HW 8 solution - EE335 Homework%Solution PM Q r”(it = Q...

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Unformatted text preview: EE335 Homework %Solution _, PM Q r” (it? = Q _ e4ms,R 311728., =0 zz+a2 41:8 z2+a (b) E=—V¢= ——2)fi]az (c) Following example 4.3. a dE2 d5; d3: d3; Pad¢ . pad¢ dE =—-—-—-L-——a ’ d5 =- t A l 4’ma(zz+“2) ' 2 4mu(z=+a2)az ad . dE4=(dEl+dEz)c°5a=E’%?%a—zj[ 12 2] _Q_Jaz Z +a _ " [2m _ _Q_ 2, E: “'LG 4 (22 + “aria d¢- 42:80 {(—22 +02)m_] Same as in (b). 0 (a) D is continuous, mercfore J‘B - ds = Q 4.6. DJJ'2 sin9d9d¢ =4mzp’ = Q =2 , 4m: r<b :1: HA! : n a: = —jEm- den! Ends—IE, rad. c cod c -AU Am Ldr -d£ +1;er = chief!!- __Q_a[i[+d+ :11?!"- +l‘fl] 479‘: r c” c 1(11) 1 1 = —--+— ——-— +———. 4m; —-Q—l:c+d.'l b 8 r: c+d a c ' 4 C:-Q-=_.._—_—————-—-—m" «5 4-1 LG- 1 p14 I c+d b e, c c+d c Using l=i+—!~-+—l-,then 1‘: cl CZ C} (a) For r<R, IsoE‘dszquPflV rt;- ' By symmetry, E=Eriir , 6,117,! L1”: sinedfldqb: p r-2 sinedrdfldtp 9-0 eaEr4mz = gin-3A _1"P ’ 3e r<R For r > R, 5:35:14!”2 = Q = gfiRapv Er: Q2 r>R 47:31" O (b) lib-Edw—s IEfdv 2 _ pr .. .Q 2 - 28 LL: [38—fl2 r 25in6drd6dj+£ [438033] 1' 5m Bdrdfldtfi] E a: 2 = —P—R or sinGdrderp-i- 92;” —sin6\drd9d¢ 182 32 I: e 2 5 '2 2 P R Q [ 1 ] 2 99 W: v 4 — + 4 —H l = 182i 7‘ 5] 321950 ”R ‘Ov 16321?“ 991 [ 12’] Q? Q: Q’ 39* 18- 15303312 = = W: 4.9. 8243 (a) From Laplace's eqfiafion, i— = 0. Therefore, 45 = A5!) + B. Applying 2 p boundary conditions, ¢(¢=¢1)=0=A¢.+B. This gives 4501b): 4!: _ 2 l 3452 ¢(¢=¢2)=100=A¢2+B' WOW-$1) __l .-.=_l 100 .. (b) E——Vd5— pa¢a‘ p[ )a, (C) P=5-D ForPlamt ¢=¢v P. =fi-D=a¢ ’[fifiafi )]3 ”44:00:90) 2 1 2 1 F9: plate at q) = 9&2, ,02 2+ ( 100 p ¢2"¢1) ...
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