Parallel Plates and potential Difference

Question

The density of the oil used to form droplets in the millikan experiment is

9.2 x 10^2 kg/m^3 and the radius of a typical oil droplets is 2.00 x 10^-6 m. When the horizontal plates are placed .018 m apart, an oil drop,later determined to have an excess of three electrons, is held in equilibrium. What potential difference must have been applied across the plates?

Answer 1

The volume of a sphere is (4/3)πr^3. The mass of the typical oil drop is:

m = dV = (920 kg/m^3) x ((4/3) x π x (2.00 x 10^-6 m)^3) = 3.08 x 10^-14 kg

The weight of the oil drop is:

F = ma = (3.08 x 10^-14 kg) x 9.8 m/s² = 3.02 x 10^-13 N

The electric field needed to keep the oil drop in equilibrium is:

E = F / Q = (3.02 x 10^-13 N) / (3 x (-1.60 x 10^-19 C) = -1.89 x 10^6 V/m

The potential difference would have to be:

∆V = -Ed = - (-1.89 x 10^6 V/m) x 0.018 m = 3.40 x 10^4 V