Question

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

• A. $1.6\times {10}^{-19}C$
• B. $3.2\times {10}^{-19}C$
• C. $4.8\times {10}^{-19}C$
• D. $8.0\times {10}^{-19}C$

Answer 1

The correct option is D $8.0\times {10}^{-19}C$

$V=\frac{2V^2}{9}\frac{(P-\sigma )}{n}g$

$\frac{2}{103}=\frac{2}{9}{V}^2\frac{(900-0)}{1.8}\times {10}^5\times 9.8$

$r=\frac{3}{7}=\times \frac{1}{{10}^5}$

For balanced oil drop.

$mg=qE$

$900\times \frac{4}{3}\pi {(\frac{3}{7}\times \frac{1}{{10}^5})}^3=q\times \frac{81}{7}\pi \times {10}^5$

$q=8\times {10}^{-19}C$