Q. An oil drop carrying a charge $q$ has a mass $m$ kg. It is falling freely in air with terminal speed $v$. The electric field required to make the drop move upwards with the same speed is
A) $\frac{mg}{q}$
B) $\frac{2mg}{q}$
C) $\frac{mgv}{q^{2}}$
D) $\frac{2mgv}{q}$
Explanation
(b) When the oil drop is falling freely under the effect of gravity is a viscous medium with terminal speed $v,$ then
$mg = 6\pi\ \eta\ r\ v$ …(i)
To move the oil drop upward with terminal velocity $v$ if $E$ is the electric field intensity applied, the
$Eq = mg + 6\pi\eta rv = mg + mg = 2mg$
So $E = 2\ mg/q$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.