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# 2_22 - PHY 5346 HW Set 4 Solutions 7 Kimel 2 2.22 a Using...

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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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