Purcell Problem 1.19

Physics · electromagnetism

Problem

The electron sphere The electric field due to the sheet is σ\sigma/2ϵ0\epsilon_0, where σ\sigma = ρ\rhox is the effective charge per area. This follows from exactly the same Gauss's law argument as in the case of a thin sheet. The field at point B on top of the sheet is the sum of this σ\sigma/2ϵ0\epsilon_0 field plus the Coulomb field due to the sphere, but at radius R+x. The field at point A underneath the sheet is the difference (because now the sheet's field points downward) between the σ\sigma/2ϵ0\epsilon_0 field and the Coulomb field due to the sphere, at radius R. The charge in the sphere is (4/3)π\piR³ρ0\rho_0, so the field is larger above the sheet if (4/3)πR3ρ04πϵ0(R+x)2+ρx2ϵ0>(4/3)πR3ρ04πϵ0R2−ρx2ϵ0\frac{(4/3)\pi R^3\rho_0}{4\pi\epsilon_0(R+x)^2} + \frac{\rho x}{2\epsilon_0} > \frac{(4/3)\pi R^3\rho_0}{4\pi\epsilon_0 R^2} - \frac{\rho x}{2\epsilon_0} ⇔ρx>Rρ03(1−1(1+x/R)2)\Leftrightarrow \rho x > \frac{R\rho_0}{3}\left(1 - \frac{1}{(1+x/R)^2}\right) ⇔ρx>Rρ03⋅2xR\Leftrightarrow \rho x > \frac{R\rho_0}{3} \cdot \frac{2x}{R} ⇔ρ>2ρ03,(12.46)\Leftrightarrow \rho > \frac{2\rho_0}{3}, \quad (12.46) where we have used 1/(1+ϵ\epsilon)² ≈\approx 1/(1+2ϵ\epsilon) ≈\approx 1−2ϵ\epsilon to go from the second to the third line.

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