Question

In Milikan's oil drop experiment, an oil drop of radius r and charge q is held in equilibrium between the plates of as charged parallell plate capacitor when the potential difference is V. To keep as drop of radius 2r and with a charge 2q in equilibrium between the plates the potential difference V required is

Answer 1

The potential difference required to keep as a drop of radius 2r and charge 2q in equillibrium is 4V

• Let us go through the given parameters in the problem

       Radius = r

      Charge = q

      Potential difference = v

• In equillibrium condition qE=mg

       q(V/d) = (4/3)πr³×pg

      Where p=density of oil drop

                 d=separation between plates

         i.e., v∝r³/g

• In case 2 r=2r and q=2q

        The potential becomes    V1 = (2r)³/2q = 8r³/2q = 4r³/2q =4(V) = 4V

• Therefore the potential difference required is 4V