Q. An oil drop of mass 50 mg and of charge$- 5\mu C$ is just balanced in air against the force of gravity. Calculate the strength of the electric field required to balance is, $(g = 9.8\ ms^{- 2})$
A) $98\ NC^{- 1}$ upwards
B) 98 $NC^{- 1}$ downwards
C) 9.8 $NC^{- 1}$ towards north
D) 9.8 $NC^{- 1}$ towards south
Explanation
(b) $qE = mg$
or $E = \frac{mg}{q} = \frac{(50 \times 10^{- 6}) \times 9.8}{5 \times 10^{- 6}} = 98\ NC^{- 1}$
Since the force due to electric field on charged particle should be opposite to the gravity pull and charge on the drop is negative, hence the electric field must act vertically downwards
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.