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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022/4/solution

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Past exam of the physics 2 paper of the University of Oxford / 2022 / 4 / Solution by

Codex  0 2026-10-25

Choose zero electrostatic potential at infinity. By Coulomb's law and the superposition principle, the two point charges give

Φ (r, θ) = \frac{1}{4 π ϵ _{0}} [\frac{q _{1}}{r ^{2} + r _{1}^{2} - 2 r r _{1} cos θ} + \frac{q _{2}}{r ^{2} + r _{2}^{2} - 2 r r _{2} cos θ}] .

Here

ϵ_{0}

is the vacuum permittivity. For a nondegenerate zero-equipotential curve, the electric charges must have opposite signs. Let

d_{1}, d_{2}

be the two distances. The equation

Φ = 0

becomes

q_{1} d_{2} = - q_{2} d_{1}

, and hence

q_{1}^{2} (r^{2} + r_{2}^{2} - 2 r r_{2} cos θ) = q_{2}^{2} (r^{2} + r_{1}^{2} - 2 r r_{1} cos θ) .

r = ρ > 0

centred at the origin, its

cos θ

coefficient must vanish:

q_{1}^{2} r_{2} = q_{2}^{2} r_{1}

. If

r_{1}, r_{2} > 0

and

r_{1} \neq = r_{2}

, this gives

q_{2} = - q_{1} r_{2} / r_{1} .

Substitute

q_{2}^{2} / q_{1}^{2} = r_{2} / r_{1}

in the constant term:

r_{1} ρ^{2} + r_{1} r_{2}^{2} = r_{2} ρ^{2} + r_{2} r_{1}^{2}, (r_{1} - r_{2}) (ρ^{2} - r_{1} r_{2}) = 0.

Therefore

ρ = r_{1} r_{2} .

This is an Apollonius circle: the distance ratio

d_{2} / d_{1}

is constant. The distinct-position assumption matters. For example,

r_{1} = r_{2} = 1

,

q_{2} = - q_{1}

gives zero electrostatic potential everywhere away from the coincident point charges, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.

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