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Unformatted text preview: PHY 5346
HW Set 4 Solutions 7 Kimel 2. 2.22 a) Using the fact that for the interior problem, the normal derivative is outward, rather than inward, we have the potenial given by the negative of Eq.
(2.21), which takes the form7 when 0 = 0 and 'y = 6/ V 2 _ 2 2 1
@(Z)Z_M/ égdfﬁ dx
477 0 (a2 + 22 — 2mm) ((12 + 2:2 + 2azm) where I’ve replaced cos 0/ by x in the integral. The integral yields ¢<z>=e(1—;—%22) <I>(z) = E + 3 +0 (36)] Z which agrees with Eq. (2.27) if 0056 = 1.
b) For 2 > a, we have, using Eq. (2.22) Ezz—gV (1—H) =Ez(z)=V—a2%(3+a—2) 2:\/a2 + 3'2 (a2 + z2) 2:2
For z <a,
2_ 2 2 3 5 2
Ez(z)=_ﬂﬂ 1_M =EZ(2)=_Z _Q_+M
82: 2 am ‘1 Z2 (a2+22)§ in agreement with the book. Expanding the second form in a Taylor series
exapansion about 2 = 0 gives 3
Ez=—— —— ——— 6
2aV—i— 8a3z 16a5z +O(z) From the ﬁrst form, on the outside, we get (I E2 (‘1) (3) First look at a plot of the ﬁeld lines: E1 an} vim Reid Line? Next, look at E in the region (2a, Ed). I will make the plot in units of 0.5
0.8 0.6 0.4 0.2 0.2 0420.6 0.8 ...
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This note was uploaded on 01/10/2012 for the course PHYSIS 506 taught by Professor Smith during the Spring '08 term at University of Arkansas Community College at Morrilton.
 Spring '08
 smith

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