Question

A charged oil drop remains stationary when situated between two parallel horizontal metal plates 25 mm apart and a p.d. of 1000 V is applied to the plates. Find the charge on the drop if it has a mass of 5 × 10^-15kg.

Draw a sketch of the electric field between the plates and state if the field is everywhere uniform.

Answer 1

Given:

Separation of plates (d) = 25mm

Mass of drop (m) = 5*10^-15kg

Potential difference (U) = 1000V

Charge on the drop (q) =?

if the oil drop remains stationary, then 

The force on oil drop due to electric field = weight of the drop

i.e. qE=mg

or, $q = \frac{mg}{E} = \frac{mg}{\frac{u}{d}} = \frac{mgd}{u}$ $= \frac{5*10^{-15}*10*25*10^{-3}}{1000} = 1.25*10^{-18}C$

The charge on the oil drop q =1.25×10^-18C

Hence the required charge on the oil drop is 1.25×10^-18C