Question

There are two charged spheres of radius $ R $ and $ 3R $. When the spheres are made to touch each other and then separate, the surface charge density becomes $ \sigma_1 $ and $ \sigma_2 $ respectively. Find the ratio $ \frac{\sigma_1}{\sigma_2} $.

• A. 5
• B. 5
• C. 3
• D. 9

Hint:

When two charged spheres touch, the charge redistributes such that the electric potential on both spheres becomes equal. The surface charge density is inversely proportional to the square of the radius of the sphere.

Answer 1

The Correct Option is C

Step 1: Charge Redistribution When two conducting spheres are in contact, charge will redistribute between them until the electric potential on both spheres is the same. Let the initial charges on the spheres be \( Q_1 \) and \( Q_2 \) for the spheres of radii \( R \) and \( 3R \) respectively. ###
Step 1: Charge Redistribution When the spheres touch, the potential on both spheres becomes equal.
The potential on a sphere is given by: \[ V = \frac{kQ}{R} \] where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere. Since the potentials are equal after they touch, we have: \[ \frac{kQ_1}{R} = \frac{kQ_2}{3R} \] Simplifying this equation: \[ Q_1 = \frac{Q_2}{3} \] ###
Step 2: Total Charge Conservation Let the total charge be \( Q_{\text{total}} \).
Since charge is conserved: \[ Q_1 + Q_2 = Q_{\text{total}} \] Substituting \( Q_1 = \frac{Q_2}{3} \) into this: \[ \frac{Q_2}{3} + Q_2 = Q_{\text{total}} \] \[ \frac{4Q_2}{3} = Q_{\text{total}} \quad \Rightarrow \quad Q_2 = \frac{3Q_{\text{total}}}{4} \] Thus, \( Q_1 = \frac{Q_{\text{total}}}{4} \). ###
Step 3: Surface Charge Densities The surface charge density \( \sigma \) on a sphere is given by: \[ \sigma = \frac{Q}{A} = \frac{Q}{4\pi r^2} \] For the sphere with radius \( R \), the surface charge density \( \sigma_1 \) is: \[ \sigma_1 = \frac{Q_1}{4\pi R^2} = \frac{\frac{Q_{\text{total}}}{4}}{4\pi R^2} = \frac{Q_{\text{total}}}{16\pi R^2} \] For the sphere with radius \( 3R \), the surface charge density \( \sigma_2 \) is: \[ \sigma_2 = \frac{Q_2}{4\pi (3R)^2} = \frac{\frac{3Q_{\text{total}}}{4}}{4\pi (9R^2)} = \frac{3Q_{\text{total}}}{36\pi R^2} = \frac{Q_{\text{total}}}{12\pi R^2} \] ###
Step 4: Ratio of Surface Charge Densities Now, the ratio of the surface charge densities is: \[ \frac{\sigma_1}{\sigma_2} = \frac{\frac{Q_{\text{total}}}{16\pi R^2}}{\frac{Q_{\text{total}}}{12\pi R^2}} = \frac{12}{16} = \frac{3}{4} \] Thus, the ratio of \( \frac{\sigma_1}{\sigma_2} \) is \( 3 \), so the correct answer is (3).