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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 $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$, 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 $\mathrm{q}$ 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
$ \frac{4}{3} \pi \mathrm{R}^3 \rho \mathrm{g}=\mathrm{qE}=6 \pi \eta \mathrm{Rv}_{\mathrm{T}} $
$ \therefore \mathrm{q}=8.0 \times 10^{-19} \mathrm{C}$
