Question

Two large vertical and parallel metal plates having a separation of $1\text{cm}$ 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

• 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 $1\times {10}^{-9}V$


Since the proton is released and it moves at an angle of ${45}^{\circ }$ with the vertical, the net force acting on the proton acts at an angle of ${45}^{\circ }$ ($\theta ={45}^{\circ }$) with the vertical.
Two forces act on the proton, downward force $mg$ and rightward force $qE$. They both must be equal in magnitude to have resultant at an angle of ${45}^{\circ }$ with them.
$F_x=F_y\Rightarrow qE=mg$
$q\frac{X}{d}=mg\Rightarrow X=\frac{mgd}{q}$
$X=\frac{1.67\times {10}^{-27}\times 10\times {10}^{-2}}{1.6\times {10}^{-19}}={10}^{-9}$