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PS2 Solutions - Physics 2217 Problem Set 2 Solutions 1...

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Physics 2217: Problem Set 2 Solutions 1. Purcell 1.5 d ~ E = k dq r 2 ˆ r and dq = Rλdθ So: d ~ E x = k dq r 2 cos θ ˆ x ~ E x = k R 2 Z π 0 cos θ ˆ x = 0 Which was expected from the symmetry of the problem and: d ~ E y = k dq r 2 sin θ ˆ y ~ E y = k R 2 Z π 0 sin θ ˆ y = - K 2 Q πR 2 ˆ y 2. Purcell 1.9 q = 4 π 3 r 3 ρ So: Q total = 4 π 3 a 3 ρ and dq = 4 πr 2 ρdr Assume that we have already brought some charge q. The work to bring in an additional shell of charge dq will be: dW = K qdq r = K r ( 4 π 3 r 3 ρ )(4 πr 2 ρdr ) = K (4 π ) 2 ρ 2 3 r 4 dr W = Z dW = (4 π ) 2 ρ 2 3 r 5 5 a 0 = K ( 4 π 3 a 3 ρ )( 4 π 3 a 3 ρ ) 3 a 5 = 3 5 Q 2 a 3. Purcell 1.19 We can use superposition to solve this problem: ~ E tot = ~ E plane + ~ E layer For the plane: (a) x < 0: ~ E plane = - 1 2 σ 0 ˆ x 1
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(b) 0 < x < d ~ E plane = + 1 2 σ 0 ˆ x (c) x > d ~ E plane = + 1 2 σ 0 ˆ x For the layer: In regions (a) and (c), one does not ”know” that there is a finite thickness of layer, ie for a Gaussian pilbox that starts in (a) and ends in (c) we get: ~ E a = - ~ E c and from Gauss law: Z ~ E · d ~ A = Q 0 = ρAd 0 E = ρd 2 0 So: ( a ) ~ E layer = -
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