Question

Àn oil drop carrying a charge of -3.2*epx-20 C is held stationary between two horizontal parallel plates 5.00mm apart when a p.d. of 500V is applied between the plates. Calculate the mass of the oil drop.

Answer 1

electric field between the two plates $\frac{\Delta V}{\Delta d}=\frac{500}{5\times10^{-3}}=10^5V/m$

as oil drop is in equilibrium hence

force due to gravity on drop = force due to electric field

$m\times9.8=3.2\times10^{-20}\times10^5\\m=3.27\times10^{-16}grams$