PS3 Solutions

PS3 Solutions - Physics 217 Problem Set 3 Solutions by Yao...

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Unformatted text preview: Physics 217: Problem Set 3 Solutions by Yao Liu Due Friday, Feb 12th, 2010 1 Purcell 2.1 E x = 6 xy E y = 3 x 2- 3 y 2 E z = 0 First integrate along the straight path from (0 , , 0) to ( x 1 , , 0) Z E · d s = Z x 1 E x ( x, , 0) dx = Z x 1 6 x (0) dx = 0 since y = 0 along the path. Then for the path from ( x 1 , , 0) to ( x 1 ,y 1 , 0) Z E · d s = Z y 1 E y ( x 1 ,y, 0) dx = Z y 1 (3 x 2 1- 3 y 2 ) dy = 3 x 2 1 y 1- y 3 1 . Now take the other route, first to (0 ,y 1 , 0) Z E · d s = Z y 1 E y (0 ,y, 0) dx = Z y 1 (- 3 y 2 ) dy =- y 3 1 then Z E · d s = Z x 1 E x ( x,y 1 , 0) dx = Z x 1 (6 xy 1 ) dx = 3 x 2 1 y 1 . Indeed it’s independent of path, and the potential function, with 0 set at the origin, is φ ( x 1 ,y 1 , 0) =- Z E · d s =- 3 x 2 1 y 1 + y 3 1 . Since E z = 0, a path along the z-axis has no contribution, so we can write φ ( x,y,z ) =- 3 x 2 y + y 3 . One can check the gradient is indeed E-∇ φ = ( 6 xy, 3 x 2- 3 y 2 , ) . 2 Purcell 2.8 Take as our Gaussian surface the cylinder with radius r < a , length l along the axis. By symmetry E is pointing radially outward, so the two ends don’t contribute, and | E | is constant along cylindrical surface. Z E · d a = E Z da = EA = E (2 πr ) l 1 Gauss’s law says that’s 4 π (or 1 / if you prefer the other units) times the charge enclosed Q enc = ρV = ρ ( πr 2 ) l, thus we find E = 2 πρr. The potential, if we take φ = 0 at r = 0, is φ =- Z E · d s =- Z r E ( r ) dr . If r < a , it’s φ ( r ) =- Z a 2 πρr dr =- πρr 2 . If r > a , we have to break it into two pieces φ ( r ) =- Z a E ( r ) dr- Z r a E ( r ) dr =- πρa 2- Z r a 2 πρa 2 r dr =- πρa 2- 2 πρa 2 ln r a So, φ ( r ) = (- πρr 2 r < a- πρa 2 ( 1 + 2ln ( r a )) r > a 3 3.1 Purcell 2.13 E x = 6 xy E y = 3 x 2- 3 y 2 E z = 0 ∇ · E = 6 y- 6 y + 0 = 0 ( ∇ × E ) x = ∂E z ∂y- ∂E y ∂z = 0- 0 = 0 ( ∇ × E ) y = ∂E x ∂z- ∂E z ∂x = 0- 0 = 0 ( ∇ × E ) z = ∂E y ∂x- ∂E x ∂y = 6 x- 6 x = 0 Since ∇ × E vanishes everywhere , there exists a potential. 3.2 Purcell 2.14 For f ( x,y ) = x 2 + y 2 , the Laplacian ∇ 2 f = 2 + 2 = 4, so it does not satisfy the Laplace’s equation. But g ( x,y ) = x 2- y 2 does, since 2- 2 = 0. I can’t sketch it here, but it is a saddle. Looking in x direction, it curves up parabolically; and the y direction curves down parabolically. The gradient ∇ g ( x,y ) = (2 x,- 2 y ) so at those four points ∇ g (0 , 1) = (0 ,- 2) ∇ g (1 , 0) = (2 , 0) ∇ g (0 ,- 1) = (0 , 2) ∇ g (- 1 , 0) = (- 2 , 0) 2 3.3 Purcell 2.15 (a) F x = x + y F y =- x + y F z =- 2 z ∇ ·...
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PS3 Solutions - Physics 217 Problem Set 3 Solutions by Yao...

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