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

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Electrostatics

An oil drop of radius 2 mm with a density

3 gcm^{- 3}

is held stationary under a constant electric field

3.55 \times 1 0^{5} Vm^{- 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}

Answer: (A)

The drop hangs in equilibrium when the electric force balances its weight,

q E = m g

. Its mass is

m = ρ \cdot \frac{4}{3} π r^{3} = 3000 \times \frac{4}{3} π (2 \times 1 0^{- 3})^{3} = 1.005 \times 1 0^{- 4} kg

so the charge it carries is

q = \frac{m g}{E} = \frac{1.005 \times 1 0 ^{- 4} \times 9.81}{3.55 \times 1 0 ^{5}} = 2.78 \times 1 0^{- 9} C

The number of excess electrons is

n = \frac{q}{e} = \frac{2.78 \times 1 0 ^{- 9}}{1.6 \times 1 0 ^{- 19}} = 1.73 \times 1 0^{10}

The drop possesses

1.73 \times 1 0^{10}

excess electrons.

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option