Question

A particular oil drop is suspended in a field with an intensity of 2.19 x 103N/C. If the apparent weight of the oil drop equal 1.4 x 10-15N. Calculate the charge of the oil drop in C?

Answer 1

$E=\frac Fq\implies q=\frac FE,$

$q=\frac{1.4\cdot10^{-15}}{2.19\cdot 10^3}=0.64\cdot 10^{-18}~\text C.$