Q. An oil drop having charge $2e$ is kept stationary between two parallel horizontal plates $2.0\ cm$ apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 $kg/m^{3},$ the radius of the drop will be
A) $2.0 \times 10^{- 6}m$
B) $1.7 \times 10^{- 6}m$
C) $1.4 \times 10^{- 6}m$
D) $1.1 \times 10^{- 6}m$
Explanation
(b) In equilibrium $QE = mg \Rightarrow Q.\frac{V}{d} = mg = \left( \frac{4}{3}\pi r^{3}\rho \right)g$
$\Rightarrow 2 \times 1.6 \times 10^{- 19} \times \frac{12000}{2 \times 10^{- 2}} = \frac{4}{3}\pi r^{3} \times 900 \times 10$
$\Rightarrow r = 1.7 \times 10^{- 6}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.