A charged oil drop is to be held stationary between two plates separated by a distance 25 mm . If th

Question

A charged oil drop is to be held stationary between two plates separated by a distance 25 mm . If the mass of the drop is $5 	imes 10^{-15} \mathrm{~kg}$ in the charge on it is $10^{-18} \mathrm{C}$, the potential to be applied between the two plates is ( $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )\

Professor PC Thomas Classes · Answer

Electrostatic force balances the weight of the drop\ $	herefore \frac{\mathrm{qV}}{\mathrm{d}}=\mathrm{mg}$\ Where V is the potential and d is the distance between the plate of capacitor.\ $\mathrm{V}=\frac{\mathrm{mgd}}{\mathrm{q}}=\frac{5 	imes 10^{-15} 	imes 10 	imes 25 	imes 10^{-3}}{10^{-18}}$\