xy 3 z 4 ˆx + 3x y z 4 ŷ + 4x y 3 z 3 ẑ c f = e x sin y ln z ˆx + e x cos y ln z ŷ + e x sin y/z ẑ Poblem h = y 6x 8 ˆx + x 8y + 8 } ŷ] h = t summit, so y 6x 8 = y 8 4y + 84 = x 8y + 8 = = 6x 4y + 84 = y = 66 = y = 3 = x 4 + 8 = = x = Top is 3 miles noth, miles west, of South Hdley b Putting in x =, y = 3: h = 36 + 36 + 84 + = 7 ft c Putting in x =, y = : h = 6 8 ˆx + 8 + 8 ŷ] = ˆx + ŷ = ˆx + ŷ h = 3 ft/mile; diection: nothwest c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 7 Poblem 3 = x x ˆx + y y ŷ + z z ẑ; = x x + y y + z z = x x +y y +z z ] ˆx+ ŷ+ ẑ = x x ˆx+y y ŷ+z z ẑ = b = x x + y y + z z ] ˆx + ŷ + ẑ = 3 x x ˆx 3 y y ŷ 3 z z ẑ = 3 x x ˆx + y y ŷ + z z ẑ] = / 3 = / ˆ c n = n n = n n Poblem 4 x = n n ˆ x, so n = n n y = +y cos φ + z sin φ; multiply by sin φ: y sin φ = +y sin φ cos φ + z sin φ z = y sin φ + z cos φ; multiply by cos φ: z cos φ = y sin φ cos φ + z cos φ Add: y sin φ + z cos φ = zsin φ + cos φ = z Likewise, y cos φ z sin φ = y So = cos φ; = sin φ; = sin φ; = cos φ Theefoe f y = f = f + f = + cos φ f } y + sin φ f z f z = f = f Poblem 5 + f = sin φ f y + cos φ f z v = x + 3xz + xz = x + x = b v b = xy + yz + 3xz = y + z + 3x c v c = y + xy + z + yz = + x + y = x + y Poblem 6 So f tnsfoms s vecto ˆ ed ] v = x + 3 y + 3 z = 3 xx + y + z 3 ] ] + yx + y + z 3 + zx + y + z 3 = 3 + x 3/ 5 x + 3 + y 3/ 5 y + 3 + z 3/ 5 z = 3 3 3 5 x + y + z = 3 3 3 3 = This conclusion is supising, becuse, fom the digm, this vecto field is obviously diveging wy fom the oigin How, then, cn v =? The nswe is tht v = eveywhee except t the oigin, but t the oigin ou clcultion is no good, since =, nd the expession fo v blows up In fct, v is infinite t tht one point, nd zeo elsewhee, s we shll see in Sect 5 Poblem 7 v y = cos φ v y + sin φ v z ; v z = sin φ v y + cos φ v z v y = vy vz cos φ + sin φ = vy + vy v z = vy = vy = vy cos φ + sin φ cos φ + vz vz sin φ + cos φ = vy vy sin φ + cos φ sin φ + vy + vy cos φ + vz sin φ vz cos φ + vz sin φ + vz + vz sin φ + vz vz sin φ + cos φ cos φ So sin φ Use esult in Pob 4: cos φ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
8 CHAPTER VECTOR ANALYSIS v y + vz = vy cos φ + vy = vy vz sin φ cos φ + vz vz sin φ cos φ + cos φ + sin φ + vz vz sin φ cos φ + cos φ sin φ + vy sin φ + cos φ = vy + vz sin φ vy sin φ cos φ Poblem 8 ˆx ŷ ẑ v = x 3xz xz = ˆx 6xz + ŷ + z + ẑ3z = 6xz ˆx + z ŷ + 3z ẑ ˆx ŷ ẑ b v b = = ˆx y + ŷ 3z + ẑ x = y ˆx 3z ŷ x ẑ xy yz 3xz ˆx ŷ ẑ c v c = = ˆxz z + ŷ + ẑy y = y xy + z yz Poblem 9 v A v B z Poblem y v v x As we go fom point A to point B 9 o clock to o clock, x inceses, y inceses, v x inceses, nd v y deceses, so v x / >, while v y / < On the cicle, v z =, nd thee is no dependence on z, so E 4 sys vy v = ẑ v x points in the negtive z diection into the pge, s the ight hnd ule would suggest Pick ny othe neby points on the cicle nd you will come to the sme conclusion I m soy, but I cnnot emembe who suggested this cute illusttion] v = y ˆx + x ŷ; o v = yz ˆx + xz ŷ + xy ẑ; o v = 3x z z 3 ˆx + 3 ŷ + x 3 3xz ẑ; o v = sin xcosh y ˆx cos xsinh y ŷ; etc Poblem i fg = fg fg ˆx + ŷ + fg = f g g ˆx + ŷ + g ẑ ẑ = + g f g + g f f f ˆx + ŷ + f ẑ ˆx + + g f f g ŷ + f g + g f ẑ = f g + g f ed iv A B = A yb z A z B y + A zb x A x B z + A xb y A y B x B = A z y + B z Ay A z By B y Az + A z Bx + B x Az A x Bz v fa = = B x Az faz +A x B y Ay A y Bx fay + B y Ax + B Ax y Bz ˆx + A y Bx Az Az By fax B z Ax B x Ay Ay + Bz Ax A Bz x By Bx = B A A B ed faz ŷ + fay fax ẑ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 9 = Poblem A B = b ˆ = = f f Az Az + A z f f Ay A y f ˆx + + f Ay + A y f f Ax A x f ˆx + A x Ay f A y A z f = f A A f ed A x B x x ˆx+y ŷ+z ẑ = x +y +z ˆ ˆ] x = = = + Az ˆx + + A z Bx ˆx + + A y Bx B A z x + A y Bz f Ax ẑ ŷ + Ay A z f A x f + A z Bz B A y x ẑ + A x f f Az Ax ŷ + + A y By ] ẑ A z f ŷ A x f A y f + A z By ŷ ] ẑ Let s just do the x component x + y + z x x +y +z { ] ] ]} x + x x + yx 3 y + zx 3 z { 3 x x 3 + xy + xz } { = x 3 x x + y + z } = x 3 x = Sme goes fo the othe components Hence: ˆ ˆ = c v v b = x + 3xz xz xy ˆx + yz ŷ + 3xz ẑ Poblem 3 = x y ˆx + ŷ + 3z ẑ + 3xz x ˆx + z ŷ + ẑ xz ˆx + y ŷ + 3x ẑ = x y + 3x z ˆx + 6xz 3 4xyz ŷ + 3x z 6x z ẑ = x y + 3z ˆx + xz 3z y ŷ 3x z ẑ ii A B] x = A xb x + A y B y + A z B z = Ax B B x + A x x By A B] x = A y B z A z B y = A y Ay B A] x = B y Ax Ax Bz Az A B] x = A x + A y + A z B Bx = A x x B A] x = B x A x + B y Ax + B z Ax So A B + B A + A B + B A] x B = A y y A y Bx A z Bx + A z Bz + B y Ay +A x B x = B x A x +B z A x + A y Bx + A z Bx + B x Ax + B y Ax Bx + A y Bx B y Ax + B z Ax + Ay B B y + A y y Az Bx + A z Bx B z Ax + B Ay y / Ax + / Ax By + A y / Bx + / Bx / + Az + / Ax + A z B / x + Bz + / Bx + A x Bx = A B] x sme fo y nd z Bz + B z Az + Az vi A B] x = A B z A B y = A xb y A y B x A zb x A x B z = Ax B B y + A y x Ay B B x A x y B A A B + A B B A] x A = B x x + B y Ax + B z Ax A x Bx A y Bx A z Bx Az B B x A x z + A Bx x + By B B z + A z z + Ax B B z + A z x + Bz Ax Bx + Ay + Az c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
