Solution

In electrostatic equilibrium, a nonzero electric field inside a conductor would move its free charges. Hence

E = 0

in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential electric field is zero. A Gaussian pillbox across the surface gives

(E_{out} - E_{in}) \cdot n = \frac{σ}{ϵ _{0}} .

(108)

Since

E_{in} = 0

, the electrostatic boundary conditions at a conductor give

σ = ϵ_{0} E_{out} \cdot n .

(109)

For the widely separated shells, let their charges be

Q_{1}, Q_{2}

. Their common potential requires

\frac{Q _{1}}{4 π ϵ _{0} R _{1}} = \frac{Q _{2}}{4 π ϵ _{0} R _{2}}, Q_{1} + Q_{2} = Q .

(110)

Thus the charge sharing between distant connected spheres is

Q_{1} = \frac{R _{1}}{R _{1} + R _{2}} Q, Q_{2} = \frac{R _{2}}{R _{1} + R _{2}} Q .

(111)

For the charge on connected concentric spherical shells, the potentials at the two radii are

Φ (R_{1}) = \frac{1}{4 π ϵ _{0}} (\frac{Q _{1}}{R _{1}} + \frac{Q _{2}}{R _{2}}), Φ (R_{2}) = \frac{Q _{1} + Q _{2}}{4 π ϵ _{0} R _{2}} .

(112)

Equality forces

Q_{1} = 0

. Hence

Q_{1} = 0, Q_{2} = Q,

(113)

so all charge lies on the exterior of the outer shell.

For the neutral sphere in the uniform field, the far-field condition gives

α = - E

. Constancy of the potential on

r = R

gives

β = E R^{3}

. Therefore

Φ (r, θ) = - E (r - \frac{R ^{3}}{r ^{2}}) cos θ .

(114)

The outward normal field at the surface is

E_{r} (R, θ) = - \frac{\partial Φ}{\partial r}_{r = R} = 3 E cos θ .

(115)

The induced charge on a conducting sphere in a uniform electric field is consequently

σ (θ) = 3 ϵ_{0} E cos θ .

(116)

Finally,

\int_{S^{2}} σ d A = 6 π ϵ_{0} E R^{2} \int_{0}^{π} cos θ sin θ d θ = 0,

(117)

confirming neutrality.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 15C / Solution

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