Question

(a) A charged oil drop is suspended in a uniform field of 300 V cm^-1 so that it neither rises nor falls. Find the charge on it, in coulomb, assuming its mass to be 9.75 * 10^-15 kg. (b) Find the number of electrons on the oil drop.​

Answer 1

Explanation:

Given : Electric field, E=3×10

4

V/m

Mass of the drop, M=9.9×10

−15

kg

Let q be the amount of the charge that the drop carries.

The coulomb force balances the gravitational force acting on the drop at equilibrium.

∴ qE=Mg ⟹q=

E

Mg

∴ q=

3×10

4

9.9×10

−15

×10

=3.3×10

−18

C