By David J. Griffiths
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Additional resources for Introduction to Electrodynamics — Instructor's Solutions Manual
Sample text
411"fOZ ' where Q = 7TR2a. 7 E is clearly in the z direction. From the diagram, dq = ada = a R2 sin 0 dO d
Da. E. L. 11 ~ GaUSsian surface: Inside: f E. da Gr Problem E. I Gaussian surface: Outside: E(41rr2) . (a41rR2) => E R2 =~ <0 2 r. } (As in Frob. L'E. da = E. Qenc= 1..! <0 <0 3 1rr3p. 1 Since Qtot = ~1rR2p, E = 4';<0j&r (as in Frob. 8). L'E. ~/ " ' = E. >"l. So <0 <0 n ,unace IE l Problem da 21r€os = ~Sl (same as Ex. 1). 14 . Qenc = 1.. <0 <0 Jp dT = 1.. } (kf)(f2 = 1.. k 41r Jorr r3dr = 411'k r4 = 1I'kr4. <0 <0 4 <0 . 'E = ~1rkr2r. 41r€o <0 sin B dr dB d4» CHAPTER 2. 1 (H) f E. 16 (i) @ (H) b a a ~ )- Gaussiansurface .
0 (if ~~ = -El. is specified). 5 Putting U = T = V3 into Green's identity: r [V3'\72V3+ VV3 . VV3] dr = Js1 V3vv3. da. lv But '\72V3= '\72V1 - '\72V2 = -P- + P- = 0,andVV3= -E3€o €o . So Iv E~ dr = - Is V2E3. da, and the rest is the same as before. 6 Place image charges +2q at z = -d and -q at z = -3d. Total force on +q is q -2q 2q -q [ 411"/;0(2d)2 + (4d)2 + (6d)2 ] z q2 ~ F = 1 ( 1 1 = 47r€0d2 -2 + 8 - 36 ) 1 ~ Z 29q2 = - 47r€0 I ( ) 72d2 ~ Z. 7 = Vr2 + a2 - 2racos8j (a) From Fig. 13: = vr2 -'l- , R = - -a -; -'l- = - q Vr2 + b2 - 2rbcos8 q (Ii) vr2 + ~- Therefore: + b2 --2rbcos-jJ-: -'l-' W (Eq.