Question

A charge of oil drop of mass 9.75 X10-16 kq is
hanging in equilibrium in uniform electric field
of 3x10 4N/c. Find the charge on drop How
many elections the drop carry

​

Answer 1

Because the oil drop is hanging in equillibrium.

Therefore,

Vertical forces on the drop will be balanced

Weight of the body = Electrostatic Force

So,

mg = qE

q = mg/E = 9.75*10^-16*10/3*10^4 Coulomb

Thus, the charge can be calculated from here further.

Now,

Total charge = No.of electrons * (Charge on one electron)

So,

No.of electrons = q/Charge on 1 Electron

= q / 1.6*10^-19

Thus, the number of electrons can be calculated from here.

Hope this helps you !