Q. A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity 20000 $V/m.$ If mass of the particle is $9.6 \times 10^{- 16}\ kg,$ the charge on it and excess number of electrons on the particle are respectively $(g = 10\ m/s^{2})$
A) $4.8 \times 10^{- 19}C,\ 3$
B) $5.8 \times 10^{- 19}C,\ 4$
C) $3.8 \times 10^{- 19}C,2$
D) $2.8 \times 10^{- 19}C,\ 1$
Explanation
(a) In equilibrium
$QE = mg \Rightarrow n = \frac{mg}{Ee} = \frac{9.6 \times 10^{- 16} \times 10}{20,000 \times 1.6 \times 10^{- 19}} = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.