Question

A charged water drop of radius r is in the equilibrium in an electric field. If charge on it equal to charge on an electron, then intensity field will be: (density of water = d, assume gravity)

• A. $\frac{8\pi r^{3}pg}{3e}$
• B. $\frac{\pi r^{3}pg}{3e}$
• C. $\frac{5\pi r^{3}pg}{3e}$
• D. $\frac{4\pi r^{3}pg}{3e}$

Answer 1

The correct option is D $\frac{4\pi r^{3}pg}{3e}$

Under equilibrium condition:-

Force due to electric field $=$ weight of the water drop

$\begin{array}{c}qE=mg\\ eE=d\times \frac{4}{3}\pi r^3\times g\end{array}$

Hence,

$E=\frac{4\pi r^{3}pg}{3e}$

so, option $D$ is correct answer.