= B y A x + A x B x CHAPTER VECTOR ANALYSIS / + / Bx + By + Bz + A y B x + Az B x = A B] x sme fo y nd z Poblem 4 f/g = = g f g = g g + Bz Ax f/g ˆx + f/g ŷ + f/g ẑ g f ˆx + g f g f ŷ + g f g f f f ˆx + g ŷ + f ẑ f + Bx Ax g g A/g = A x/g + A y/g + A z/g = g Ax = g g Ay Ax g g + g Ax + Ay Ay g g + Az A/g] x = A z/g A y/g = g Az = g g Az g g Az g Ay Ay + g Az Ay g g ẑ / / Ax Ay ˆx + g ŷ + g ẑ Az g g ] g A x + A y g + A z g g A z A y g ] = g Ax+A gx g sme fo y nd z ed Poblem 5 ˆx ŷ ẑ A B = x y 3z 3y x = ˆx6xz + ŷ9zy + ẑ x 6y Az = g f f g g ed ] = g A A g g ed A B = 6xz + 9zy + x 6y = 6z + 9z + = 5z A = ˆx 3z y + ŷ x 3z + ẑ y x = ; B A = B = ˆx x + ŷ 3y + ẑ x 3y = 5 ẑ; A B = 5z A B? = B A A B = 5z = 5z b A B = 3xy 4xy = xy ; A B = xy = ˆx xy + ŷ xy = y ˆx x ŷ ˆx ŷ ẑ A B = x y 3z = ˆx y + ŷ5x; B A = 5 A B = x + y + 3z 3y ˆx x ŷ = ˆx6y + ŷ x B A = 3y x x ˆx + y ŷ + 3z ẑ = ˆx3y + ŷ 4x A B + B A + A B + B A = y ˆx + 5x ŷ + 6y ˆx x ŷ + 3y ˆx 4x ŷ = y ˆx x ŷ = A B c A B = ˆx x 6y 9zy + ŷ 6xz x 6y + ẑ 9zy 6xz = ˆx y 9y + ŷ6x + 4x + ẑ = y ˆx + x ŷ A = x + y + 3z = + + 3 = 6; B = 3y + x = c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS B A A B + A B B A = 3y ˆx 4x ŷ 6y ˆx + x ŷ 8y ˆx + x ŷ = y ˆx + x ŷ = A B Poblem 6 T = ; T = T = T = b T b = T b = T b = T b T b = 3T b = 3 sin x sin y sin z c T c d v x v y v z = 5T c ; T c = ; v x = v y = v z Poblem 7 v = = v z v z = 6T c ; T c = 9T c T c = = v x = v x = = ; v y = 6x v y = 6x = v z v = ˆx + 6x ŷ = v z = vz vy + vx + v x v x vz + + vy vx v y v y =, by eulity of coss-deivtives Fom Pob 8: v = 6xz ˆx+z ŷ+3z ẑ v = 6xz+ z+ 3z = 6z+6z = Poblem 8 ˆx ŷ ẑ t = = ˆx t t t t t =, by eulity of coss-deivtives + ŷ t t + ẑ t t In Pob b, f = xy 3 z 4 ˆx + 3x y z 4 ŷ + 4x y 3 z 3 ẑ, so ˆx ŷ ẑ f = xy 3 z 4 3x y z 4 4x y 3 z 3 = ˆx3 4x y z 3 4 3x y z 3 + ŷ4 xy 3 z 3 4xy 3 z 3 + ẑ 3xy z 4 3 xy z 4 = Poblem 9,,,, x :, y = z = ; dl = dx ˆx; v dl = x dx; v dl = x dx = x 3 /3 = /3,,,, x =, y :, z = ; dl = dy ŷ; v dl = yz dy = ; v dl =,,,, x = y =, z : ; dl = dz ẑ; v dl = y dz = dz; v dl = dz = z = Totl: v dl = /3 + + = 4/3 b,,,, x = y =, z : ; dl = dz ẑ; v dl = y dz = ; v dl =,,,, x =, y :, z = ; dl = dy ŷ; v dl = yz dy = y dy; v dl = y dy = y =,,,, x :, y = z = ; dl = dx ˆx; v dl = x dx; v dl = x dx = x 3 /3 = /3 Totl: v dl = + + /3 = 4/3 c x = y = z : ; dx = dy = dz; v dl = x dx + yz dy + y dz = x dx + x dx + x dx = 4x dx; v dl = 4x dx = 4x 3 /3 = 4/3 d v dl = 4/3 4/3 = c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS Poblem 3 x, y :, z = ; d = dx dy ẑ; v d = yz 3 dx dy = 3y dx dy; v d = 3 dx y dy = 3x y = 3 = In Ex 7 we got, fo the sme boundy line the sue in the xy-plne, so the nswe is no: the sufce integl does not depend only on the boundy line The totl flux fo the cube is + = 3 Poblem 3 T dτ = z dx dy dz You cn do the integls in ny ode hee it is simplest to sve z fo lst: z ] dx dy dz The sloping sufce is x+y +z =, so the x integl is y z dx = y z Fo given z, y nges fom to z, so the y integl is z y z dy = zy y /] z = z z /] = z / = / z + z / Finlly, the z integl is z z + z dz = 6 4 + = /6 Poblem 3 T b = + 4 + = 7; T = T b T = 7 z z3 + z4 z3 dz = 6 z4 4 + z5 = T = x + 4yˆx + 4x + z 3 ŷ + 6yz ẑ; T dl = x + 4ydx + 4x + z 3 dy + 6yz dz Segment : x :, y = z = dy = dz = T dl = x dx = x = Segment : y :, x =, z =, dx = dz = T dl = 4 dy = 4y = 4 b Segment 3: z :, x = y =, dx = dy = T dl = T dl = 7 6z dz = z 3 = b Segment : z :, x = y = dx = dy = T dl = dz = Segment : y :, x =, z =, dx = dz = T dl = dy = y = Segment 3: x :, y = z =, dy = dz = T dl = b T dl = 7 x + 4 dx = x + 4x = + 4 = 5 c x :, y = x, z = x, dy = dx, dz = x dx T dl = x + 4xdx + 4x + x 6 dx + 6xx 4 x dx = x + 4x 6 dx b T dl = x + 4x6 dx = 5x + x 7 = 5 + = 7 Poblem 33 v = y + z + 3x { } vdτ = y + z + 3x dx dy dz = y + z + 3x dx dy dz y + zx + 3 x] = y + z + 6 = { } y + 4z + 6dy dz y + 4z + 6y ] = 4 + 4z + 6 = 8z + 6 = 8z + 6dz = 4z + 6z = 6 + 3 = 48 Numbeing the sufces s in Fig 9: c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 3 i d = dy dz ˆx, x = v d = y dy dz v d = y dy dz = y = 8 ii d = dy dz ˆx, x = v d = v d = iii d = dx dz ŷ, y = v d = 4z dx dz v d = 4z dx dz = 6 iv d = dx dz ŷ, y = v d = v d = v d = dx dy ẑ, z = v d = 6x dx dy v d = 4 vi d = dx dy ẑ, z = v d = v d = v d = 8 + 6 + 4 = 48 Poblem 34 v = ˆx y + ŷ 3z + ẑ x = y ˆx 3z ŷ x ẑ d = dy dz ˆx, if we gee tht the pth integl shll un counteclockwise So v d = y dy dz { } z v d = ydy dz y z = z = 4 4z + z dz = 4z z + z3 3 = 8 8 + 3 8 = 8 3 y Menwhile, v dl = xydx + yzdy + 3zxdz Thee e thee segments z 3 y z y = z x = z = ; dx = dz = y : v dl = x = ; z = y; dx =, dz = dy, y : v dl = yz dy v dl = y ydy = 4y y dy = y 3 y3 = 8 3 8 = 8 3 3 x = y = ; dx = dy = ; z : v dl = v dl = So v dl = 8 3 Poblem 35 By Coolly, v d should eul 4 3 v = 4z xˆx + z ẑ i d = dy dz ˆx, x = ; y, z : v d = 4z dy dz; v d = 4z dz = 4 3 z3 z = 4 3 = 3 ii d = dx dy ẑ, z = ; x, y : v d = ; v d = iii d = dx dz ŷ, y = ; x, z : v d = ; v d = iv d = dx dz ŷ, y = ; x, z : v d = ; v d = v d = dx dy ẑ, z = ; x, y : v d = dx dy; v d = v d = 3 + = 4 3 c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
4 CHAPTER VECTOR ANALYSIS Poblem 36 Use the poduct ule fa = f A A f : f A d = fa d + A f] d = S I used Stokes theoem in the lst step S S P fa dl + A f] d ed S b Use the poduct ule A B = B A A B : B Adτ = A B dτ + A B dτ = V V I used the divegence theoem in the lst step Poblem 37 = x + y + z ; θ = cos z ; φ = tn y x +y +z x V S A B d + A B dτ V ed Poblem 38 Thee e mny wys to do this one pobbly the most illuminting wy is to wok it out by tigonomety fom Fig 36 The most systemtic ppoch is to study the expession: = x ˆx + y ŷ + z ẑ = sin θ cos φ ˆx + sin θ sin φ ŷ + cos θ ẑ If I only vy slightly, then d = d is shot vecto pointing in the diection of incese in To mke it unit vecto, I must divide by its length Thus: ˆ = ; ˆθ = θ θ ; ˆφ = = sin θ cos φ ˆx + sin θ sin φ ŷ + cos θ ẑ; = sin θ cos φ + sin θ sin φ + cos θ = θ = cos θ cos φ ˆx + cos θ sin φ ŷ sin θ ẑ; θ = cos θ cos φ + cos θ sin φ + sin θ = φ = sin θ sin φ ˆx + sin θ cos φ ŷ; φ = sin θ sin φ + sin θ cos φ = sin θ ˆ = sin θ cos φ ˆx + sin θ sin φ ŷ + cos θ ẑ ˆθ = cos θ cos φ ˆx + cos θ sin φ ŷ sin θ ẑ ˆφ = sin φ ˆx + cos φ ŷ φ φ Check: ˆ ˆ = sin θcos φ + sin φ + cos θ = sin θ + cos θ =, ˆθ ˆφ = cos θ sin φ cos φ + cos θ sin φ cos φ =, etc sin θ ˆ = sin θ cos φ ˆx + sin θ sin φ ŷ + sin θ cos θ ẑ cos θ ˆθ = cos θ cos φ ˆx + cos θ sin φ ŷ sin θ cos θ ẑ Add these: sin θ ˆ + cos θ ˆθ = + cos φ ˆx + sin φ ŷ; ˆφ = sin φ ˆx + cos φ ŷ Multiply by cos φ, by sin