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Unformatted text preview: 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 Rλ R 2 Z π dθ 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 Rλ R 2 Z π dθ 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 = 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 ˆ x 1 (b) 0 < x < d ~ E plane = + 1 2 σ ˆ x (c) x > d ~ E plane = + 1 2 σ 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:...
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This note was uploaded on 06/10/2011 for the course PHYS 2217 taught by Professor Leclair, a during the Spring '06 term at Cornell.
 Spring '06
 LECLAIR, A
 Physics

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