hw5solution - Problem 2.31 Problem 2.32(a w = fdeT From...

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Unformatted text preview: Problem 2.31 Problem 2.32 (a) w = ; fdeT. From Prob. 2.21 (or Prob. 2.28): v = if; (R2 — r?) = 4 "1‘0 59,? (3 — l 1 q R r2 2 qp r3 1 r z- _ —— 4 = ————— W 2p47r50212/0 (3 122 7”" dr 45012 33 R2 5 _qu2__q_Ifi q — 1 _ 560 _ 560 g—nR3 _ 47reo 5 ' 41110 00 1 2 R r 2 2 A 130‘ 41rdr) +I/o (47rr dr)} ”+;(i"_;lil+; _1§£¢ R R6 5 _41r602 R SR "@6051? o (c) W = f5 VE - (13 + fv Ezdr}, where V is large enough to enclose all the charge, but otherwise arbitrary. Let’s use a sphere of radius a > R. Here V 2' “1‘02. 1 g 1 i 2. f“ 2 /° _1_l 2 2 {/ (linear) (41m) r2)r sm0d6dd>+ o E dT+ R 47m) T2 (47rr dr) (I2 1 q2 411' 1 2 1 = —- —4 —— 4 _... 2 {(47r60)2 a F + (471’60)2 5H + (41mg)2 7”] ( 1') 1 i{iri_i+1}=_L§£, (b) W = iz‘lesz. Outside (r > R) E = 1 £7? ; Inside (r < R) E 2 “1‘0 72951‘1“. 60 W:— 2 60 2 As a ——> 00, the contribution from the surface integral (473(05213) goes to zero, while the volume integral ( 1 fi(% _ 1)) picks up the slack. 4n<o 2a Problem 2.35 (b) no) = — fiE-dl = — Jim, Mr ~ mow — Jimmie — mow = (g + % — g). (C) (the charge “drains off”); V(O) = — f;(0)dr - Labia) f2)dr — fg(0)dr 2 1 — g). a Problem 2.36 (In Qb (Ia + (10 ' a : ‘ i a = “ . ; : - 1 r + . (b) E0,” = 47“: J“ T2 qb r, where r 2 vector from center of large sphere. i ‘0 r , where ra (rb) is the vector from center of cavity a (b). (d) 22$. (e) on changes (but not 0,, or 05); Eomside changes (but not E, or Eb); force on qa and qb still zero. Problem 2.37 Between the plates, E : 0; outside the plates E = (7/60 : Q/coA. So P : 6—0E’2 : 2 Q2 2 Q2 I 2 2 csgA2 260A2 Problem 2.39 Say the charge on the inner cylinder is Q, for a length L. The field is given by Gauss‘s law: f E - da = E - 21rs . L = éQenc = id) é E = —Q—1 s. Potential difference between the cylinders is — b _ Q b1 — Q b V(b)—V(a)—-/a E-dl——2MOL —ds_—2mLin As set up here, a is at the higher potential, so V = V(a) — V(b) = —% 1n 27reo C = 3‘ = [2773-]? so capacitance per unit length is Problem 2.40 (a) W = (force)><(distance) = (pressure)x(area)x(distance) = (b) W = (energy per unit volume)x(decrease in volume) = (60372) (As). Same as (a), confirming that the energy lost is equal to the work done. ...
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hw5solution - Problem 2.31 Problem 2.32(a w = fdeT From...

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