A = π (0.10 m)^{2}

, the initial capacitance and opposite plate electric charges are

C_{i} = \frac{ϵ _{0} A}{0.010 m} = 2.78163 \times 1 0^{- 11} F, Q_{\pm} = \pm C_{i} (100 V) = \pm 2.78163 \times 1 0^{- 9} C .

After the electric battery is disconnected, the isolated plate electric charges stay fixed. Doubling the separation halves the capacitance. The quasistatic pulling work is the increase of electrostatic energy:

W = \frac{Q ^{2}}{2 C _{f}} - \frac{Q ^{2}}{2 C _{i}} = \frac{1}{2} C_{i} (100 V)^{2}, W = 1.39081 \times 1 0^{- 7} J .

Δ V_{f} = \frac{Q}{C _{f}} = 200 V .

An electron released at rest gains kinetic energy

e Δ V_{f}

, where

e

is the positive elementary charge. Neglecting radiation and the negligible perturbation caused by this one electron, conservation of energy gives

\frac{1}{2} m_{e} v^{2} = e Δ V_{f}, v = \frac{2 e ( 200 V )}{m _{e}} = 8.39 \times 1 0^{6} m s^{- 1} .

This is

0.0280 c

, so the nonrelativistic kinetic energy is adequate. The circular plates have finite radii; the numerical results are leading parallel-plate capacitor values rather than exact finite-plate capacitances.