Q. A charged oil drop of mass $9.75 \times 10^{- 15}$kg and charge $30 \times 10^{- 16}$ C is suspended in a uniform electric field existing between two parallel plates. The field between the plates, taking (g =${\ ms}^{- 2})\ is\ $
A) $3.25\ {Vm}^{- 1}$
B) $300\ {Vm}^{- 1}$
C) $325\ {Vm}^{- 1}$
D) $32.5\ {Vm}^{- 1}$
Explanation
(d) In equilibrium
$eE = mg$
$E = \frac{mg}{e} = \frac{9.75 \times 10^{- 15} \times 10}{30 \times 10^{- 16}}$
$\ \ \ \ = 32.5\ {Vm}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.