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: {Calculating 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: {Balancing forces}
The oil drop is stationary, so the electric force balances 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: {Calculating the number of excess electrons}
\[ q = ne \Rightarrow n = \frac{q}{e} = \frac{1.44 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 9 \] Thus, the number of excess electrons is approximately 9.

Answer 2

Step 1: Understand the physical condition
The oil drop is stationary, so the electric force balances the gravitational force:
\[ qE = mg \]

Step 2: Volume and mass of the oil drop
Radius of the drop: \( r = 1\, \mu m = 1 \times 10^{-6}\, m \)
Density: \( \rho = 1.26\, g/cm^3 = 1260\, kg/m^3 \)
Volume of a sphere: \( V = \frac{4}{3} \pi r^3 \)
\[ V = \frac{4}{3} \pi (1 \times 10^{-6})^3 = \frac{4}{3} \pi \times 10^{-18} \, m^3 \]
Mass: \( m = \rho \cdot V = 1260 \cdot \frac{4}{3} \pi \times 10^{-18} \approx 5.28 \times 10^{-15} \, kg \)

Step 3: Calculate the charge on the drop
\[ q = \frac{mg}{E} = \frac{5.28 \times 10^{-15} \cdot 10}{3.65 \times 10^4} \approx \frac{5.28 \times 10^{-14}}{3.65 \times 10^4} \approx 1.45 \times 10^{-18} \, C \]

Step 4: Calculate number of excess electrons
Charge of one electron: \( e = 1.6 \times 10^{-19} \, C \)
\[ n = \frac{q}{e} = \frac{1.45 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 9.06 \approx 9 \]

Final Answer: 9