Q. 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 $45^{o}$ 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) $1 \times 10^{- 10}V$
Explanation
(c) [image:1_image373.emf]
$mg = qE$
$1.67 \times 10^{- 27} \times 10 = 1.6 \times 10^{- 19} \times \frac{X}{0.01}$
$X = \frac{1.67}{1.6} \times 10^{- 9}V \Rightarrow X = 1 \times 10^{- 9}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.