q^{⋆} = - \frac{a}{r _{0}} q, r^{⋆} = \frac{a ^{2}}{r _{0}^{2}} r_{0} = (0, 0, \frac{a ^{2}}{r _{0}}) .

It lies inside the electrical conductor, outside the region where it represents the physical electric field. The electric potential is

ϕ (r) = ⎩ ⎨ ⎧ 0, \frac{1}{4 π ϵ _{0}} (\frac{q}{∣ r - r _{0} ∣} - \frac{q a / r _{0}}{∣ r - r ^{⋆} ∣}), r \leq a, r > a, r \neq = r_{0} .

At the physical point charge, the electric potential has its usual Coulomb's law singularity. There is no physical singularity at the image charge: the interior expression is identically zero.

For

∣ r ∣ = a

, direct expansion of squared distances gives

∣ r - r^{⋆} ∣^{2} = a^{2} + \frac{a ^{4}}{r _{0}^{2}} - \frac{2 a ^{2}}{r _{0}^{2}} r \cdot r_{0} = \frac{a ^{2}}{r _{0}^{2}} ∣ r - r_{0} ∣^{2} .

Consequently the two terms cancel exactly, proving the zero Dirichlet boundary condition on the sphere. At infinity each term is

O (r^{- 1})

, and more explicitly

ϕ (r) = \frac{q ( 1 - a / r _{0} )}{4 π ϵ _{0} r} + O (r^{- 2}) ⟶ 0.

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.

The equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.