309-2008-Solutions4

309-2008-Solutions4 - ECE 309 Electromagnetic Fields...

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ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 4 Solutions 1. Energy of a uniformly charged sphere (a) The charge density is given by, ρ = q 4 3 πR 3 = 3 q 4 πR 3 ( * ) so that the energy stored can be calculated as, U = 1 2 Z V ρ Φ dv = ρ 2 Z π 0 sin θdθ Z 2 π 0 Z R 0 r 2 Φ( r ) dr = 2 πρ Z R 0 r 2 Φ( r ) dr From Gauss’s Law, the field due to a uniform sphere of charge is ~ E = q 4 π 0 ˆ r r 2 if r R ρr 3 0 ˆ r if r R Thus, the potential at position r inside the sphere is, Φ( r ) = - Z R ~ E outside ( r 0 ) · d~ r 0 - Z r R ~ E inside ( r 0 ) · d~ r 0 = - q 4 π 0 Z R dr 0 r 0 2 - ρ 3 0 Z r R r 0 dr 0 Φ( r ) = q 4 π 0 1 R - ρ 6 0 R 2 - r 2 = ρ 3 0 n 3 R 2 2 - r 2 2 o Using relation ( * ) above. Thus the energy stored is, U = 2 πρ 2 3 0 Z R 0 3 R 2 2 r 2 - r 4 2 dr = 2 πρ 2 3 0 R 5 2 - R 5 10 Thus, U = 4 πρ 2 R 5 15 0 = 3 q 2 20 π 0 R
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(b) Using the expression for U in terms of the electric field, U = 0 2 Z V ~ E · ~ E dv = 4 π 0 2 n Z R 0 | E inside | 2 dv Z R | E outside | 2 dv o U = 2 π 0 n ρ 3 0 2 Z R 0 r 4 dr + q 4 π 0 2 Z R dr r 2 o U = 2 π 0 n ρ 3 0 2 n R 5 5 + R 5 o = 4 πρ 2 R 5 15 0 , as before (c) Finally, consider a sphere of radius R and total charge Q R . The potential at radius R is given by, Φ R = Q R 4 π 0 R , where Q R = 4 3 ρπR 3 The infinitesimal chage in a spherical layer of thickness dR is, dQ R = 4 πR 2 ρdR and the work required to bring this infinitesimal charge, dQ R from infinity to radius R is,
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dU = Φ R dQ R = 4 π 2 0 ρ 2 R 4 dR Thus, the total work needed to assemble the sphere, layer-by-layer is U = Z R 0 dU = 4 π 2 0 ρ 2 Z R 0 R 0 4 dR 0 = 4 πρ 2 R 5 15 0 , again as expected!
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