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**Unformatted text preview: **ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 6 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 π 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 π Z 2 π dφ Z π R 2 sin θ dθ √ s 2 + R 2- 2 sR cos θ = q 4 π 1 2 sR p s 2 + R 2- 2 sR cos θ π = q 4 π 1 2 sR ( p ( s + R ) 2- p ( s- R ) 2 ) Φ avg = 1 4 π q s which is the potential due to the point charge evaluated at the center of the sphere. 2. Potential between two conducting planes forming a wedge. From the geometry (symmetry) of the problem, we immediately see that the potential is inde- pendent of the cylindrical coordinates r and z (because the potential on the given boundaries are independent of those coordinates). Thus, Laplace’s equation reduces to, ∇ 2 Φ = 1 r 2 d 2 Φ dφ 2 = 0 ⇒ d 2 Φ dφ 2 = 0 This can be solved by direct integration, giving Φ( φ ) = Aφ + B where A and B are constants to be determined from the boundary conditions. For < φ < α Φ(0) = 0 = B and Φ( α ) = Aα = V Thus, Φ( φ ) = V α φ, < φ < α For α < φ < 2 π Φ( α ) = Aα + B = V , and Φ(2 π ) = 2 πA + B = 0 A ( α- 2 π ) = V , and B =- 2 πA Thus, Φ( φ ) = V α- 2 π ( φ- 2 π ) , α < φ < 2 π 3. Potential outside a conducting sphere held at potential V in presence of a point charge. This is similar to the example worked in class. We already know the solution when the sphere is grounded. In this case, the image charge is placed at position d = R 2 /s and has magnitude q i =- q ( R/s ) : The total potential in this case (found in class) is, Φ grounded sphere = q 4 π 1 r q- R sr i = q 4 π 1 √ r 2 + s 2- 2 rs cos θ- R s p r 2 + ( R 2 /s ) 2- 2 r ( R 2 /s )cos θ where we have used the Law of Cosines, r 2 q = r 2 + s 2- 2 rs cos θ and r 2 i = r 2 +( R 2 /s ) 2- 2 r (...

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