φ, nd subtct: ˆx = sin θ cos φˆ + cos θ cos φ ˆθ sin φ ˆφ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 5 Multiply by sin φ, by cos φ, nd dd: ŷ = sin θ sin φˆ + cos θ sin φ ˆθ + cos φ ˆφ cos θ ˆ = sin θ cos θ cos φ ˆx + sin θ cos θ sin φ ŷ + cos θ ẑ sin θ ˆθ = sin θ cos θ cos φ ˆx + sin θ cos θ sin φ ŷ sin θ ẑ Subtct these: ẑ = cos θ ˆ sin θ ˆθ Poblem 39 v = = 4 3 = 4 v dτ = 4 sin θ d dθ dφ = 4 R v d = ˆ sin θ dθ dφˆ = 4 sin θ dθ = v dτ = 3 d sin θ dθ dφ = 4 R 4 4 = 4R 4 dφ = 4R 4 Note: t sufce of sphee = R b v = v d = ˆ sin θ dθ dφˆ = sin θ dθ dφ = 4 They don t gee! The point is tht this divegence is zeo except t the oigin, whee it blows up, so ou clcultion of v is incoect The ight nswe is 4 Poblem 4 v = cos θ + sin θ θ sin θ sin θ + sin θ φ sin θ cos φ = 3 cos θ + sin θ sin θ cos θ + sin θ sin θ sin φ = 3 cos θ + cos θ sin φ = 5 cos θ sin φ vdτ = 5 cos θ sin φ sin θ d dθ dφ = R d θ ] 5 cos θ sin φ dφ dθ sin θ R = 3 3 = 5 3 R3 sin θ cos θ dθ sin θ = 5 cos θ Two sufces one the hemisphee: d = R sin θ dθ dφˆ; = R; φ :, θ : v d = cos θr sin θ dθ dφ = R 3 sin θ cos θ dθ dφ = R 3 = R 3 othe the flt bottom: d = d sin θ dφ+ˆθ = d dφ ˆθ hee θ = : R, φ : R v d = sin θ d dφ = d dφ = R3 3 Totl: v d = R 3 + 3 R3 = 5 3 R3 Poblem 4 t = cos θ + sin θ cos φˆ + sin θ + cos θ cos φˆθ + / sin θ sin/ θsin φ ˆφ t = t = cos θ + sin θ cos φ + sin θ θ sin θ sin θ + cos θ cos φ + sin θ = cos θ + sin θ cos φ + sin θ sin θ cos θ + cos θ cos φ sin θ cos φ = sin θ sin θ cos θ + sin θ cos φ sin θ cos θ + cos θ cos φ sin θ cos φ cos φ] sin θ + cos θ cos φ cos φ ] = = sin θ φ sin φ sin θ cos φ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
6 CHAPTER VECTOR ANALYSIS t = Check: cos θ = z, sin θ cos φ = x in Ctesin coodintes t = x + z Obviously Lplcin is zeo Gdient Theoem: b t dl = tb t Segment : θ =, φ =, : dl = d ˆ; t dl = cos θ + sin θ cos φd = + d = d t dl = d = Segment : θ =, =, φ : dl = sin θ dφ ˆφ = dφ ˆφ t dl = sin φ dφ = sin φ dφ t dl = sin φ dφ = cos φ = Segment 3: =, φ = ; θ : dl = dθ ˆθ = dθ ˆθ; t dl = sin θ + cos θ cos φ dθ = sin θ dθ t dl = sin θ dθ = cos θ = Totl: b t dl = + = Menwhile, tb t = + ] ] = Poblem 4 Fom Fig 4, ŝ = cos φ ˆx + sin φ ŷ; ˆφ = sin φ ˆx + cos φ ŷ; ẑ = ẑ Multiply fist by cos φ, second by sin φ, nd subtct: ŝ cos φ ˆφ sin φ = cos φ ˆx + cos φ sin φ ŷ + sin φ ˆx sin φ cos φ ŷ = ˆxsin φ + cos φ = ˆx So ˆx = cos φ ŝ sin φ ˆφ Multiply fist by sin φ, second by cos φ, nd dd: ŝ sin φ + ˆφ cos φ = sin φ cos φ ˆx + sin φ ŷ sin φ cos φ ˆx + cos φ ŷ = ŷsin φ + cos φ = ŷ So ŷ = sin φ ŝ + cos φ ˆφ ẑ = ẑ Poblem 43 v = s s s s + sin φ + s φ s sin φ cos φ + 3z = s s + sin φ + s scos φ sin φ + 3 = 4 + sin φ + cos φ sin φ + 3 = 4 + sin φ + cos φ + 3 = 8 b vdτ = 8s ds dφ dz = 8 s ds dφ 5 dz = 8 5 = 4 Menwhile, the sufce integl hs five pts: top: z = 5, d = s ds dφ ẑ; v d = 3z s ds dφ = 5s ds dφ v d = 5 s ds dφ = 5 bottom: z =, d = s ds dφ ẑ; v d = 3z s ds dφ = v d = bck: φ =, d = ds dz ˆφ; v d = s sin φ cos φ ds dz = v d = left: φ =, d = ds dz ˆφ; v d = s sin φ cos φ ds dz = v d = font: s =, d = s dφ dz ŝ; v d = s + sin φs dφ dz = 4 + sin φdφ dz v d = 4 + sin φdφ 5 dz = 4 + 4 5 = 5 So v d = 5 + 5 = 4 c v = s φ 3z s sin φ cos φ ŝ + s + sin φ s 3z ˆφ + s s s sin φ cos φ φ s + sin φ ẑ s sin φ cos φ s sin φ cos φ ẑ = = s c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 7 Poblem 44 33 3 = 7 6 = b cos = - c zeo d ln + 3 = ln = zeo Poblem 45 x + 3 3 δx dx = 3 + 3 = b By E 94, δ x = δx, so + 3 + = 6 c 9x 3 δx + 3 dx = 9 3 3 = 3 d if > b, if < b Poblem 46 fx x d dx δx] dx = x fxδx The fist tem is zeo, since δx = t ± ; d So the integl is So, x d dxδx = δx ed x df dx + f dx d dx x fx δx dx x fx = x df dx + dx dx f = x df δx dx = f = f = dx + f fxδx dx b dθ fx dx dx = fxθx df dx θxdx = f df dx dx = f f f = f = dθ fxδx dx So dx = δx ed Poblem 47 ρ = δ 3 Check: ρdτ = δ 3 dτ = b ρ = δ 3 δ 3 c Evidently ρ = Aδ R To detemine the constnt A, we euie Q = ρ dτ = Aδ R4 d = A 4R So A = Q 4R ρ = Q 4R δ R Poblem 48 + + = 3 b b 5 3 δ 3 dτ = 5 b = 5 4 + 3 = 5 c c = 5 + 9 + 4 = 38 > 36 = 6, so c is outside V, so the integl is zeo d e ˆx + ŷ + ẑ = ˆx + ŷ + ẑ = + = < 5 = 5, so e is inside V, nd hence the integl is e d e = 3,,,, = 6 + + = -4 Poblem 49 Fist method: use E 99 to wite J = e 4δ 3 dτ = 4e = 4 Second method: integting by pts use E 59 c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
8 CHAPTER VECTOR ANALYSIS J = = V ˆ e dτ + e 4 d + S ˆ e d But e = e ˆ = e ˆ ˆ e sin θ dθ dφˆ = 4 R e d + e R sin θ dθ dφ = 4 e R + 4e R = 4 e R + e + 4e R = 4 Hee R =, so e R = Poblem 5 F = + + x = ; F = + + = + + = 3 ˆx ŷ ẑ F = x = ŷ x ˆx ŷ ẑ = xŷ ; F = x y z = F is gdient; F is cul U = Ay Fo A, we wnt Az = A x x 3 + y + z would do F = U Az = ; A y Ax = x A y = x3 3, A x = A z = would do it A = 3 x ŷ F = A But these e not uniue b F 3 = yz + xz + xy = ; F ˆx ŷ ẑ 3 = = ˆx x x + ŷ y y + ẑ z z = yz xz xy So F 3 cn be witten s the gdient of scl F 3 = U 3 nd s the cul of vecto F 3 = A 3 In fct, U 3 = xyz does the job Fo the vecto potentil, we hve A z Ay A x Az A y Ax = yz, which suggests A z = 4 y z + fx, z; A y = 4 yz + gx, y = xz, suggesting = xy, so A x = 4 z x + hx, y; A z = 4 zx + jy, z y = 4 x y + ky, z; A x = 4 xy + lx, z Putting this ll togethe: A 3 = 4 { x z y ˆx + y x z ŷ + z y x ẑ } gin, not uniue Poblem 5 d : F = U = E 44 cul of gdient is lwys zeo c: F dl = F d = E 57 Stokes theoem c b: b F dl b I F dl = b II F dl + I b F dl = F dl =, so II b I F dl = b II F dl b c: sme s c b, only in evese; c : sme s c Poblem 5 d : F = W = E 46 divegence of cul is lwys zeo c: F d = F dτ = E 56 divegence theoem c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 9 c b: I F d II F d = F d =, so I F d = II F d Note: sign chnge becuse fo F d, d is outwd, whees fo sufce II it is inwd b c: sme s c b, in evese; c : sme s c Poblem 