= Solution
Use the <parallel-plate capacitor> approximation, neglecting the <fringing field of a capacitor>. With plate area $A=\pi(0.10\,\mathrm m)^2$, the initial <capacitance> and opposite plate <electric charges> are
$$
C_i=\frac{\epsilon_0A}{0.010\,\mathrm m}=2.78163\times10^{-11}\,\mathrm F,
\qquad
\boxed{Q_\pm=\pm C_i(100\,\mathrm V)=\pm2.78163\times10^{-9}\,\mathrm C.}
$$
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>:
$$
W=\frac{Q^2}{2C_f}-\frac{Q^2}{2C_i}
=\frac12C_i(100\,\mathrm V)^2,
\qquad
\boxed{W=1.39081\times10^{-7}\,\mathrm J.}
$$
The final <electric potential difference> is therefore
$$
\boxed{\Delta V_f=\frac Q{C_f}=200\,\mathrm V.}
$$
An <electron> released at rest gains <kinetic energy> $e\Delta V_f$, where $e$ is the positive <elementary charge>. Neglecting radiation and the negligible perturbation caused by this one <electron>, <conservation of energy> gives
$$
\frac12m_ev^2=e\Delta V_f,
\qquad
\boxed{v=\sqrt{\frac{2e(200\,\mathrm V)}{m_e}}
=8.39\times10^6\,\mathrm{m\,s^{-1}}.}
$$
This is $0.0280c$, 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>.
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