Two large vertical and parallel metal plates having a separation of $1 \ cm$ are connected to a $DC$ voltage source of potential difference $X$. A proton is released at rest midway between the two plates. It is found to move at $45^{\circ}$ to the vertical $JUST$ after release. Then $X$ is nearly
IIT 2012, Advanced
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$mg = qE $
$1.67 \times 10^{-27} \times 10=1.6 \times 10^{-19} \times \frac{ X }{0.01} $
$X=\frac{1.67}{1.6} \times 10^{-9} V $
$X=1 \times 10^{-9} V$
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