53 In Pob 5 we found tht v = ; in Pob 8 we found tht v c = So v c cn be witten s the gdient of scl; v cn be witten s the cul of vecto To find t: t = y t = y x + fy, z t = xy + z 3 t = yz f Fom & 3 we get = yz f = yz + gy t = y x + yz + gy, so t = xy + z + g = xy + z fom g = We my s well pick g = ; then t = xy + yz b To find W: W z Pick W x = ; then Wy = x ; W x Wz = 3z x; W y Wx = xz W z W z = 3xz W z = 3 x z + fy, z W y = xz W y = x z + gy, z Wy = f + x g = x f g = My s well pick f = g = W = x z ŷ 3 x z ẑ ˆx ŷ ẑ Check: W = x z 3 x z = ˆx x + ŷ 3xz + ẑ xz You cn dd ny gdient t to W without chnging its cul, so this nswe is f fom uniue Some othe solutions: W = xz 3 ˆx x z ŷ; W = xyz + xz 3 ˆx + x y ẑ; W = xyz ˆx x z ŷ + x y 3z ẑ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS Poblem 54 v = cos θ + sin θ sin θ cos φ + θ = 43 cos θ + sin θ cos θ cos φ + sin θ = cos θ sin θ 4 sin θ + cos φ cos φ] = 4 cos θ cos θ sin φ sin θ φ cos θ cos φ v dτ = 4 cos θ sin θ d dθ dφ = 4 = R 4 = R4 4 R 3 d / cos θ sin θ dθ / dφ Sufce consists of fou pts: Cuved: d = R sin θ dθ dφˆ; = R v d = R cos θ R sin θ dθ dφ v d = R 4 / cos θ sin θ dθ / dφ = R 4 = R4 4 Left: d = d dθ ˆφ; φ = v d = cos θ sin φ d dθ = v d = 3 Bck: d = d dθ ˆφ; φ = / v d = cos θ sin φ d dθ = 3 cos θ d dθ v d = R 3 d / cos θ dθ = 4 R4 + = 4 R4 4 Bottom: d = sin θ d dφ ˆθ; θ = / v d = cos φ d dφ R v d = 3 d / cos φ dφ = 4 R4 Totl: v d = R 4 /4 + 4 R4 + 4 R4 = R4 4 Poblem 55 ˆx ŷ ẑ v = y bx = ẑ b So v d = b R v dl = y ˆx + bx ŷ dx ˆx + dy ŷ + dz ẑ = y dx + bx dy; x + y = R x dx + y dy =, so dy = x/y dx So v dl = y dx + bx x/y dx = y y bx dx c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS Fo the uppe semicicle, y = R x, so v dl = R x bx R x dx v dl = R R R + bx { dx = R x = R b sin x/r = R b R +R R sin x R + b x R x + R x ]} R sin R +R = R b sin sin + = R b And the sme fo the lowe semicicle y chnges sign, but the limits on the integl e evesed so v dl = R b Poblem 56 x = z = ; dx = dz = ; y : v dl = yz dy = ; v dl = x = ; z = y; dz = dy; y : v dl = yz dy +3y +z dz = y y dy 3y + y dy; v dl = y y 3 4y 4 + y dy = 4y3 3 + y y = 4 3 3 x = y = ; dx = dy = ; z : v dl = 3y + z dz = z dz; v dl = z dz = z = Totl: v dl = + 4 3 = 8 3 Menwhile, Stokes theeom sys v dl = v d Hee d = dy dz ˆx, so ll we need is v x = 3y + z yz = 3 yz Theefoe v d = 3 yz dy dz = Poblem 57 Stt t the oigin = y 3 y y y] dy = = y 4 + 8 3 y3 5y + 6y ] 3 yz dz dy = + 8 3 5 + 6 = 8 3 4y 3 + 8y y + 6 dy θ =, φ = ; : v dl = cos θ d = v dl = =, θ = ; φ : / v dl = 3 sin θ dφ = 3 dφ / v dl = 3 dφ = 3 c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 3 φ = ; sin θ = y =, so = sin θ, d = sin θ cos θ dθ, θ : θ tn / v dl = cos θ d cos θ sin θ dθ = cos θ cos θ sin θ sin dθ θ cos 3 θ = sin 3 θ + cos θ dθ = cos θ cos θ + sin θ sin θ sin θ sin θ cos θ sin θ sin θ dθ = cos θ sin 3 θ dθ dθ Theefoe v dl = θ / cos θ sin 3 θ dθ = sin θ θ / = /5 = 5 = 4 θ = θ, φ = ; : 5 v dl = cos θ d = 4 5 d v dl = 4 d = 4 5 5 5 = 4 5 5 5 = Totl: v dl = + 3 + = 3 Stokes theoem sys this should eul v d v = ] sin θ 3 sin θ cos θ ˆ + cos θ ] sin θ θ φ sin θ φ 3 ˆθ + cos θ sin θ cos θ ] ˆφ θ = sin θ 3 cos θ]ˆ + 6] ˆθ + cos θ sin θ + cos θ sin θ] ˆφ = 3 cot θ ˆ 6 ˆθ Bck fce: d = d dθ ˆφ; v d = v d = Bottom: d = sin θ d dφ ˆθ; v d = 6 sin θ d dφ θ =, so v d = 6 d dφ v d = 6 d / dφ = 6 = 3 Poblem 58 v dl = y dz Left side: z = x; dz = dx; y = Theefoe v dl = Bottom: dz = Theefoe v dl = c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 3 3 Bck: z = y; dz = / dy; y : v dl = y dy = y = 4 4 = Menwhile, v = ˆx, so v d is the pojection of this sufce on the x y plne = = Poblem 59 v = sin θ + sin θ sin θ 4 cos θ + θ = 43 sin θ + sin θ 4 cos θ sin θ = = 4 cos θ sin θ 4 sin θ tn θ sin θ φ sin θ + cos θ sin θ v dτ = 4 cos θ sin θ d dθ dφ R = sin θ = R 4 sin 6 + = R4 3 + 3 4 6 4 3 d /6 = R4 cos θ dθ dφ = R 4 + 3 3 ] θ sin θ /6 + 4 Sufce coinsists of two pts: The ice cem: = R; φ : ; θ : /6; d = R sin θ dθ dφˆ; v d = R sin θ R sin θ dθ dφ = R 4 sin θ dθ dφ v d = R 4 /6 sin θ dθ dφ = R 4 θ ] /6 sin θ = R 4 4 4 sin 6 = R4 3 6 The cone: θ = 6 ; φ : ; : R; d = sin θ dφ d ˆθ = 3 d dφ ˆθ; v d = 3 3 d dφ Theefoe v d = R4 v d = 3 R 3 3 + 3 3 d = R4 dφ = 3 R4 4 = + 3 3 3 R4 Poblem 6 Coolly sys T dl = Stokes theoem sys T dl = T ] d So T ] d =, nd since this is tue fo ny sufce, the integnd must vnish: T =, confiming E 44 b Coolly sys v d = Divegence theoem sys v d = v dτ So v dτ =, nd since this is tue fo ny volume, the integnd must vnish: v =, confiming E 46 Poblem 6 Divegence theoem: v d = v dτ Let v = ct, whee c is constnt vecto Using poduct ule #5 in font cove: v = ct = T c + c T But c is constnt so c = Theefoe we hve: c T dτ = T c d Since c is constnt, tke it outside the integls: c T dτ = c T d But c 3 c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
4 CHAPTER VECTOR ANALYSIS is ny constnt vecto in pticul, it could be be ˆx, o ŷ, o ẑ so ech component of the integl on left euls coesponding component on the ight, nd hence T dτ = T d ed b Let v v c in divegence theoem Then v cdτ = v c d Poduct ule #6 v c = c v v c = c v Note: c =, since c is constnt Menwhile vecto indentity sys d v c = c d v = c v d Thus c v dτ = c v d Tke c outside, nd gin let c be ˆx, ŷ, ẑ then: v dτ = v d ed c Let v = T U in divegence theoem: T U dτ = T U d Poduct ule #5 T U = T U + U T = T U + U T Theefoe T U + U T dτ = T U d ed d Rewite c with T U : U T + T U dτ = U T d Subtct this fom c, noting tht the U T tems cncel: T U U T dτ = T U U T d ed e Stokes theoem: v d = v dl Let v = ct By Poduct Rule #7: ct = T c c T = c T since c is constnt Theefoe, c T d = T c dl Use vecto indentity # to ewite the fist tem c T d = c T d So c T d = c T dl Pull c outside, nd let c ˆx, ŷ, nd ẑ to pove: T d = T dl ed Poblem 6 d = R sin θ dθ dφˆ Let the sufce be the