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)

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