Q. Two parallel plates separated by a distance of $5mm$ are kept at a potential difference of $50V$ A particle of mass $10^{- 15}kg$ and charge $10^{- 11}C$ enters in it with a velocity $10^{7}m/s.$ The acceleration of the particle will be
A) $10^{8}\ m/s^{2}$
B) $5 \times 10^{5}\ m/s^{2}$
C) $10^{5}\ m/s^{2}$
D) $2 \times 10^{3}\ m/s^{2}$
Explanation
(a) $a = \frac{qE}{m} = \frac{q}{m}\left( \frac{V}{d} \right) = \frac{10^{- 11}}{10^{- 15}} \times \frac{50}{5 \times 10^{- 3}} = 10^{8}m/\sec^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.