An oil drop of the radius 2 mm with a density 3 g cm$$^{-3}$$ is held stationary under a constant electric field $$3.55 \times 10^5$$ V m$$^{-1}$$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider $$g = 9.81$$ m s$$^{-2}$$).

For the oil drop to be held stationary, the electric force must balance the gravitational force: $$qE = mg$$, where $$m$$ is the mass of the oil drop and $$E$$ is the electric field.

The mass of the spherical oil drop is $$m = \rho \times \frac{4}{3}\pi r^3$$. With $$r = 2$$ mm $$= 2 \times 10^{-3}$$ m and $$\rho = 3$$ g/cm$$^3$$ $$= 3000$$ kg/m$$^3$$: $$m = 3000 \times \frac{4}{3}\pi (2 \times 10^{-3})^3 = 3000 \times \frac{4}{3}\pi \times 8 \times 10^{-9} = 1.005 \times 10^{-4}$$ kg.

The charge on the drop is $$q = \frac{mg}{E} = \frac{1.005 \times 10^{-4} \times 9.81}{3.55 \times 10^5} = \frac{9.86 \times 10^{-4}}{3.55 \times 10^5} = 2.776 \times 10^{-9}$$ C.

The number of excess electrons is $$n = \frac{q}{e} = \frac{2.776 \times 10^{-9}}{1.6 \times 10^{-19}} = 1.735 \times 10^{10} \approx 1.73 \times 10^{10}$$.