309-2008-Solutions2

# 309-2008-Solutions2 - ECE 309 — Electromagnetic Fields...

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Unformatted text preview: ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 2 Solutions 1. By symmetry, the electric field at Q must be directed perpendicular to the plane of the disc (away from the disc if it is positively charged, toward the disc if it is negatively charged). The field is found, as usual, through superposition — by integrating over the charge distribution ~ E ( Q ) = 1 4 π Z S σ r 2 da ˆ r Referring to the geometry below, and noting that the superposition integral above must be a vector sum , we need to single out the component of ~ E in the z-direction (that is the only component that we have at Q ): Thus, we have ~ E ( Q ) = σ 4 π Z R Z 2 π ρ h 2 + ρ 2 h p h 2 + ρ 2 | {z } cos θ dρdφ ˆ z = σh 2 Z R ρdρ ( h 2 + ρ 2 ) 3 2 ˆ z =- σh 2 1 p h 2 + ρ 2 R ˆ z = σh 2 1 h- 1 p h 2 + R 2 ˆ z ~ E ( Q ) = σ 2 1- h p h 2 + R 2 ˆ z ~ F ( Q ) = Q ~ E ( Q ) = σQ 2 1- h p h 2 + R 2 ˆ z In the limiting case, h R , we have (using a Taylor series approximation): h R , ~ F ( Q ) = σQ 2 1- 1 p 1 + ( R /h ) 2 ˆ z ≈ σQ 2 1- h 1- 1 2 h R h i 2 i ˆ z ~ F ( Q ) ≈ σR 2 Q 4 h 2 ˆ z = 1 4 π σπR 2 Q h 2 The final expression above is simply Coulomb’s law for a test charge Q and a source charge q = σπR 2 , which is the net charge of the disc. This makes sense, because as one gets further away from the disc, it begins to look more-and-more like a point charge....
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309-2008-Solutions2 - ECE 309 — Electromagnetic Fields...

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