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 r . First, lets find the bound charge densities, b = ~ P n = - ka, inner surface kb, outer surface b =- ~ ~ P =- 1 r 2 d dr r 2 P r =- 3 k Using these charge densities, lets 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). Lets do this by finding ( z ) first, and then determine ~ E by taking its gradient: ( z ) = 1 4 I S b da r + 1 4 Z V b dv r = 1 4 n- ka Z 2 d Z a 2 sin d a 2 + z 2- 2 az cos + kb Z 2 d Z b 2 sin d b 2 + z 2- 2 bz cos - 3 k Z 2 d Z b a Z r 2 sin drd r 2 + z 2- 2 rz cos o = 1 4 n- 2 ka 2 z p a 2 + z 2- 2 az cos fl fl fl + 2 kb 2 z p b 2 + z 2- 2 bz cos fl fl fl - 6 k z Z b a p r 2 + z 2- 2 rz cos fl fl fl o For z > b , ( z > b ) = 1 4 n- 2 ka 2 z ( a + z- z + a )+ 2 kb 2 z ( b + z- z + b )- 6 k z Z b a r ( r + z- z + r ) dr o = k z ( b 3- a 3 )- 1 4 6 k z h 2 r 3 3 i b a = 0 Thus, for r > b , ~ E ( r > b ) =- ~ ( z ) = 0 For a < z < b , ( z ) = 1 4 n- 2 ka 2 z ( a + z- z + a ) + 2 kb 2 z ( b + z- b + z )- 6 k z Z z a r ( r + z- z + r ) dr- 6 k z Z b z r ( r + z- r + z ) dr o = 1 4 n 4 kb 2- 4 ka 3 z- 6 k z 2 r 3 3 fl fl fl z a + zr 2 fl fl fl b z o Thus, ( a < z < b ) = 1 4 n 2 kz 2- 2 kb 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 ) =- ~ ( r ) =- d dr k 2 ( r 2- b 2 ) r =- kr r For r < a , ( z < a ) = 1 4 n- 2 ka 2 z ( a + z- a + z )+ 2 kb 2 z ( b + z- b + z )- 6 k z Z b a r ( r + z- r + z ) dr = 1 4 n 4 kb 2- 4 ka 2- 12 k Z b a rdrBigl } =- k 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 Gausss 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 = ~ E + ~ P ~ E =- ~ P For r > b , there is no polarization ( ~ P = 0 ). Thus,)....
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3209-2009-Solutions5 - ECE 3209 Electromagnetic Fields...

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