• # question_answer A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength$\,\frac{81\pi }{7}\times {{10}^{5}}V{{m}^{-1}}$. When the field is switched off, the drop is observed to fall with terminal velocity $2\times {{10}^{-3}}m{{s}^{-1}}$ Given $g=9.8\text{ }{{\operatorname{ms}}^{-2}}$, viscosity of the air $=1.8\times \text{ }$${{10}^{-5}}Ns\,{{m}^{-2}}$ and the density of oil $=900\text{ }kg\,{{\operatorname{m}}^{-3}}$ the magnitude of $q$ is A) $1.6\times {{10}^{-19}}C$ B) $2.2\times {{10}^{-19}}C$ C) $4.8\times {{10}^{-19}}C$ D) $8.0\times {{10}^{-19}}C$

 When the electric field is on Force due to electric field= weight $qE=mg$ $qE=\frac{4}{3}\pi {{R}^{3}}\rho g$ $qE=\frac{4\pi {{R}^{3}}\rho g}{3E}...(i)$ When the electric field is switched off
 Weight=viscous drag force $\operatorname{mg}=6\pi \eta R{{v}_{t}}$ $\frac{4}{3}\pi {{R}^{3}}\rho g=6\pi \eta R{{v}_{t}}$$\therefore \,\,\,\,R=\sqrt{\frac{9\nu \eta {{v}_{t}}}{2\rho g}}...\left( \operatorname{ii} \right)$ From (i) and (ii), $q=\frac{4}{3}\pi {{\left[ \frac{9\nu \eta {{v}_{t}}}{2\rho g} \right]}^{\frac{3}{2}}}\times \frac{\rho g}{E}$$=\frac{4}{3}\times \pi {{\left[ \frac{9\times 1.8\times {{10}^{-5}}\times 2\times {{10}^{-3}}}{2\times 900\times 9.8} \right]}^{\frac{3}{2}}}$$\times \frac{900\times 9.8\times 7}{81\pi \times {{10}^{5}}}$