nothen hemisphee The ˆx nd ŷ components clely integte to zeo, nd the ẑ component of ˆ is cos θ, so = R sin θ cos θ dθ dφ ẑ = R ẑ / sin θ cos θ dθ = R ẑ sin θ / = R ẑ b Let T = in Pob 6 Then T =, so d = ed c This follows fom b Fo suppose ; then if you put them togethe to mke closed sufce, d = d Fo one such tingle, d = dl since dl is the e of the pllelogm, nd the diection is pependicul to the sufce, so fo the entie conicl sufce, = dl e Let T = c, nd use poduct ule #4: T = c = c + c But =, nd c = c x + c y + c z x ˆx + y ŷ + z ẑ = c x ˆx + c y ŷ + c z ẑ = c So Pob 6e sys T dl = c dl = T d = c d = c d = c = c ed c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER VECTOR ANALYSIS 5 Poblem 63 Fo sphee of dius R: v = = = v d = R ˆ R sin θ dθ dφˆ = R sin θ dθ dφ = 4R v dτ = sin θ d dθ dφ R = Evidently thee is no delt function t the oigin d sin θ dθ dφ = 4R n ˆ = n = n+ = n + n+ = n + n So divegence theoem checks except fo n =, fo which we ledy know E 99 tht the divegence is 4δ 3 Geometiclly, it should be zeo Likewise, the cul in the spheicl coodintes obviously gives zeo To be cetin thee is no luking delt function hee, we integte ove sphee of dius R, using Pob 6b: If n ˆ =, then v dτ =? = v d But v = n ˆ nd d = R sin θ dθ dφˆ e both in the ˆ diections, so v d = Poblem 64 Since the gument is not function of ngle, E 73 sys D = d 4 ] = d d + ɛ 3/ 4 d = 3 4 + ɛ 3 3 ] 3/ + ɛ 5/3 b Setting : 3 + ɛ 3/ = 4 3 + ɛ 5/ + ɛ = D, ɛ = 3ɛ 4ɛ 5 = 3 4ɛ 3, which goes to infinity s ɛ c Fom it is cle tht D, = fo d D, ɛ 4 d = 3ɛ d = + ɛ 3ɛ 5/ 3ɛ = I looked up the integl Note tht b, c, nd d e the defining conditions fo δ 3 ] 3ɛ 4 + ɛ 5/ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
6 CHAPTER ELECTROSTATICS Chpte Electosttics Poblem Zeo b F = Q, whee is the distnce fom cente to ech numel F points towd the missing 4ɛ Explntion: by supeposition, this is euivlent to, with n ext t 6 o clock since the foce of ll twelve is zeo, the net foce is tht of only c Zeo Q d, pointing towd the missing Sme eson s b Note, howeve, tht if you explined b s 4ɛ cncelltion in pis of opposite chges o clock ginst 7 o clock; ginst 8, etc, with one unpied doing the job, then you ll need diffeent explntion fo d Poblem This time the veticl components cncel, leving E = 4ɛ sin θ ˆx, o E = 4ɛ z + d d 3/ ˆx E θ z x Fom f wy, z d, the field goes like E z ẑ, which, s we shll see, is the field of dipole If we 3 set d, we get E =, s is ppopite; to the extent tht this configution looks like single point chge fom f wy, the net chge is zeo, so E 4ɛ d c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER Poblem 3 ELECTROSTATICS 7 Poblem 3 Poblem 3 z E z = L λ dx 4ɛ η cos θ; η = z + x ; cos θ = z z η E z E= z = L L λ dx λ dx 4ɛ = 4ɛ λz cos θ; = z + x ; cos θ = zθη 4ɛ z η cos θ; η = z + x ; cos θ = z = λdx η d L L } x {{} x dx = z θ +x = 3/ ] z 4ɛ d = λdx = λz L 4ɛ λz L dx z dx z x 4ɛ λz z +x L +x 3/ ] 3/ ] = λ L z +x 4ɛ z z +L = 4ɛ L E x = L λ dx 4ɛ }{{} η sin θ = 4ɛ λ λz z x L = z = λ L 4ɛ +x 4ɛ z x dx λz z x L = λ L z +x 4ɛ z z +L x +z E ] 3/ x = L λ dx x x = 4ɛ λ L ] 4ɛ sin θ = = 4ɛ λ z +L E x = L λ dx x dx 4ɛ η sin θ = 4ɛ λ x dx x x x +z 4ɛ λ +z z ] +z 3/ z +L = 4ɛ E = λ L ] 3/ ] = x = +z ] 4ɛ λ z 4ɛ λ L ] = x +z 4ɛ z λ z z +L +L λ z L + ˆx + ẑ E = 4ɛ ] E = ] λ λ z z + z z L + + L ˆx + ˆx + z L + L ẑ ẑ 4ɛ 4ɛ z Fo z L you z z expect it to z + look + L L z like point z + chge + L L = λl: E λl 4ɛ z ẑ It checks, fo with z L the ˆx Fo Fo z tem z L Lyou, youexpect nd expect it the ẑ tem it to tolook looklike like λ L point 4ɛ z z ẑ point chge chge = = λl: λl: EE λl λl 4ɛ 4ɛ z ẑ z ẑ It It checks, checks, fo fo with withz z L Lthe the ˆx ˆx tem tem,, nd ndthe the ẑ tem ẑ tem λ L λ L 4ɛ 4ɛ Poblem 4 z ẑ z z ẑ Poblem Poblem4 Fom Ex 4, with L z nd z + distnce fom cente of edge to P, field of one edge is: Fom FomEx Ex,, with withl L z nd z + z nd z + distnce E = distnce fom fomcente cente of of edge edge to to P, P, field fieldof of one oneedge edge is: is: λ E E 4ɛ = = λ λ z + 4 z + 4ɛ 4 + 4ɛ 4 z z + + 4 z 4 z + + 4 + 4 z Thee e 4 sides, nd we wnt veticl components only, so multiply + 4 4 by 4 cos θ = 4 : Thee Thee e e 4 sides, 4 sides, nd nd we we wnt wnt veticl veticl components components only, only, so so multiply multiply by by 4 cos 4 cos θ = θ = 4 z 4 z + : 4 : E = z + z 4λz + 4 4ɛ 4 ẑ E E = = 4λz 4λz 4ɛ z + 4 z + 4ɛ ẑ ẑ z z + + z z + + Poblem 5 Poblem 5 Poblem 5 θ θ z z { Hoizontl components cncel, leving: E = { { λdl 4ɛ η cos ẑ Hee, η = + z, cos θ = z η both constnts, E = λdl } } Hoizontl components cncel, leving: E = 4ɛ while λdl η cos θ ẑ 4ɛ dl = So η = + z, cos θ = z η both while cos θ ẑ Hee, = + z, cos θ = z both constnts, while dl = dl = So So E E = = λz λz 4ɛ 4ɛ 4ɛ z + + z z 4 4 ẑ 3/ ẑ ẑ 3/ 3/ Poblem 6 Bek it into ings of dius, nd thickness d, nd use Pob 5 to expess the field of ech ing Totl chge of ing is σ d = λ, so λ = σd is the line chge of ech ing E ing = σdz 4ɛ + z ; E 3/ disk = R σz 4ɛ + z E disk = ] σz 4ɛ z ẑ R + z c 5 Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is c 5 potected Peson unde Eduction, ll copyight Inc, lws Uppe s Sddle they cuently Rive, NJ exist AllNo ights potion eseved of this This mteil mteil myis be potected epoduced, unde inll ny copyight fom o lws by ny s mens, they cuently withoutexist pemission No potion in witing of this fom mteil the publishe my be 3/ d
