A tiny spherical oil drop carrying a net charge \(q\) is balanced in still air, with a vertical uniform electric field of strèngth \(\frac{81}{7} \pi \times 10^5 \mathrm{~V} / \mathrm{m}\). When the field is switched OFF, the drop is observed to fall with terminal velocity \(2 \times 10^{-3} \mathrm{~ms}^{-1}\). Here \(g=9.8 \mathrm{~m} / \mathrm{s}^2\), viscosity of air is \(1.8 \times 10^{-5} \mathrm{Ns} / \mathrm{m}^2\) and density of oil is \(900 \mathrm{~kg}\) \(\mathrm{m}^{-3}\). The magnitude of \(q\) is

Given, \(E=\frac{81 \pi}{7} \times 10^5 \mathrm{~V} / \mathrm{m}\)
Terminal velocity , \(v=2 \times 10^{-3} \mathrm{~m} / \mathrm{s}\)
\(g=9.8 \mathrm{~m} / \mathrm{s}^2\)
Viscosity, \(\eta=1.8 \times 10^{-5} \mathrm{~N}-\mathrm{s} / \mathrm{m}^2\)
Density, \(\rho=900 \mathrm{~kg} / \mathrm{m}^3\)
'Since, \(q E=m g\)
In the absence of electric field,
\(m g =6 \pi \eta r v \)
\(\Rightarrow q E =6 \pi \eta r \nu \)
\(\Rightarrow r=\frac{q E}{6 \pi \eta v}\)
[from Eq. (i)]
From Eq. (i), we get
\(m=\frac{q E}{g} \)
\(\Rightarrow \frac{4}{3} \pi r^3 d=\frac{q E}{g} \Rightarrow \frac{4}{3} \pi\left(\frac{q E}{6 \pi \eta v}\right)^3 d=\frac{q E}{g} \)
\(\Rightarrow q=\sqrt{\frac{3 \times 6^3 \pi^2 \eta^3 v^3}{4 E^2 g}}\)
\(=\sqrt{\frac{3 \times 6^3 \times(314)^2 \times\left(1.8 \times 10^{-5}\right)^3 \times\left(2 \times 10^{-3}\right)^3}{4 \times\left(\frac{81 \pi}{7} \times 10^5\right)^2 \times 9.8}} \)
\(=8 \times 10^{-19} \mathrm{C}\)