Two spherical conductors A and B of radii R and 2R respectively , are separated by a large distance . If some charge is given to both the spheres and later they are connected by a conducting wire , then in equilibrium condition , the ratio of the magnitude of the electric fields at the surface of spheres A and B is

Options

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

Answer

To solve the problem, we need to find the ratio of the electric fields at the surfaces of two spherical conductors A and B with radii R and 2R respectively, after they are connected by a conducting wire.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two spherical conductors: Sphere A with radius \( R \) and Sphere B with radius \( 2R \).
- Initially, they have charges \( q_1 \) and \( q_2 \) respectively.
- When connected by a conducting wire, they will reach an equilibrium state where their electric potentials are equal.

2. **Electric Potential of Spheres**:
- The electric potential \( V \) of a charged sphere is given by:
\[ V = \frac{q}{4\pi \epsilon_0 R} \]
- For Sphere A:
\[ V_A = \frac{q_1}{4\pi \epsilon_0 R} \]
- For Sphere B:
\[ V_B = \frac{q_2}{4\pi \epsilon_0 (2R)} = \frac{q_2}{8\pi \epsilon_0 R} \]

3. **Setting the Potentials Equal**:
- At equilibrium, \( V_A = V_B \):
\[ \frac{q_1}{4\pi \epsilon_0 R} = \frac{q_2}{8\pi \epsilon_0 R} \]
- Simplifying this gives:
\[ 2q_1 = q_2 \]

4. **Surface Charge Densities**:
- The surface charge density \( \sigma \) is given by:
\[ \sigma = \frac{q}{4\pi R^2} \]
- For Sphere A:
\[ \sigma_A = \frac{q_1}{4\pi R^2} \]
- For Sphere B:
\[ \sigma_B = \frac{q_2}{4\pi (2R)^2} = \frac{q_2}{16\pi R^2} \]

5. **Finding the Ratio of Surface Charge Densities**:
- Using \( q_2 = 2q_1 \):
\[ \sigma_B = \frac{2q_1}{16\pi R^2} = \frac{q_1}{8\pi R^2} \]
- Now, the ratio of surface charge densities is:
\[ \frac{\sigma_A}{\sigma_B} = \frac{\frac{q_1}{4\pi R^2}}{\frac{q_1}{8\pi R^2}} = \frac{8}{4} = 2 \]

6. **Electric Field at the Surface of the Spheres**:
- The electric field \( E \) at the surface of a charged conductor is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
- For Sphere A:
\[ E_A = \frac{\sigma_A}{\epsilon_0} \]
- For Sphere B:
\[ E_B = \frac{\sigma_B}{\epsilon_0} \]

7. **Finding the Ratio of Electric Fields**:
- The ratio of the electric fields is:
\[ \frac{E_A}{E_B} = \frac{\sigma_A}{\sigma_B} = 2 \]

### Conclusion:
The ratio of the magnitudes of the electric fields at the surfaces of spheres A and B is \( 2:1 \).