By David J. Griffiths

**Read or Download Introduction to Electrodynamics — Instructor's Solutions Manual PDF**

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**Sample text**

E = 100tx (ax) = The same charge density foa (constant everywhere). would be compatible (as far as Gauss's I I law is concerned) with E = ayy, for instance, or E = (~)r, etc. The point is that Gauss's law (and VxE = 0) by themselves do not determine the field-like any differential equations, they must be supplemented by appropriate boundary conditions. Ordinarily, these are so "obvious" that we impose them almost subconsciously ("E must go to zero far from the source charges" )-or we appeal to symmetry to resolve the ambiguity ("the field must be the same-in magnitude-on both sides of an infinite plane of surface charge").

L. rz (vr2 x - 27r 'f' - + Z2 - . l. 0 rz 2/z, (vr2 + Z2+ 2rz - vr2 + Z2- 2rz ) ifr < z, = rlz(r + z -IT - zl) = { 2/r, if r > z. 211'. 2 411"<0 ... P... 29 = 4;<0'V2f(*)dr = -L- f 411"<0 Problem z lr2dr + z {f = 4;<0f p(r') ('V2k)dr (since p is a function air', (r ) . 30. (a) Ex. ft. ,f <0 Ex. £. £.. <0 ,f Prob. 11: 0 (b) Eout : : 'IF" '. f. <0 <0 ,f §E. § <0 <0 s <0 (at surface = O. :. §. £. Ji.. 1. '. W4 = qV (b) WI - q2 = 1-41I'foa + - -L- (-2 -L V2a. 1. 1) V2 - 1 (-2 + ) V (1). - M 2 = 0, W2= 4;<0(=f); W3= 4;<0(:;a - ~ ; W4= (see(a)).

2 = 411"r 1 dW (q = charge on sphere of radius r). r3 q = 311"r3 p = 411"r2 dr p = 43qdr 311"R qr3 411"100R3 1 3q 2 R3 r dr ( ) ( 1 3q2 {R Jo r4dr = 411"100 R6 } a - 1)) picks up the slack. 33 dW 2 (411"r2dr) ;: 3q = R3 r )= 1 3q2 R5 = 411"100R65 2 dr. 34 =~f (a) W E2 dr. -rr (r > a), E2 = 4;fO::;ir (r > b). So (r > b), and hencef E1. r471T2dr = - 4::0b' 2 . E2 dr = 4;R2; aa = ~;-q q (b)V(O)= - J. 35 r - zero elsewhere. -IL (~- ~b ) . -r (a = ~q2 (~+! ,( ab= 4;b2' q I E. 36 I I (a ) aa I = -~; qa I (b) Eout II ab = 4- 1 aR I ~ (t ~ -~) ~ (~ - ~) )dr - f~(O)dr = + I = qa41rR2 + qb .