8 CHAPTER ELECTROSTATICS Fo R z the second tem, so E plne = 4ɛ σẑ = / Fo z R, = R +z z + R z z nd E = R σ 4ɛ z = Q 4ɛ z, whee Q = R σ Poblem 7 E is clely in the z diection Fom the digm, d = σd = σr sin θ dθ dφ, = R + z Rz cos θ, z R cos θ cos ψ = So σ ɛ ẑ R z, so ] z z + R z = R 3 z, 3 z E z = σr sin θ dθ dφz R cos θ 4ɛ R + z Rz cos θ dφ = 3/ = 4ɛ R σ = 4ɛ R σ = 4ɛ R σ z z x z R cos θ sin θ dθ Let u = cos θ; du = sin θ dθ; R + z Rz cos θ 3/ zu R R + z Rzu Fo z > R outside the sphee, E z = 4ɛ 4R σ z = 4ɛ Fo z < R inside, E z =, so E = φ ψ θ R y { } θ = u = + θ = u = z Ru du Integl cn be done by ptil fctions o look it up R + z Rzu 3/ ] = R { } σ z R z R 4ɛ z z R z + R z, so E = 4ɛ z ẑ Poblem 8 Accoding to Pob 7, ll shells inteio to the point ie t smlle contibute s though thei chge wee concentted t the cente, while ll exteio shells contibute nothing Theefoe: E = Q int 4ɛ ˆ, whee Q int is the totl chge inteio to the point Outside the sphee, ll the chge is inteio, so E = 4ɛ Q ˆ Inside the sphee, only tht fction of the totl which is inteio to the point counts: Q int = 4 3 3 4 3 3 3 Q = Q, so E = R3 R3 4ɛ R 3 Q ˆ = Q 4ɛ R 3 Poblem 9 ρ = ɛ E = ɛ k 3 = ɛ k5 4 = 5ɛ k c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
3 CHAPTER ELECTROSTATICS Poblem 5 i Q enc =, so E = ii E d = E4 = ɛ Q enc = ɛ ρ dτ = ɛ k sin θ d dθ dφ ˆ = 4k ɛ 4k d = ɛ E = k ɛ E iii E4 = 4k ɛ E = k ɛ b b ˆ 4k d = ɛ b, so b Poblem 6 i l Gussin sufce E d = E s l = ɛ Q enc = ɛ ρs l; E = ρs ɛ ŝ ii s l Gussin sufce E d = E s l = ɛ Q enc = ɛ ρ l; E = ρ ɛ sŝ iii s Gussin sufce E d = E s l = ɛ Q enc = ; E = l E Poblem 7 On the x z plne E = by symmety Set up Gussin pillbox with one fce in this plne nd the othe t y b s y Gussin pillbox E d = E A = ɛ Q enc = ɛ Ayρ; E = ρ ɛ y ŷ fo y < d c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER ELECTROSTATICS 3 Q enc = ɛ Adρ E = ρ ɛ d ŷ fo y > d d E ρd ɛ d y Poblem 8 Fom Pob, the field inside the positive sphee is E + = ρ 3ɛ +, whee + is the vecto fom the positive cente to the point in uestion Likewise, the field of the negtive sphee is ρ 3ɛ So the totl field is But see digm + = d So E = Poblem 9 E = 4ɛ ˆ ρ dτ = = since ˆ E = ρ 3ɛ d ρ 3ɛ + + d + ] ˆ ρ dτ since ρ depends on, not 4ɛ =, fom Pob 63 Poblem ˆx ŷ ẑ E = k = k ˆx y + ŷ 3z + ẑ x], xy yz 3zx so E is n impossible electosttic field ˆx ŷ ẑ E = k = k ˆxz z + ŷ + ẑy y] =, y xy + z yz so E is possible electosttic field Let s go by the indicted pth: z E dl = y dx + xy + z dy + yz dzk Step I: y = z = ; dy = dz = E dl = ky dx = Step II: x = x, y : y, z = dx = dz = E dl = kxy + z dy = kx y dy II E dl = kx y y dy = kx y Step III : x = x, y = y, z : z ; dx = dy = E dl = kyz dz = ky z dz I II x x, y, z III y c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
3 CHAPTER ELECTROSTATICS III E dl = y k z z dz = ky z x,y,z V x, y, z = E dl = kx y + y z, o V x, y, z = kxy + yz Check: V =k xy +yz ˆx+ xy +yz ŷ+ xy +yz ẑ]=ky ˆx+xy+z ŷ+yz ẑ]=e Poblem V = E dl So fo > R: V = nd fo < R: V = R When > R, V = When < R, V = Outside the sphee > R : E = 4ɛ Inside the sphee < R : E = 4ɛ = 4ɛ R 4ɛ 4ɛ R 4ɛ ˆ d = 4ɛ = 4ɛ, 4ɛ 3 d R R ˆ = 4ɛ 3 R ˆ = 4ɛ R 3 d = ˆ, so E = V = 4ɛ R R 3 ˆ 4ɛ R R 3 4ɛ ˆ R ˆ = 4ɛ ] R R 3ˆ; so E = V = 4ɛ R 3 ˆ V 6 4 8 6 In the figue, is in units of R, nd V is in units of /4ɛ R 4 5 5 5 3 Poblem E = λ 4ɛ s ŝ Pob 3 In this cse we cnnot set the efeence point t, since the chge itself extends to Let s set it t s = Then V s = s λ 4ɛ s d s = s λ ln 4ɛ In this fom it is cle why = would be no good likewise the othe ntul point, = V = 4ɛ λ s ln s ŝ = 4ɛ λ s ŝ = E Poblem 3 V = E dl = b k b ɛ d k b ɛ d d = k b ɛ b k ɛ ln b + b { = k ɛ b ln b + b } = k ɛ ln b Poblem 4 Using E nd the fields fom Pob 6: V b V = b E dl = E dl b E dl = ρ ρ ɛ s ds ɛ b s ds c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER ELECTROSTATICS 33 = ρ s ɛ + ρ ɛ Poblem 5 V = 4ɛ b V = 4ɛ L L = λ 4ɛ ln z + d λ dx = λ z +x 4ɛ ln s b = ρ 4ɛ + ln b lnx + z + x L L L + ] z + L L + = λ L + z ln + L z + L ɛ z z c V = R σ d 4ɛ + z = σ 4ɛ + z R = σ R + z ɛ z x In ech cse, by symmety V = V = E = V ẑ E = 4ɛ z z + d 3/ ẑ = 4ɛ z + d z 3/ ẑ gees with Ex b E = λ { } 4ɛ L + z + L z + L z L + z + L z + L z ẑ { = λ z L + z + L L } z + L 4ɛ z + L z + L L ẑ = Lλ 4ɛ z ẑ gees with Ex z + L c E = σ ɛ { } R + z z ẑ = σ ɛ ] z ẑ gees with Pob 6 R + z If the ight-hnd chge in is, then V =, which, nively, suggests E = V =, in contdiction with the nswe to Pob The point is tht we only know V on the z xis, nd fom this we cnnot hope to compute E x = V o E y = V Tht ws OK in pt, becuse we knew fom symmety tht E x = E y = But now E points in the x diection, so knowing V on the z xis is insufficient to detemine E Poblem 6 b h h V = 4ɛ h σ d = σ h = σh 4ɛ ɛ whee = / c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
34 CHAPTER ELECTROSTATICS V b = 4ɛ h = σ 4ɛ h = = σ ɛ σ ɛ σ d whee = h + h h + h d h + h + h ln h + h + h] h + h lnh + h h h h lnh ] h ] = σh + ln 4ɛ = σ ɛ h lnh + h lnh h = σh ɛ ln + V V b = σh ln + ] ɛ = σh ln 4ɛ h + Poblem 7 V = Cut the cylinde into slbs, s shown in the figue, nd use esult of Pob 5c, with z x nd σ ρ dx: ρ z+l/ ɛ z L/ = ρ ɛ 8 < = ρ 4ɛ R + x x dx x R + x + R lnx + R + x x ] z+l/ : z+ L R +z+ L z L R +z L +R ln z L/ 4 z+ L + R +z+ L z L + R + z L 3 9 = 5 zl ; z {}} L { L {}}{ }{{ } x dx Note: z + L + z L = z zl L 4 + z zl + L 4 = zl { E = V = ẑ V = ẑρ R 4ɛ + z + L + E = ẑρ 4ɛ R + z + L R + z L L z + L R + R + z + L z L z L R + z L z+ + L z + L ] } + R R +z+ L R z + L R + + +z L z + L z L R + + L z L }{{} R + z + L R + z L c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
Hello CHAPTER ELECTROSTATICS 35 = ρ L R ɛ + z + L + R + z L ẑ Poblem 8 z Oient xes so P is on z xis { P V = ρ Hee ρ is constnt, dτ = sin θ d dθ dφ, 4ɛ dτ z = z + z cos θ θ V = ρ 4ɛ sin θ d dθ dφ ; z dφ = + z cos θ sin θ dθ = z + z cos θ z + z z cos θ = z V = ρ 4ɛ { z = z + z z = { /z, if < z, /, if > z R z d + z d } x + z + z + z z } { } = ρ z 3 ɛ z 3 + R z = ρ ɛ R z 3 φ y But ρ = Poblem 9 4, so V z = 3 R3 ɛ 3 4R 3 R z 3 = 8ɛ R 3 z R ; V = 3 8ɛ R R V = 4ɛ ρ dτ = 4ɛ ρ dτ since ρ is function of, not = 4ɛ ρ 4δ 3 ] dτ = ɛ ρ Poblem 3 Ex 5: E bove = σ ɛ ˆn; E below = σ ɛ ˆn ˆn lwys pointing up; E bove E below = σ ɛ ˆn Ex 6: At ech sufce, E = one side nd E = σ ɛ othe side, so E = σ ɛ Pob : E out = σr ɛ ˆ = σ ɛ ˆ ; E in = ; so E = σ ɛ ˆ b s R Outside: E d = Esl = ɛ Q enc = σ ɛ Rl E = σ ɛ R s ŝ = σ ɛ ŝ t sufce c 5 Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is potected }{{ unde } ll copyight Inside: lws Q enc s they =, cuently so E = exist No E potion = σ ɛ of ŝ this mteil my be l c V out = R σ ɛ = Rσ ɛ t sufce; V in = Rσ ɛ ; so V out = V in V out = R σ ɛ = σ ɛ t sufce; Vin = ; so Vout Vin = σ ɛ c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
