Question

There are ${}^{\prime }{n}^{\prime }$ electrons of charge ${}^{\prime }{e}^{\prime }$ in a drop of oil of density $\rho$ . It is in equilibrium in an electric field $E$. Then the radius of drop is :

• A. ${\left(\frac{2neE}{4\pi \rho g}\right)}^{1/2}$
• B. ${\left(\frac{neE}{\rho g}\right)}^{1/2}$
• C. ${\left(\frac{3neE}{4\pi \rho g}\right)}^{1/3}$
• D. ${\left(\frac{2neE}{\pi \rho g}\right)}^{1/3}$

Answer 1

The correct option is C ${\left(\frac{3neE}{4\pi \rho g}\right)}^{1/3}$

The drop is in equilibrium due to cancelling out of gravitational force and the electric force acting on it
Electric Force $=E\times ne$
Gravitational Force $=\frac{4}{3}\pi r^3\rho g$

$\frac{4}{3}\pi r^3\rho g=Ene\Rightarrow r=\sqrt[3]{3\frac{3Ene}{4\pi \rho g}}$