A 100 eV proton is projected directly towards a large metal plate that has surface charge density of $2.2\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2$. From what distance (in mm) must the proton be projected, if it is to just fail to strike that plate?

Applying conservation of mechanical energy

$\mathrm{\Delta K}+\mathrm{\Delta U}=0\text{ or }\left({\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}\right)+\left({\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}\right)=0$

$\left(0-100\times 1.6\times {10}^{-19}\right)+1.6\times {10}^{-19}\cdot \mathrm{\Delta V}=0$

$\mathrm{\Delta V}=\mathrm{Ed}=\left(\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0}\right)\mathrm{d}=\left(\frac{2.2\times {10}^{-6}}{8.8\times {10}^{-12}}\right)\mathrm{d}$

$\left(0-100\times 1.6\times {10}^{-19}\right)+1.6\times {10}^{-19}\cdot \left(\frac{2.2\times {10}^{-6}}{8.8\times {10}^{-12}}\right)\mathrm{d}=0$

$\mathrm{d}=100\times \frac{8.8\times {10}^{-12}}{2.2\times {10}^{-6}}=4\times {10}^{-4} \mathrm{m}=0.4 \mathrm{mm}$