Classical Electrodynamics · Problem 3.25

Multipoles, Electrostatics of Macroscopic Media, Dielectrics — John David Jackson

Physics — E&M · mixed-boundary-conditions-conducting-plane

Problem

Consider the surface-charge densities for the problem of Section 3.13 of the conducting plane with a circular hole of radius aa. (a) Show that the surface-charge densities on the top and bottom of the plane for ρ≥a\rho \geq a are σ+(ρ)=−ϵ0E0+Δσ(ρ)\sigma_+(\rho) = -\epsilon_0 E_0 + \Delta\sigma(\rho) σ−(ρ)=ϵ0E1+Δσ(ρ)\sigma_-(\rho) = \epsilon_0 E_1 + \Delta\sigma(\rho) where Δσ(ρ)=−ϵ0(E0−E1)π[aρ2−a2−sin⁡−1(aρ)]\Delta\sigma(\rho) = -\epsilon_0 \frac{(E_0 - E_1)}{\pi}\left[\frac{a}{\sqrt{\rho^2 - a^2}} - \sin^{-1}\left(\frac{a}{\rho}\right)\right] How does Δσ(ρ)\Delta\sigma(\rho) behave for large ρ\rho? Is Δσ(ρ)\Delta\sigma(\rho), defined in terms of Φ(1)\Phi^{(1)}, zero for ρ<a\rho < a? Explain. (b) Show by direct integration that lim⁡R→∞[2π∫aRdρ ρ(σ++σ−)+2πϵ0∫0Rdρ ρ(E0−E1)]=0\lim_{R\to\infty}\left[2\pi\int_a^R d\rho\,\rho(\sigma_+ + \sigma_-) + 2\pi\epsilon_0\int_0^R d\rho\,\rho(E_0 - E_1)\right] = 0 Interpret.

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