Question

An oil drop of radius 2 mm with a density 3 g cm⁻³ is held stationary under a constant electric field 3.55 × 10⁵ 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/s²

• A. 17.3 × 10¹⁰
• B. 1.73 × 10¹⁰
• C. 1.73 × 10¹²
• D. 48.8 × 10¹¹

Hint:

In Millikan's experiment, balance the upward force ($qE$) against gravity ($mg$). Always ensure density is in $\text{kg/m}^3$ and radius in meters.

Answer 1

The Correct Option is B

Step 1: For the drop to be stationary, Electric force = Weight. $qE = mg \implies (ne)E = (\text{Volume} \times \text{Density})g$.
Step 2: $n \times (1.6 \times 10^{-19}) \times (3.55 \times 10^5) = \frac{4}{3} \pi (2 \times 10^{-3})^3 \times (3000) \times 9.81$. Note: $3 \text{ g cm}^{-3} = 3000 \text{ kg m}^{-3}$.
Step 3: $n \times 5.68 \times 10^{-14} \approx 9.85 \times 10^{-4}$. $n = \frac{9.85 \times 10^{-4}}{5.68 \times 10^{-14}} \approx 1.73 \times 10^{10}$.