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 $${45^ \circ }$$ 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 :
$$1 \times {10^{ - 9}}V\,$$
Solution :
The two forces acting on the proton just after the release are shown in the figure. In this situation
$$\eqalign{ & qE = mg\,\,\,\,\,\,\left[ {\therefore \theta = {{45}^ \circ }} \right] \cr & \therefore q\left( {\frac{V}{d}} \right) = mg \cr} $$
$$\therefore V = \frac{{mgd}}{q} = \frac{{1.67 \times {{10}^{ - 27}} \times 10 \times {{10}^{ - 2}}}}{{1.6 \times {{10}^{ - 19}}}} = {10^{ - 9}}V$$
The two forces acting on the proton just after the release are shown in the figure. In this situation
$$\eqalign{ & qE = mg\,\,\,\,\,\,\left[ {\therefore \theta = {{45}^ \circ }} \right] \cr & \therefore q\left( {\frac{V}{d}} \right) = mg \cr} $$
$$\therefore V = \frac{{mgd}}{q} = \frac{{1.67 \times {{10}^{ - 27}} \times 10 \times {{10}^{ - 2}}}}{{1.6 \times {{10}^{ - 19}}}} = {10^{ - 9}}V$$
