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Homework_3

# Homework_3 - Homework 3 solutions 1.3(a The charge density...

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Homework 3 solutions 1.3 (a) The charge density will be of form ( ) , k r R δ - where k is a constant. Integrating over a spherical volume with radius greater than R , we find that ( ) 2 4 . k Q R π = Hence ( ) ( ) 2 . 4 Q r R R ρ δ π = - x (b) Similarly to part (a) by integrating over a cylinder of unit length and radius greater than b , we find ( ) ( ) . 2 r b b λ ρ δ π = - x (c) Assume the disk is centered on the origin of coordinates and lies in the z = 0 plane. For , r R < the surface charge density is ( ) 2 . Q R σ π = The charge density is zero everywhere else. By integrating the charge density over the volume between two concentric cylinders of radii r and r r + Δ where r < R , we find ( ) ( ) 2 , , . 0, . Q z r R R r R δ ρ π < = > x (d) For , r R < the charge density will be proportional to ( ) 2 . δ θ π - By integrating the charge density over the volume between two spheres of radii r and r r + Δ where r < R , we find ( ) 2 , , . 2 0, . Q r R R r r R π δ θ ρ π - < = > x 2.3 (a) The line charge will have 3 image line charges. Images of charge density λ - need to be placed at ( ) 0 0 , x y - and ( ) 0 0 , , x y - and an image of charge density λ needs to be placed at ( ) 0 0 , . x y - - The potential from the 4 line charges is ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 0 , ln . 4 x x y y x x y y x y x x y y x x y y λ πε + + - - + + Φ = - + - + + + The potential on y = 0 is

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( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 0 ,0 ln 0. 4 x x y x x y x x x y x x y λ πε + + - - + Φ = = - + - + + Similarly ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 0 0, ln 0. 4 x y y x y y y x y y x y y λ πε + - - + + Φ = = - + - + + We conclude that the potential is zero on the conducting surfaces. The electric field has components ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 0 0 0 0 2 2 2 2 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 2 2 0 0 0 0 ln 4 2 2 2 2 , 4 ln 4 x y x x y y x x y y E x x x y y x x y y x x x x x x x x x x y y x x y y x x y y x x y y x x y y x x y y E y x x y y x x λ πε λ πε λ πε + + - - + + = - - + - + + + + - - + = - - + - + + - - + - - + + + + + + + - - + + = - - + - + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 0 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 2 2 2 2 . 4 y y y y y y y y y y x x y y x x y y x x y y x x y y λ πε + + - - + + = - - + - + + - - + - - + + + + + On y = 0, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 2 2 2 2 0. 4 x x x x x E x y y x y y x y y x y y λ πε - - = - - + - = + - - + - - + + + + With a similar result for E y on x = 0, we see that the tangential component of the electric field is zero on the conductors. (b) On the half plane y = 0, x > 0, the normal component of the electric field is ( ) ( ) 0 0 2 2 2 2 0 0 0 0 0 4 4 . 4 y y y E x x y x x y λ πε = - - - + + + The charge density is
( ) ( ) 0 0 0 0 2 2 2 2 0 0 0 0 4 4 . 4 y y y E x x y x x y λ σ ε ε π = = = - - - + + + n E The plot below shows σ λ at ( x 0 , y 0 ) = (2,1), (1,1) and (1,2). (c) Integrating the charge density with respect to x , we obtain ( ) ( ) ( ) ( ) 0 2 2 2 2 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 1 1 tan tan 2 2 2 tan . x y Q dx x x y x x y y y dx dx x x y x x y x x y y x y λ π λ λ π π λ π λ π π π λ π - - - = - - - + + + = - + + - + = - - - - = -

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Homework_3 - Homework 3 solutions 1.3(a The charge density...

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