Question

An oil drop, carrying 6 electronic charges and having a mass of $1.6\times {10}^{-12}g$, falls with terminal velocity in a medium. What magnitude of vertical electric field $\left(\frac{kN}{C}\right)$ is required to move the drop upward with the same speed as it was formerly moving downward with? Ignore buoyancy.

Answer 1

In the above diagram, two situations are depicted i.e in first its moving downwards and in second situation going upward.
By force balancing we get
In first situation
$F_v=mg$
where $F_v$ is the viscous force
and in second situation
$F_E=F_v+mg$
So replacing $F_v$ in second situation from first
$F_E=2mg\Rightarrow QE=2mg$
where $Q=6\times \text{charge of an electon}$
$\Rightarrow E=\frac{2mg}{Q}$
$\Rightarrow E=\frac{2\times 1.6\times {10}^{-15}\times 10}{6\times 1.6\times {10}^{-19}}=33.33\text{kN/C}$