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Unformatted text preview: ECE 3209 Electromagnetic Fields University of Virginia Fall 2009 Homework # 6 Solutions 1. Cheng P.4-7. For a point charge Q placed a distance d above a large grounded conducting plane, we found the potential in class to be, ( x,y,z ) = Q 4 1 p x 2 + y 2 + ( z- d ) 2- 1 p x 2 + y 2 + ( z + d ) 2 From the boundary conditions on the electrostatic field, we know that at the surface of the conductor, ~ E = n =- ~ n =- Thus, the surface charge on the conductor is given by, =- z fl fl fl z =0 =- 1 2 Qd ( x 2 + y 2 + d 2 ) 3 / 2 The total charge on the conducting plane is found by integrating over the charge density: Q ind = Z S ( x,y ) dxdy =- 1 2 Z Z 2 Qdr ( r 2 + d 2 ) 3 / 2 drd where we have used the rectangular-cylindrical coordinate relations, r 2 = x 2 + y 2 , and da = rdrd Performing the integrals, we get Q ind =- Qd 2 (2 ) - 1 r 2 + d 2 fl fl fl =- Q So the total induced charge is equal an opposite to the point charge that induced it (as expected). 2. Cheng P.4-13. Consider a two-wire transmission line made of parallel conducting cylinders of radii a 1 and a 2 separated by distance...
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- Fall '08