36 CHAPTER ELECTROSTATICS Poblem 3 V = 4ɛ i ij = W 4 = V = { 4ɛ + + 4ɛ + } = 4ɛ + 4 + b W =, W = 4ɛ W tot = 4ɛ ; W 3 = 4ɛ { } + + = 4ɛ ; W 4 = see + + 3 Poblem 3 Consevtion of enegy kinetic plus potentil: At elese v A = v B =, =, so m AvA + m BvB + A B 4ɛ E = 4ɛ A B When they e vey f pt the potentil enegy is zeo, so = E m AvA + m BvB = A B 4ɛ Menwhile, consevtion of momentum sys m A v A = m B v B, o v B = m A /m B v A So m Av A + m B v A = ɛ ma m B A B m A + m B Poblem 33 Fom E 4, the enegy of one chge is v A = ma m B ma m B ; v B = m A + m B v A = 4ɛ A B ɛ A B m A + m B mb m A W = V = n= n = 4ɛ n 4ɛ n n The fcto of out font counts the chges to the left s well s to the ight of The sum is ln you cn get it fom the Tylo expnsion of ln + x: ln + x = x x + 3 x3 4 x4 + with x = Evidently α = ln c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER ELECTROSTATICS 37 Poblem 34 W = ρv dτ Fom Pob o Pob 8: V = W = ρ 4ɛ R R = ρ 5ɛ R = R 5ɛ b W = ɛ W = ɛ 3 R 4 = 3 R3 4 d = 4ɛ 3 5 R ρ 4ɛ R E dτ Outside > R E = 4ɛ { 4ɛ = 4ɛ = { R 4 4 d + R + } 5 R R 6 5 R ρ ɛ R 3 = 4ɛ 3 3 3 ] 5 R R = ρ 5 4ɛ R ˆ ; Inside < R E = 4ɛ } 4 R 3 d R 3 R R 3 R3 5 R 3 ˆ = 4ɛ R + = 3 5R 4ɛ 5 R { c W = ɛ S V E d + V E dτ }, whee V is lge enough to enclose ll the chge, but othewise bity Let s use sphee of dius > R Hee V = 4ɛ { W = ɛ R 4ɛ 4ɛ sin θ dθ dφ + E dτ + 4ɛ 4 d} = ɛ { 4ɛ 4 + 4 4ɛ 5R + 4ɛ 4 = { 4ɛ + 5R + } = 3 R 4ɛ 5 R As, the contibution fom the sufce integl 4ɛ 6 5R picks up the slck Poblem 35 dw = d V = d 4ɛ, = 4 3 3 ρ = 3 R 3 } R 4ɛ = chge on sphee of dius = totl chge on sphee d = 4 d ρ = 4 3 4 d = 3R3 R 3 d dw = 3 3 4ɛ R 3 R 3 d = 3 4ɛ W = 3 4ɛ R 6 R R 6 4 d 4 d = 4ɛ 3 R 6 R 5 5 = 4ɛ 3 5 R R goes to zeo, while the volume integl d c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
38 CHAPTER ELECTROSTATICS Poblem 36 W = ɛ W = ɛ b W = E E = E dτ E = 4ɛ b 4ɛ 4 d = 8ɛ 4ɛ 8ɛ < < b, zeo elsewhee = 8ɛ b 8ɛ b W tot = W + W + ɛ E E dτ = 8ɛ + b b Poblem 37, W = b, E = 4ɛ ˆ >, E = ˆ > b So, > b, nd hence E 4 E dτ = 4ɛ b = 8ɛ b 4ɛ 4 4 d = 4ɛ b z b x y whee fom the figue Theefoe E = 4ɛ ˆ; E = 4ɛ ˆ ; W i = ɛ 4ɛ = + cos θ, cos β = cos β sin θ d dθ dφ, cos θ W i = cos θ 4 ɛ 3 sin θ d dθ It s simplest to do the integl fist, chnging vibles to : As :, :, so The integl is /, so d = cos θ d cos θ d = d W i = 8ɛ W i = 8ɛ d sin θ dθ sin θ dθ = 4ɛ Of couse, this is pecisely the intection enegy of two point chges Poblem 38 σ R = 4R ; σ = 4 ; σ b = 4b c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER ELECTROSTATICS 39 b V = E dl = b 4ɛ d b d R 4ɛ d R d = 4ɛ b + R c σ b the chge dins off ; V = d R 4ɛ d R d = 4ɛ R Poblem 39 σ = 4 ; σ b = b 4b ; σ R = + b 4R b E out = + b 4ɛ ˆ, whee = vecto fom cente of lge sphee c E = 4ɛ ˆ, E b = b ˆ b, whee b is the vecto fom cente of cvity b 4ɛ d Zeo b e σ R chnges but not σ o σ b ; E outside chnges but not E o E b ; foce on nd b still zeo Poblem 4 No Fo exmple, if it is vey close to the wll, it will induce chge of the opposite sign on the wll, nd it will be ttcted b No Typiclly it will be ttctive, but see footnote fo extodiny counteexmple Poblem 4 Between the pltes, E = ; outside the pltes E = σ/ɛ = Q/ɛ A So P = ɛ E = ɛ Q ɛ = A Q ɛ A Poblem 4 Inside, E = ; outside, E = 4ɛ Q ˆ; so E ve = 4ɛ Q R ˆ; f z = σe ve z ; σ = Q 4R z θ E F z = f z d = Q Q 4R 4ɛ R cos θ R sin θ dθ dφ = Q / ɛ 4R sin θ cos θ dθ = Q ɛ 4R sin θ / = Q Q ɛ 4R = 3R ɛ Poblem 43 Sy the chge on the inne cylinde is Q, fo length L The field is given by Guss s lw: E d = E s L = ɛ Q enc = ɛ Q E = Q ɛ L s ŝ Potentil diffeence between the cylindes is V b V = b E dl = Q ɛ L As set up hee, is t the highe potentil, so V = V V b = b s ds = Q b ɛ L ln Q ɛ L ln b c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
4 CHAPTER ELECTROSTATICS C = Q V ln, so cpcitnce pe unit length is ɛ b ln b = ɛl Poblem 44 W = foce distnce = pessue e distnce = ɛ E Aɛ b W = enegy pe unit volume decese in volume = enegy lost is eul to the wok done Poblem 45 ɛ E Aɛ Sme s, confiming tht the Fom Pob 4, the field t height z bove the cente of sue loop side is Hee λ σ d E = 4ɛ z + 4 4λz z + see figue, nd we integte ove fom to ā: ẑ d +d d E = ā σz 4ɛ = ā /4 4σz 4ɛ = σ ɛ {tn z + 4 d z + du u + z u + z = σz ɛ ā + } z tn ; z Let u =, so d = du 4 z tn ]ā /4 u + z z E = σ ] tn + ɛ z ẑ = σ tn 4 ɛ 4z ẑ z + / ] infinite plne: E = σ ɛ tn 4 = σ ɛ 4 = σ ɛ z point chge: Let fx = tn + x 4, nd expnd s Tylo seies: Hee f = tn 4 = 4 4 = ; f x = Thus since z = x, E σ fx = f + xf + x f + ɛ 4 z ++x +x = fx = 4 x + x + x 3 + = 4ɛ σ z = 4ɛ z +x +x, so f = 4, so c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER ELECTROSTATICS 43 The double integl is pue numbe; Mthemtic sys it is So V = σr ɛ Poblem 5 Potentil of +λ is V + = λ ɛ ln s +, whee s+ is distnce fom λ + Pob Potentil of λ is V = + λ ɛ ln s, whee s is distnce fom λ Totl V = λ ɛ ln s s + Now s + = y + z, nd s = y + + z, so V x, y, z = λ y+ +z ɛ ln = λ y + + z ] ln y +z 4ɛ y + z s s + y λ λ b Euipotentils e given by y+ +z y +z = e 4ɛV/λ = k = constnt Tht is: y + y + + z = ky y + + z y k + z k + k yk + =, o y + z + y = The eution fo cicle, with cente t y, nd dius R, is k+ k y y + z = R, o y + z + y R yy = Evidently the euipotentils e cicles, with y = = y R R = y = k+ k R = k k ; o, in tems of V : z k+ k nd = k +k+ k +k k = 4k k, o y = e4ɛv/λ + eɛv/λ e 4ɛV/λ = + e ɛv/λ ɛ V = coth e ɛv/λ e ɛv/λ λ R = eɛv/λ e 4ɛV/λ = e ɛv/λ e ɛv/λ = sinh ɛ V ɛ V = csch λ z λ x, y, z λ λ y R y c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
