Question

An oil drop of radius \(1 \, \mu{m}\) is held stationary under a constant electric field of \(3.65 \times 10^4 \, {N/C}\) due to some excess electrons present on it. If the density of the oil drop is \(1.26 \, {g/cm}^3\), then the number of excess electrons on the oil drop approximately is:
[Take, g = 10 m/s2]

• A. 7
• B. 12
• C. 9
• D. 8

Hint:

The charge on an oil drop in Millikan's experiment can be used to determine the number of excess electrons by balancing electric and gravitational forces.

Answer 1

The Correct Option is C

Step 1: Calculate the mass of the oil drop.
\[ {Density} = 1.26 \, {g/cm}^3 = 1.26 \times 10^3 \, {kg/m}^3 \] \[ {Volume} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1 \times 10^{-6} \, {m})^3 = \frac{4}{3}\pi \times 10^{-18} \, {m}^3 \] \[ {Mass} = {Density} \times {Volume} = 1.26 \times 10^3 \times \frac{4}{3}\pi \times 10^{-18} = 1.68\pi \times 10^{-15} \, {kg} \] Step 2: Balance forces.
The oil drop is stationary, indicating that the electric force equals the gravitational force: \[ qE = mg \] \[ q = \frac{mg}{E} = \frac{1.68\pi \times 10^{-15} \times 10}{3.65 \times 10^4} \approx 1.44 \times 10^{-18} \, {C} \] Step 3: Determine the number of excess electrons.
\[ q = ne \Rightarrow n = \frac{q}{e} = \frac{1.44 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 9 \] Consequently, the number of excess electrons is approximately 9.