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 its independent of path, and the potential function, with 0 set at the origin, is ( 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 ( x,y,z ) =- 3 x 2 y + y 3 . One can check the gradient is indeed E- = ( 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 dont contribute, and | E | is constant along cylindrical surface. Z E d a = E Z da = EA = E (2 r ) l 1 Gausss law says thats 4 (or 1 / if you prefer the other units) times the charge enclosed Q enc = V = ( r 2 ) l, thus we find E = 2 r. The potential, if we take = 0 at r = 0, is =- Z E d s =- Z r E ( r ) dr . If r < a , its ( r ) =- Z a 2 r dr =- r 2 . If r > a , we have to break it into two pieces ( r ) =- Z a E ( r ) dr- Z r a E ( r ) dr =- a 2- Z r a 2 a 2 r dr =- a 2- 2 a 2 ln r a So, ( r ) = (- r 2 r < a- a 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 Laplaces equation. But g ( x,y ) = x 2- y 2 does, since 2- 2 = 0. I cant 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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