Question

A charged oil drop is suspended in uniform field of 3×10^4 Vm^-1 so that it neither falls nor arises. The charge on the drop will be

Answer 1

for this we require mass of that oil drop

by equating qE = mg

we can easily get the q

here E is 3×10^-4V/m and g is 10m/s^2

Answer 2

ANSWER:

Since the ball is hanging, force by gravity is balanced by the electric force.

qE = mg

q = mg/E

= (9.9 x 10-15 x 10)/(3 x 104)

q = 3.3 x 10^18 C