3209-2009-Solutions3

3209-2009-Solutions3 - ECE 3209 Electromagnetic Fields...

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ECE 3209 — Electromagnetic Fields University of Virginia Fall 2009 Homework # 3 Solutions 1. Cheng, P.3-11 Consider a spherical distribution of charge, ρ = ρ 0 1 - R 2 b 2 · in the region 0 R b , surrounded by a conducting shell with inner radius R i and outer radius R o . Using symmetry We can exploit Gauss’s Law to find the charge in each region: For 0 R b , I S ~ E · d~a = Q enc ² 0 From symmetry, ~ E must be directed in the radial ( ˆ r ) direction and is not a function of the spherical coordinates, θ or φ . Choosing a sphere of radius r , concentric with the origin, as the Gaussian surface, we have Z π 0 Z 2 π 0 E r R 2 sin θdθdφ R = r = 4 πr 2 E r = 1 ² 0 Z r 0 ρ 0 1 - R 2 b 2 · R 2 sin θdrdθdφ | {z } Q enc Thus, = 4 π ρ 0 ² 0 h R 3 3 - R 5 5 b 2 i r 0 = 4 π ρ 0 ² 0 r 3 h 1 3 - r 2 5 b 2 i ~ E = ρ 0 ~ r 3 ² 0 h 1 - 3 r 2 5 b 2 i , for 0 R b For b r R i , all the charge is enclosed and so the field is, ~ E = ˆ r ρ 0 b 3 3 ² 0 r 2 h 1 - 3 5 i = ˆ r 2 ρ 0 b 3 15 ² 0 r 2 , for b r R i Inside a conductor, electrostatic fields are zero. Thus, we have
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~ E = 0 , for R i r < R 0 , and ~ E = ˆ r 2 ρ 0 b 3 15 ² 0 r 2 , for r R 0 2. Cheng, P.3-15 Three charges ( q , - 2 q , and q ) sit along the z -axis at z = d/ 2 ,z = 0 , and z = - d/ 2 , respectively: We can determine the potential at P using superposition: Φ( P ) = 1 4 π² 0 h q r + - 2 q r + q r - i To express this in terms of the spherical coordinates, let’s write r + and r - in terms of r and θ (using the law of cosines): r + = r r 2 + d 2 · 2 - rd cos θ, r - = v u u t r 2 + d 2 · 2 - rd cos( π - θ ) | {z } - cos θ With point P distant from the charges, we can approximate r + and r - using the Taylor series for r ± d :
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3209-2009-Solutions3 - ECE 3209 Electromagnetic Fields...

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