Q. Two metal pieces having a potential difference of $800V$ are $0.02m$ apart horizontally. A particle of mass $1.96 \times 10^{- 15}kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge, then charge on the particle is
A) $e$
B) $3e$
C) $6e$
D) $8e$
Explanation
(b) For equilibrium $mg = qE$
$\therefore 1.96 \times 10^{- 15} \times 9.8 = q \times \left( \frac{800}{0.02} \right)$
$\Rightarrow q = \frac{1.96 \times 10^{- 15} \times 9.8 \times 0.02}{800}$
$\Rightarrow n \times 1.6 \times 10^{- 19} = \frac{1.96 \times 10^{- 15} \times 9.8 \times 0.02}{800} \Rightarrow n = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.