Q. Two metal pieces having a potential difference of 800 V are 0.2 m apart horizontally. A particle of mass $1.96 \times 10^{- 15}$kg is suspended in equilibrium between the plates. If $e$ is an elementary charge, then charge on the particle is
A) 8 $e$
B) 6 $e$
C) 3 $e$
D) $e$
Explanation
(c) In equilibrium, $F = qE = (ne)\frac{v}{d} = mg$
$n = \frac{mgd}{eV} = \frac{1.96 \times 10^{- 15} \times 9.8 \times 0.02}{1.6 \times 10^{- 19} \times 800} = 3\ $
Therefore, $\ q = 3\ e$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.