Question

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. $\frac{m g}{q}$
• B. $\frac{2 m g}{q}$
• C. $\frac{m g v}{q^{2}}$
• D. $\frac{2 m g v}{q}$

Answer 1

Answer: (B)

$\frac{\text{QV}}{\ell } = \frac{4}{3} \pi \text{r}^{3} \rho \text{g}$
When the oil drop is falling freely under the effect of gravity is a viscous medium with terminal speed $v,$ then
$mg=6\pi \, η \, r \, v$ ...(i)
To move the oil drop upward with terminal velocity $v$ if $E$ is the electric field intensity applied, the
$Eq=mg+6\pi ηrv=mg+mg=2mg$