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

To solve this problem, we need to understand how charge distribution occurs when two conducting spheres are brought into contact and then separated.

1. Given two metal spheres with radii \( R \) and \( 3R \) and the same initial surface charge density \( \sigma \), the charges \( Q_1 \) and \( Q_2 \) on these spheres can be calculated as:
   • The surface area of the smaller sphere = \( 4\pi R^2 \)
   • The surface area of the larger sphere = \( 4\pi (3R)^2 = 36\pi R^2 \)
   • Initial charge on smaller sphere \( Q_1 = \sigma \times 4\pi R^2 \)
   • Initial charge on larger sphere \( Q_2 = \sigma \times 36\pi R^2 \)
2. Total initial charge before contact:
   • \( Q_{\text{total}} = Q_1 + Q_2 = \sigma \times 4\pi R^2 + \sigma \times 36\pi R^2 = 40\pi R^2 \sigma \)
3. When both spheres are brought into contact, charge will redistribute based on the spheres' capacitances (which are proportional to their radii). The smaller sphere has radius \( R \) and the larger sphere has radius \( 3R \).
4. Total capacitance is \( C_{\text{total}} = R + 3R = 4R \). Charge is distributed in direct proportion to capacitance:
   • Charge on smaller sphere after separation \( Q_1' = \frac{R}{4R} \times 40\pi R^2 \sigma = 10\pi R^2 \sigma \)
   • Charge on larger sphere after separation \( Q_2' = \frac{3R}{4R} \times 40\pi R^2 \sigma = 30\pi R^2 \sigma \)
5. Surface charge density after separation:
   • \( \sigma_1 = \frac{Q_1'}{4\pi R^2} = \frac{10\pi R^2 \sigma}{4\pi R^2} = 2.5\sigma \)
   • \( \sigma_2 = \frac{Q_2'}{36\pi R^2} = \frac{30\pi R^2 \sigma}{36\pi R^2} = \frac{5\sigma}{6} \)
6. Hence, the ratio \( \frac{\sigma_1}{\sigma_2} \) is calculated as:
   • \( \frac{\sigma_1}{\sigma_2} = \frac{2.5\sigma}{\frac{5\sigma}{6}} = \frac{2.5 \times 6}{5} = 3 \)

The correct answer is therefore the ratio \( \frac{\sigma_1}{\sigma_2} = 3 \).

Answer 2

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.