3209-2009-Solutions5

3209-2009-Solutions5 - ECE 3209 — Electromagnetic Fields...

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Unformatted text preview: ECE 3209 — Electromagnetic Fields University of Virginia Fall 2009 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 π² I S σ b da r + 1 4 π² Z V ρ b dv r = 1 4 π² 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 π² n- 2 πka 2 z p a 2 + z 2- 2 az cos θ fl fl fl π + 2 πkb 2 z p b 2 + z 2- 2 bz cos θ fl fl fl π- 6 kπ z Z b a p r 2 + z 2- 2 rz cos θ fl fl fl π o For z > b , Φ( z > b ) = 1 4 π² 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 ² z ( b 3- a 3 )- 1 4 π² 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 π² 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 π² n 4 πkb 2- 4 πka 3 z- 6 kπ z ‡ 2 r 3 3 fl fl fl z a + zr 2 fl fl fl b z ·o Thus, Φ( a < z < b ) = 1 4 π² n 2 πkz 2- 2 πkb 2 o- k 2 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 ² ( r 2- b 2 ) · ˆ r =- kr ² ˆ r For r < a , Φ( z < a ) = 1 4 π² 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 π² n 4 πkb 2- 4 πka 2- 12 πk Z b a rdrBigl } =- k 2 ² ( b 2- a 2 ) This is constant (not a function of z ), so its gradient is zero. Thus, ~ E ( r < a ) = 0 These results are a little easier to obtain from Gauss’s Law, I S ~ D · d~a = Q f = 0 where Q f is the total free charge enclosed within surface S . Since we have no free charge in this problem, Q f = 0 . Moreover, from symmetry, we see that ~ D is constant over any spherical surface centered at the origin. Thus, ~ D = 0 everywhere ! This gives us, ~ D = ² ~ E + ~ P ⇒ ~ E =- ~ P ² For r > b , there is no polarization ( ~ P = 0 ). Thus,)....
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This note was uploaded on 10/16/2010 for the course ECE 309 taught by Professor Weikle during the Fall '08 term at UVA.

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3209-2009-Solutions5 - ECE 3209 — Electromagnetic Fields...

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