Question

Two conducting solid spheres (A & B) are placed at a very large distance with charge densities and radii as shown:

• A. 4 : 1
• B. 1 : 2
• C. 2 : 1
• D. 1 : 4

Answer 1

The Correct Option is C

Given:

The final potential is the same on two conducting spheres of radii \( R \) and \( 2R \), with charges \( Q_1 \) and \( Q_2 \) respectively.

Step-by-Step Solution:

1. Potential Equality:

The potential on each sphere is given by: \[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R} \] Equating the potentials of the two spheres: \[ \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1}{R} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_2}{2R} \] Simplifying: \[ Q_1 = \frac{Q_2}{2} \tag{1} \]

2. Total Charge Distribution:

The total charge distributed between the spheres is: \[ Q_1 + Q_2 = \sigma \cdot 4 \pi R^2 + \sigma \cdot 4 \pi (2R)^2 \] Simplifying: \[ Q_1 + Q_2 = \sigma \cdot 4 \pi R^2 + \sigma \cdot 16 \pi R^2 \] \[ Q_1 + Q_2 = \sigma \cdot 20 \pi R^2 \tag{2} \]

3. Surface Charge Densities:

The surface charge densities are related to the charges and surface areas of the spheres: \[ \sigma_1 = \frac{Q_1}{4 \pi R^2}, \quad \sigma_2 = \frac{Q_2}{4 \pi (2R)^2} \] From \( Q_1 = \frac{Q_2}{2} \): \[ \frac{\sigma_2}{\sigma_1} = \frac{\frac{Q_2}{4 \pi (2R)^2}}{\frac{Q_1}{4 \pi R^2}} \] Substituting \( Q_1 = \frac{Q_2}{2} \): \[ \frac{\sigma_2}{\sigma_1} = \frac{Q_2}{4 \pi \cdot 4R^2} \cdot \frac{4 \pi R^2}{Q_1} \] Simplifying: \[ \frac{\sigma_2}{\sigma_1} = 2 \tag{3} \]

Final Answer:

The ratio of surface charge densities \( \sigma_2 : \sigma_1 \) is:

\( \sigma_2 : \sigma_1 = 2 : 1 \)

Thus, the correct answer is (C) : 2 : 1.