3209-2009-Solutions7

# 3209-2009-Solutions7 - ECE 3209 Electromagnetic Fields...

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ECE 3209 — Electromagnetic Fields University of Virginia Fall 2009 Homework # 7 Solutions 1. Average potential over a sphere with point charge q outside. Consider the geometry of the problem shown below: The average potential over the spherical surface is, Φ avg 1 4 πR 2 I sphere Φ da, where Φ( r ) = 1 4 π² 0 q r is the potential at position ~ r from a point charge. From the Law of Cosines, r 2 = s 2 + R 2 - 2 sR cos θ , so Φ avg = 1 4 πR 2 q 4 π² 0 Z 2 π 0 Z π 0 R 2 sin θ dθ s 2 + R 2 - 2 sR cos θ = q 4 π² 0 1 2 sR p s 2 + R 2 - 2 sR cos θ fl fl fl π 0 = q 4 π² 0 1 2 sR ( p ( s + R ) 2 - p ( s - R ) 2 ) Φ avg = 1 4 π² 0 q s which is the potential due to the point charge evaluated at the center of the sphere. 2. Conductive medium between concentric spheres. (a) For a conducting medium, ~ J = σ c ~ E , where σ c is the conductivity. For a uniform conductor and steady current, ~ ∇ · ~ E = 0 , within the medium, and thus, 2 Φ = 0 . The

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potential between the conductors satisfies Laplace’s equation which, in spherical coordinates, is written, 2 Φ = 1 r 2 d dr r 2 d Φ dr · = 0 Since the problem is spherically symmetric and independent of the θ and φ coordinates. The region of interest does not contain the point r = 0 , so we can recast Laplace’s equation as, ∂r r 2 Φ · = 0 r 2 d Φ dr = - C d Φ dr = - C r 2 where C is an integration constant. Integrating this equation, we obtain Φ( r ) = C r + B where B is the second integration constant. Setting the boundary conditions to be V ( a ) = 0 and V ( b ) = V 0 we get C a + B = 0 B = - C a C b + B = C b - C a = C a - b ab = V 0 C = ab a - b V 0 and thus the potential between the conductors is, Φ( r ) = ab a - b V 0 1 r - 1 a · and the electric field is, ~ E ( r ) = - ~ Φ = ˆ r d Φ dr = ab b - a V 0 r 2 ˆ r The current density is,
~ J = σ c ~ E = σ c ab b - a · V 0 r 2 ˆ r We find the total current by integrating over the cross-sectional area (a spherical surface) between the conductors: I = Z S ~ J · d~a = σ c ab b - a · V 0 Z 2 π 0 Z π 0 1 r 2 r 2 sin θdθ = 4 π ab b - a · σ c V 0 Notice that as b → ∞ , I

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