Question

Two large vertical and parallel metal plates having a separation of $1cm$ 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 ${45}^o$ to the vertical JUST after release. Then $X$ is nearly

• A. $1\times {10}^{-5}V$
• B. $1\times {10}^{-7}V$
• C. $1\times {10}^{-9}V$
• D. $1\times {10}^{-10}V$

Answer 1

The correct option is C $1\times {10}^{-9}V$

For proton to move at ${45}^o$

$qE=mg$

so that the net force is along ${45}^o$

Now $qE=mg$ and $E=\frac{V}{d}$

$\frac{1.6}{{10}^{19}}\times \frac{V\times 100}{1}=\frac{1.6}{{10}^{27}}\times 10$

$V=1\times {10}^{-9}V$