2013S_721_HW4_sol

Of the electric eld near the center of the sphere

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Unformatted text preview: 0 dl+1 q lAl Pl (cos θ) = a 4πǫ0 l (l + 1)Bl l al − 1 Pl (cos θ) − dl+1 a al l+1 Al = l+1 − Bl d l 3 Solving, this gives al 2l + 1 l + ǫ0 /ǫ(l + 1) dl+1 (ǫ0 /ǫ − 1)l al Bl = l + ǫ0 /ǫ(l + 1) dl+1 al r l q 2l + 1 Pl (cos θ) Φin = 4πǫ l + ǫ0 /ǫ(l + 1) dl+1 al Al = l Φout = q 4πǫ0 l l r< (ǫ0 /ǫ − 1)l al + l+1 l + ǫ0 /ǫ(l + 1) dl+1 r> a r l+1 Pl (cos θ) (b) Calculate the rectangular components of the electric field near the center of the sphere When r << a, the high l terms drop out, the potential becomes r r2 3 cos2 θ − 1 q 1 3 5 + cos θ + + ··· 4πǫd ǫ0 /ǫ 1 + 2ǫ0 /ǫ d 2 + 3ǫ0 /ǫ d2 2 3 5 z 3z 2 − r2 q 1+ + + ··· = 4πǫ0 d 2 + ǫ/ǫ0 d 3 + 2ǫ/ǫ0 2d2 3 5 z ˆ 2z z − xx − y y ˆ ˆ ˆ q + + ··· =− 4πǫ0 d 2 + ǫ/ǫ0 d 3 + 2ǫ/ǫ0 2d2 Φin = Ein (c) Verify that, in the limit ǫ/ǫ0 → ∞, your result is the same as that for the conduc...
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