= Solution
Put the charged sheet at $z=0$ and the grounded plates at $z=\pm d/2$. By symmetry the <electric field> points away from the positive sheet with equal magnitude in the two gaps. A <Gaussian pillbox> crossing the sheet gives
$$
E_z(0^+)-E_z(0^-)=\frac\sigma{\varepsilon_0},\qquad
|E|=\frac\sigma{2\varepsilon_0}.
$$
The plate boundary conditions fix their <electric potentials> to zero. For the <charged sheet between grounded parallel plates>, the central potential is
$$
V_s=|E|\frac d2=\frac{\sigma d}{4\varepsilon_0},\qquad
\sigma=1.0\times10^{-7}\,\mathrm{C\,m^{-2}},\quad d=1.0\times10^{-2}\,\mathrm m.
$$
With <vacuum permittivity> $\varepsilon_0=8.85419\times10^{-12}\,\mathrm{F\,m^{-1}}$, $V_s\simeq28.24\,\mathrm V$. An <Electron> has <electric charge> $-e$, so moving from a grounded plate to this positive potential decreases its <potential energy> by $eV_s$. <Conservation of energy> then gives
$$
\frac12m_ev^2=eV_s=\frac{e\sigma d}{4\varepsilon_0},
\qquad
\boxed{v=\sqrt{\frac{e\sigma d}{2\varepsilon_0m_e}}
\simeq3.2\times10^6\,\mathrm{m\,s^{-1}}.}
$$
The unrounded value is $3.15\times10^6\,\mathrm{m\,s^{-1}}$, about $0.0105c$. Either plate gives the same <speed>. The ideal calculation neglects the electron's disturbance of the prescribed field and image-force effects at its release point, consistently with treating it as a <test particle>.
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