Question

Two metal spheres of radius $ R $ and $ 3R $ have same surface charge density $ \sigma $. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes $ \sigma_1 $ and $ \sigma_2 $, respectively. The ratio $ \frac{\sigma_1}{\sigma_2} $ is:

• A. \( 9 \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{9} \)
• D. \( 3 \)

Hint:

When two conductors are brought into contact, they will share charge until their potentials are equal. Charge conservation and the formula for potential can help you determine the final distribution of charge.

Answer 1

The Correct Option is D

Given that the two spheres have the same surface charge density \( \sigma \), we know the total charge on each sphere is given by: \[ Q = \sigma \times \text{Surface area} \] For the smaller sphere, with radius \( R \), the surface area is \( 4 \pi R^2 \).
Hence, the total charge on the smaller sphere is: \[ Q_{\text{small}} = \sigma \times 4 \pi R^2 \] For the larger sphere, with radius \( 3R \), the surface area is \( 4 \pi (3R)^2 = 36 \pi R^2 \).
Hence, the total charge on the larger sphere is: \[ Q_{\text{large}} = \sigma \times 36 \pi R^2 \] When the spheres are brought into contact, charge will flow between them until they reach the same potential.
Since the potential of a sphere is given by \( V = \frac{Q}{4 \pi \epsilon_0 r} \), the potentials of the two spheres must be equal when they are in contact. Let the charge on the smaller sphere after contact be \( Q_1 \) and on the larger sphere be \( Q_2 \).
Using the condition for equal potentials: \[ \frac{Q_1}{4 \pi \epsilon_0 R} = \frac{Q_2}{4 \pi \epsilon_0 (3R)} \] Simplifying: \[ Q_1 = \frac{Q_2}{3} \] Since charge is conserved: \[ Q_1 + Q_2 = Q_{\text{small}} + Q_{\text{large}} = \sigma \times 4 \pi R^2 + \sigma \times 36 \pi R^2 = 40 \pi R^2 \sigma \] Substituting \( Q_1 = \frac{Q_2}{3} \) into this: \[ \frac{Q_2}{3} + Q_2 = 40 \pi R^2 \sigma \] Solving for \( Q_2 \): \[ \frac{4 Q_2}{3} = 40 \pi R^2 \sigma \quad \Rightarrow \quad Q_2 = 30 \pi R^2 \sigma \] Now, using \( Q_2 = 30 \pi R^2 \sigma \) to find the surface charge density on the larger sphere after separation: \[ \sigma_2 = \frac{Q_2}{36 \pi R^2} = \frac{30 \pi R^2 \sigma}{36 \pi R^2} = \frac{5 \sigma}{6} \] For the smaller sphere: \[ \sigma_1 = \frac{Q_1}{4 \pi R^2} = \frac{10 \pi R^2 \sigma}{4 \pi R^2} = \frac{5 \sigma}{2} \] Finally, the ratio \( \frac{\sigma_1}{\sigma_2} \) is: \[ \frac{\sigma_1}{\sigma_2} = \frac{\frac{5 \sigma}{2}}{\frac{5 \sigma}{6}} = 3 \] Thus, the correct answer is Option 4.