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 Incoming links: Electrostatic reciprocity identity

\nabla^{2} ϕ_{i} = - ρ_{i} / ε_{0}

F = ϕ_{1} \nabla ϕ_{2} - ϕ_{2} \nabla ϕ_{1} .

The two cross terms in its divergence cancel, leaving

\nabla \cdot F = ϕ_{1} \nabla^{2} ϕ_{2} - ϕ_{2} \nabla^{2} ϕ_{1} = \frac{ϕ _{2} ρ _{1} - ϕ _{1} ρ _{2}}{ε _{0}} .

This electrostatic reciprocity identity follows from the divergence theorem, and is exactly Green's second identity for these potentials. Integrating and rearranging gives

\frac{1}{ε _{0}} \int_{V} ϕ_{1} ρ_{2} d V + \int_{\partial V} ϕ_{1} \nabla ϕ_{2} \cdot d S = \frac{1}{ε _{0}} \int_{V} ϕ_{2} ρ_{1} d V + \int_{\partial V} ϕ_{2} \nabla ϕ_{1} \cdot d S .

The boundary orientation is outward from the chosen region. For smooth charge densities this follows directly from classical differentiation. For isolated point charges, one may excise small spheres and take their radii to zero, provided the other potential is regular at each charge.

Use the auxiliary point charge of strength

q \neq = 0

at the origin, with potential

ϕ_{2} = q / (4 π ε_{0} r)

from part (a). Apply the electrostatic reciprocity identity of part (b) to the exterior region

a < r < R

, then let

R \to \infty

. Its inner boundary has outward normal

- e_{r}

, and

ρ_{2} = 0

everywhere in this exterior region. The outer boundary terms vanish under the stipulated decay assumption.

r = a

ϕ_{1} = Φ

and

\nabla ϕ_{2} \cdot (- e_{r}) = \frac{q}{4 π ε _{0} a ^{2}} .

Thus the surviving boundary term on the left is

Φ q / ε_{0}

. On the right,

ϕ_{2}

is constant on the inner sphere, so its boundary term is

\frac{q}{4 π ε _{0} a} \int_{r = a} \nabla ϕ_{1} \cdot (- e_{r}) d S = 0.

Indeed,

\nabla ϕ_{1} = - E_{1}

, and this integral is the outward electric flux from the interior ball, whose enclosed charge is zero by the hypothesis and Gauss's law. No symmetry of the exterior charge distribution is needed.

\frac{q Φ}{ε _{0}} = \frac{1}{ε _{0}} \int_{r > a} \frac{q ρ _{1} ( x )}{4 π ε _{0} r} d V .

Canceling

q / ε_{0}

gives

Φ = \frac{1}{4 π ε _{0}} \int_{r > a} \frac{ρ _{1} ( x )}{r} d V .

The inward normal on the inner boundary is essential to the positive sign in this result.