3209-2009-Solutions7

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

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Unformatted text preview: 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, avg 1 4 R 2 I sphere da, where ( r ) = 1 4 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 , so avg = 1 4 R 2 q 4 Z 2 d Z R 2 sin d s 2 + R 2- 2 sR cos = q 4 1 2 sR p s 2 + R 2- 2 sR cos fl fl fl = q 4 1 2 sR ( p ( s + R ) 2- p ( s- R ) 2 ) avg = 1 4 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 = c ~ E , where c is the conductivity. For a uniform conductor and steady current, ~ ~ E = 0 , within the medium, and thus, 2 = 0 . The potential between the conductors satisfies Laplaces equation which, in spherical coordinates, is written, 2 = 1 r 2 d dr r 2 d dr = 0 Since the problem is spherically symmetric and independent of the and coordinates. The region of interest does not contain the point r = 0 , so we can recast Laplaces equation as, r r 2 = 0 r 2 d dr =- C d dr =- C r 2 where C is an integration constant. Integrating this equation, we obtain ( r ) = C r + B where B is the second integration constant. Setting the boundary conditions to be V ( a ) = 0 and V ( b ) = V we get C a + B = 0 B =- C a C b + B = C b- C a = C a- b ab = V C = ab a- b V and thus the potential between the conductors is, ( r ) = ab a- b V 1 r- 1 a and the electric field is, ~ E ( r ) =- ~ = r d dr = ab b- a V r 2 r The current density is, ~ J = c ~ E = c ab b- a V r 2 r We find the total current by integrating over the cross-sectional area (a spherical surface)...
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3209-2009-Solutions7 - ECE 3209 Electromagnetic Fields...

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