Question

In Millikan's oil drop experiment an oil drop of radius r and change Q is held in equilibrium between the plates of a charged parallel plate capacitor when the potential change is V. To keep a drop radius 2r and with a change 2Q is equilibriu between the plates the potential difference V' required is:

Answer 1

In equilibrium , qE=mg

or q(V/d)=(4/3)πr 3

ρg where d= separation between plates, ρ= density of oil drop.

or V∝ qr 3.

So when r=2r and q=2q , the potential becomes V′ = 2q8r 3=4(r 3 /q)=4V