44 CHAPTER ELECTROSTATICS Poblem 53 V = ρ ɛ E 4, so d V dx = ɛ ρ b V = mv v = V m c d = Aρ dx ; d dx dt = ρ dt = Aρv = I constnt Note: ρ, hence lso I, is negtive d d V dx = ɛ ρ = I ɛ Av = I m ɛ A V d V dx = βv /, whee β = I m ɛ A Note: I is negtive, so β is positive; is positive e Multiply by V = dv dx : V dv dx = βv / dv dx V dv = β V / dv V = βv / + constnt But V = V = cthode is t potentil zeo, nd field t cthode is zeo, so the constnt is zeo, nd V = 4βV / dv dx = β V /4 V /4 dv = β dx; V /4 dv = β dx 4 3 V 3/4 = β x + constnt But V =, so this constnt is lso zeo V 3/4 = 3 β x, so V x = 3 4/3 /3 9 8I β x 4/3, o V x = 4 β x 4/3 /3 m = 3ɛ x 4/3 A Intems of V insted of I: V x = V x d 4/3 see gph Without spce-chge, V would incese linely: V x = V x d V V d V ρ = ɛ dx = ɛ 4 V d 4/3 3 3 x /3 = 4ɛ V 9d x /3 without with v = f V d = V = x /3 V = V /m m d 8I m 3ɛ A /3d 4/3 V 3 = 8md4 I = 4 ɛ A 9 m d V 3/ = KV 3/, whee K = 4ɛ A 9d Poblem 54 E = 4ɛ ρ ˆ + e /λ dτ λ 8md V 3 4 ; 3ɛ A I ; I = 3ɛ A m d x c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER ELECTROSTATICS 45 b Yes The field of point chge t the oigin is dil nd symmetic, so E =, nd hence this is lso tue by supeposition fo ny collection of chges c V = E dl = 4ɛ + e /λ d λ = 4ɛ + e /λ d = { λ 4ɛ e /λ d + } λ e /λ d d e /λ Now e /λ d = e /λ λ d exctly ight to kill the lst tem Theefoe { V = } e /λ 4ɛ = e /λ 4ɛ S V E d = 4ɛ R V dτ = 4ɛ + R e R/λ 4 R = + R λ ɛ λ R e /λ = λ { e R/λ + R ɛ λ S E d + λ V e R/λ 4 d = R e /λ d = e /λ ɛ ɛ + } /λ λ ]R V dτ = { + R e R/λ + R } e R/λ + ɛ λ λ ] } = ed ɛ e Does the esult in d hold fo nonspheicl sufce? Suppose we mke dent in the sphee pushing ptch e R sin θ dθ dφ fom dius R out to dius S e S sin θ dθ dφ R S E d = 4ɛ = 4ɛ { S + S e S/λ S sin θ dθ dφ R λ + R } e R/λ R sin θ dθ dφ λ e S/λ + R ] e R/λ sin θ dθ dφ λ + S λ λ V dτ = e /λ λ sin θ d dθ dφ = 4ɛ λ = sin θ dθ dφ e /λ + S 4ɛ λ R = + S e S/λ + R 4ɛ λ λ S sin θ dθ dφ e /λ d 4ɛ R e R/λ ] sin θ dθ dφ So the chnge in λ V dτ exctly compenstes fo the chnge in E d, nd we get ɛ fo the totl using the dented sphee, just s we did with the pefect sphee Any closed sufce cn be built up by successive distotions of the sphee, so the esult holds fo ll shpes By supeposition, if thee e mny chges inside, the totl is ɛ Q enc Chges outside do not contibute in the gument bove we found tht t fo this volume E d + λ V dτ = nd, gin, the sum is not chnged by distotions of the sufce, s long s emins outside So the new Guss s Lw holds fo ny chge configution c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
46 CHAPTER ELECTROSTATICS f In diffeentil fom, Guss s lw eds: E + λ V = ɛ ρ, o, putting it ll in tems of E: E λ E dl = ɛ ρ Since E = V, this lso yields Poisson s eution : V + λ V = ɛ ρ ρ V = 4ɛ R ρ e /λ dτ ɛ λ V = ρ E E= V E= 4ɛ R ρ + λ e /λ dτ ˆ R λ E dl= ρ ɛ ; E= V V = R E dl E g Refe to Guss s lw in diffeentil fom f Since E is zeo, inside conducto othewise chge would move, nd in such diection s to cncel the field, V is constnt inside, nd hence ρ is unifom, thoughout the volume Any ext chge must eside on the sufce The fction t the sufce depends on λ, nd on the shpe of the conducto Poblem 55 ρ = ɛ E = ɛ x = ɛ constnt eveywhee The sme chge density would be comptible s f s Guss s lw is concened with E = yŷ, fo instnce, o E = 3, etc The point is tht Guss s lw nd E = by themselves do not detemine the field like ny diffeentil eutions, they must be supplemented by ppopite boundy conditions Odinily, these e so obvious tht we impose them lmost subconsciously E must go to zeo f fom the souce chges o we ppel to symmety to esolve the mbiguity the field must be the sme in mgnitude on both sides of n infinite plne of sufce chge But in this cse thee e no ntul boundy conditions, nd no pesusive symmety conditions, to fix the nswe The uestion Wht is the electic field poduced by unifom chge density filling ll of spce? is simply ill-posed: it does not give us sufficient infomtion to detemine the nswe Incidentlly, it won t help to ppel to Coulomb s lw the integl is hopelessly indefinite, in this cse E = 4ɛ ρ ˆ dτ Poblem 56 Compe Newton s lw of univesl gvittion to Coulomb s lw: F = G m m ˆ; F = 4ɛ ˆ Evidently 4ɛ G nd m The gvittionl enegy of sphee tnslting Pob 34 is theefoe W gv = 3 5 GM R c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
CHAPTER ELECTROSTATICS 47 Now, G = 667 N m /kg, nd fo the sun M = 99 3 kg, R = 696 8 m, so the sun s gvittionl enegy is W = 8 4 J At the cuent te this enegy would be dissipted in time t = W P = 8 4 386 6 = 59 4 s = 87 7 yes Poblem 57 Fist eliminte z, using the fomul fo the ellipsoid: σx, y = Q 4b c x / 4 + c y /b 4 + x / y /b Now fo pts nd b set c, sushing the ellipsoid down to n ellipse in the x y plne: σx, y = Q b x/ y/b I multiplied by to count both sufces Fo the cicul disk, set = b = R nd let x + y σ = Q R R b Fo the ibbon, let Q/b Λ, nd then tke the limit b : σx = Λ x c Let b = c, y + z, mking n ellipsoid of evolution: x + Q =, with σ = c 4c x / 4 + /c 4 The chge on ing of width dx is d = σ ds, whee ds = dx + d = dx + d/dx Now x dx + d c = d dx = x c, so ds = dx + c4 x 4 = dxc x / 4 + /c 4 Thus λx = d dx = Q c x 4c / x / 4 + /c 4 4 + /c 4 = Q Constnt! c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is
48 CHAPTER ELECTROSTATICS Poblem 58 y _, 3 _ b,- x _,- 3 _ One such point is on the x xis see digm t x = Hee the field is cos θ Ex = =, 4 + b o cos θ = b + Now, / cos θ = ; b b = +! 3 = + Theefoe / ] To simplify, let = + + 3/ u =, 3/ + u u + u o u: u + u4 = u + u 3 c Peson Eduction, Inc, Uppe Sddle Rive, NJ All ights eseved This mteil is