{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}