Question

An oil drop of radius $10^{-6}m$ carries charge equal to that of $3$ electrons. If density of the oil is $2\times 10^{3}kg-m^3$, then electric field required to keep the drop stationary is:

• A. $1.71\times 10^{5}V{m}^{-1}$
• B. $1.71\times 10^{3}V{m}^{-1}$
• C. $1.71\times 10^{6}V{m}^{-1}$
• D. $1.71V{m}^{-1}$

Answer 1

Answer: (A)

In order to keep the drop stationary.
For due to electric field = weight of the drop
$\therefore qE=mg$
$\Rightarrow E=\frac{mg}{q}=\frac{\text{ volume }\times \text{ density }\times g}{q}$
Given, $V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times {\left(10^{-6}\right)}^3$
$\rho =2\times 10^{3}kg{m}^{-3},$
$g=10m/s^2,q=ne=3\times 1.6\times 10^{-19}C$
$\therefore E=\frac{4\pi \times 10^{-18}}{3}\times \frac{2\times 10^3\times 10}{3\times 1.6\times 10^{-19}}$
$\Rightarrow E=\frac{8\times 3.14\times 10^5}{9\times 1.6}$
$\Rightarrow E=1.74\times 10^{5}V{m}^{-1}$