Question

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

• A. $1.6\times 10^{-19}C$
• B. $3.2\times 10^{-19}C$
• C. $0.8\times 10^{-19}C$
• D. $8.0\times 10^{-19}C$

Answer 1

Answer: (D)

Here,
$<br/>E=\frac{81\pi }{7}\times 10^{5}V{\text{m }}^{-1}<br/>$
$<br/>v=2\times 10^{-3}m{s}^{-1}<br/>$
$<br/>\eta =1.8\times 10^{5}Ns{m}^{-2}<br/>$
$<br/>\rho =900kg{m}^{-3}<br/>$
When the electric field is switched off, let the drop falls with terminal velocity v , then
$<br/>v=\frac{2r^2(\rho -\sigma )g}{9\eta }\text{ or }r={\left[\frac{9v\eta }{2(\rho -\sigma )g}\right]}^{\frac{1}{2}}<br/>$
$<br/>\therefore q=\frac{1}{E}\times \frac{4}{3}\pi \rho g\left[\frac{9v\eta }{2(\rho -\sigma )g}\right]<br/>$
$<br/>=\frac{7}{81\pi \times 10^5}\times \frac{4}{3}\times \pi \times 900\times 9.8\times {\left[\frac{9\times 8\times 10^{-5}\times 2\times 10^{-3}}{2\times 900\times 9.8}\right]}^{\frac{3}{2}}<br/>$
On solving we get, $q=8\times 10^{-19}C$