 Solution

ID: past-exam-of-the-electromagnetism-course-of-the-university-of-oxford/2023/1/a/ii/solution

Let

n = r / a

E_{out} = \frac{σ}{ϵ _{0}} n, σ = ϵ_{0} E_{out} \cdot n = - ϵ_{0} \partial_{r} ϕ ∣_{a^{+}} .

Write

u = cos θ

D = ∣ r - r_{0} ∣

r = a

, and

z^{⋆} = a^{2} / r_{0}

. Differentiating the electric potential at fixed angles gives

\partial_{r} ϕ ∣_{a^{+}} = \frac{q}{4 π ϵ _{0} D ^{3}} [- (a - r_{0} u) + \frac{r _{0}^{2}}{a ^{2}} (a - \frac{a ^{2}}{r _{0}} u)] = \frac{q}{4 π ϵ _{0} D ^{3}} (\frac{r _{0}^{2}}{a} - a) .

Here the distance identity in the preceding answer was used before simplifying. Hence the induced surface charge density is

σ (r) = \frac{q ( a - r _{0}^{2} / a )}{4 π ∣ r - r _{0} ∣ ^{3}}, ∣ r ∣ = a .

For positive

q

this surface charge density is negative everywhere, with largest magnitude nearest the point charge.

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