309-2008-Solutions5

309-2008-Solutions5 - ECE 309 — Electromagnetic Fields...

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Unformatted text preview: ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 5 Solutions 1. Spherical shell with “frozen-in” polarization ~ P = kr ˆ r . First, let’s find the bound charge densities, σ b = ~ P · ˆ n =- ka, inner surface kb, outer surface ρ b =- ~ ∇ · ~ P =- 1 r 2 d dr r 2 P r =- 3 k Using these charge densities, let’s integrate over the sphere to find the field. Without loss of generality, we can find the field on the z-axis (because the problem is spherically symmetric). Let’s do this by finding Φ( z ) first, and then determine ~ E by taking its gradient: Φ( z ) = 1 4 π I S σ b da r + 1 4 π Z V ρ b dv r = 1 4 π n- ka Z 2 π dφ Z π a 2 sin θ dθ √ a 2 + z 2- 2 az cos θ + kb Z 2 π dφ Z π b 2 sin θ dθ √ b 2 + z 2- 2 bz cos θ- 3 k Z 2 π dφ Z b a Z π r 2 sin θ drdθ √ r 2 + z 2- 2 rz cos θ o = 1 4 π n- 2 πka 2 z p a 2 + z 2- 2 az cos θ π + 2 πkb 2 z p b 2 + z 2- 2 bz cos θ π- 6 kπ z Z b a p r 2 + z 2- 2 rz cos θ π o For z > b , Φ( z > b ) = 1 4 π n- 2 πka 2 z ( a + z- z + a )+ 2 πkb 2 z ( b + z- z + b )- 6 kπ z Z b a r ( r + z- z + r ) dr o = k z ( b 3- a 3 )- 1 4 π 6 kπ z h 2 r 3 3 i b a = 0 Thus, for r > b , ~ E ( r > b ) =- ~ ∇ Φ( z ) = 0 For a < z < b , Φ( z ) = 1 4 π n- 2 πka 2 z ( a + z- z + a ) + 2 πkb 2 z ( b + z- b + z )- 6 kπ z Z z a r ( r + z- z + r ) dr- 6 kπ z Z b z r ( r + z- r + z ) dr o = 1 4 π n 4 πkb 2- 4 πka 3 z- 6 kπ z 2 r 3 3 z a + zr 2 b z o Thus, Φ( a < z < b ) = 1 4 π n 2 πkz 2- 2 πkb 2 o- k 2 ( z 2- b 2 ) Now, let z = r and take the negative gradient to find the field: ~ E ( a < r < b ) =- ~ ∇ Φ( r ) =- d dr k 2 ( r 2- b 2 ) ˆ r =- kr ˆ r For r < a , Φ( z < a ) = 1 4 π n- 2 πka 2 z ( a + z- a + z )+ 2 πkb 2 z ( b + z- b + z )- 6 kπ z Z b a r ( r + z- r + z ) dr = 1 4 π n 4 πkb 2- 4 πka 2- 12 πk Z b a rdrBigl } =- k 2 ( b 2- a 2 ) This is constant (not a function of z ), so its gradient is zero....
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309-2008-Solutions5 - ECE 309 — Electromagnetic Fields...

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