Physics - Electrostatics Question with Solution | TestHub

According to the question, a proton is released at rest mid-way between the two plates and is found to move at $45^{\circ }$, so net force is at $45^{\circ }$ from vertical and two forces acting on the proton just after the release are as shown in the figure. $qE=mg$
$\therefore V=\frac{mgd}{q}=\frac{1.67\times 10^{-27}\times 10\times 10^{-2}}{1.6\times 10^{-19}}=10^{-9}\mathrm{V}$