There are n electrons on a drop of oil. It is in equilibrium in an electric field of intensity E. If the density of the oil is $\rho$, then radius of the drop will be

• A

  ${\left(\frac{4\mathrm{\pi eE}}{3\mathrm{n\rho g}}\right)}^{1/3}$

• B

  ${\left(\frac{3\mathrm{n\rho g}}{4\mathrm{\pi eE}}\right)}^{1/3}$

• C

  ${\left(\frac{4\mathrm{\pi \rho g}}{3\mathrm{neE}}\right)}^{1/3}$

• D

  ${\left(\frac{3\mathrm{neE}}{4\mathrm{\pi \rho g}}\right)}^{1/3}$