Two large vertical and parallel metal plates having a separation of $1 \mathrm{~cm}$ are connected to a DC…
Two large vertical and parallel metal plates having a separation of $1 \mathrm{~cm}$ 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
$1 \times 10^{-5} \mathrm{~V}$
$1 \times 10^{-7} V$
$1 \times 10^{-9} V$
$1 \times 10^{-10} V$
Solution
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.
$q E=m g$
$\therefore \quad V=\frac{m g d}{q}=\frac{1.67 \times 10^{-27} \times 10 \times 10^{-2}}{1.6 \times 10^{-19}}=10^{-9} \mathrm{~V}$
