3209-2009-Solutions6

# 3209-2009-Solutions6 - ECE 3209 — Electromagnetic Fields...

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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, Φ( x,y,z ) = Q 4 π² ‡ 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 = σ ² ˆ n =- ~ ∇ Φ ⇒ ∂ Φ ∂n =- σ ² Thus, the surface charge on the conductor is given by, σ =- ² ∂ Φ ∂z fl fl fl z =0 =- 1 2 π 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 σ ( x,y ) dxdy =- 1 2 π Z ∞ Z 2 π Qdr ( r 2 + d 2 ) 3 / 2 drdφ where we have used the rectangular-cylindrical coordinate relations, r 2 = x 2 + y 2 , and da = rdrdφ Performing the integrals, we get Q ind =- Qd 2 π (2 π ) ‡- 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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3209-2009-Solutions6 - ECE 3209 — Electromagnetic Fields...

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