Question

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

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

Answer 1

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

$qE=mg$
In absence of electric field,
$mg=6\pi \eta rv$
$\Rightarrow qE=6\pi \eta rv\Rightarrow r=\frac{qE}{6\pi \eta v}$
Since, $m=\frac{4}{3}\pi r^{3}d=\frac{qE}{g}$
$\Rightarrow \frac{4}{3}\pi {\left(\frac{qE}{6\pi \eta v}\right)}^{3}d=\frac{qE}{g}$
$\Rightarrow \frac{4}{3}\pi .\frac{q^2{E}^{2}d}{(6{)}^3{\pi }^3{\eta }^3{v}^3}=\frac{1}{g}$
$\Rightarrow q^2=\frac{(6{)}^3{\pi }^2{\eta }^3{v}^3\times 3}{4\times E^2\times d\times g}$
$q=8\times {10}^{-19}\text{C}$