A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform…
A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 \pi}{7} \times 10^5 \mathrm{Vm}^{-1}$. When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \mathrm{~ms}^{-1}$. Given $g=9.8 \mathrm{~ms}^{-2}$, viscosity of the air $=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2}$ and the density of oil $=900 \mathrm{~kg} \mathrm{~m}^{-3}$, the magnitude of $q$ is
$1.6 \times 10^{-19} \mathrm{C}$
$3.2 \times 10^{-19} \mathrm{C}$
$4.8 \times 10^{-19} \mathrm{C}$
$8.0 \times 10^{-19} \mathrm{C}$
Solution
$
\begin{aligned}
& q E=m g \\
& \quad 6 \pi \eta r v=m g \\
& \frac{4}{3} \pi r^3 \rho g=m g \\
& \therefore \quad r=\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}
\end{aligned}
$
Substituting the value of $r$ in Eq. (ii) we get,
$
6 \pi \eta v\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}=m g
$
or $\quad(6 \pi \eta v)^3\left(\frac{3 m g}{4 \pi \rho g}\right)=(m g)^3$
Again substituting $m g=q E$ we get,
$
(q E)^2=\left(\frac{3}{4 \pi \rho g}\right)(6 \pi \eta v)^3
$
or $\quad q E=\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}$
$
\therefore \quad q=\frac{1}{E}\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}
$
Substituting the values we get ,
$
\begin{aligned}
q= & \frac{7}{81 \pi \times 10^5} \sqrt{\frac{3}{4 \pi \times 900 \times 9.8} \times 216 \pi^3} \\
& \times \sqrt{\left(1.8 \times 10^{-5} \times 2 \times 10^{-3}\right)^3} \\
& =8.0 \times 10^{-19} \mathrm{C}
\end{aligned}
$
$\therefore$ correct option is $(\mathrm{d})$