Question

An oil drop having charge $2\, e$ is kept stationary between two parallel horizontal plates $2.0\, cm$ apart when a potential difference of $12000$ volts is applied between them. If the density of oil is $900\, kg / m ^{3}$, the radius of the drop will be

• A. $2.0 \times 10^{-6} m$
• B. $1.7 \times 10^{-6} m$
• C. $1.4 \times 10^{-6} m$
• D. $1.1 \times 10^{-6} m$

Answer 1

Answer: (B)

In equilibrium, $Q E=m g$
$\Rightarrow Q \cdot \frac{V}{d}=m g=\left(\frac{4}{3} \pi r^{3} \rho\right) g$
$\Rightarrow 2 \times 1.6 \times 10^{-19} \times \frac{12000}{2 \times 10^{-2}}$
$=\frac{4}{3} \pi r^{3} \times 900 \times 10$
$\Rightarrow r=1.7 \times 10^{-6} m$