Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

Options

A. (a) `4:1`
B. (b) `1:2`
C. (c) `2:1`
D. (d) 2 volt

Explanation

To solve the problem, we need to find the ratio of the magnitudes of the electric fields at the surfaces of two spherical conductors A and B when they are connected by a conducting wire.

### Step-by-Step Solution:

1. **Identify the Radii of the Spheres:**
- Radius of sphere A, \( R_A = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Radius of sphere B, \( R_B = 2 \, \text{mm} = 0.002 \, \text{m} \)

2. **Understand the Concept of Electric Field:**
- The electric field \( E \) at the surface of a charged spherical conductor is given by the formula:
\[ E = \frac{KQ}{R^2} \]
where \( K \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is the radius of the sphere.

3. **Establish the Relationship Between Charges:**
- When the two spheres are connected by a conducting wire, they will reach the same electric potential. The potential \( V \) of a sphere is given by:
\[ V = \frac{KQ}{R} \]
- For spheres A and B in equilibrium:
\[ V_A = V_B \implies \frac{KQ_A}{R_A} = \frac{KQ_B}{R_B} \]
- This simplifies to:
\[ \frac{Q_A}{Q_B} = \frac{R_A}{R_B} \]

4. **Substituting the Values of Radii:**
- Substituting the values of \( R_A \) and \( R_B \):
\[ \frac{Q_A}{Q_B} = \frac{0.001}{0.002} = \frac{1}{2} \]
- Thus, \( Q_A = \frac{1}{2} Q_B \).

5. **Calculate the Electric Fields at the Surfaces:**
- Electric field at the surface of sphere A:
\[ E_A = \frac{KQ_A}{R_A^2} \]
- Electric field at the surface of sphere B:
\[ E_B = \frac{KQ_B}{R_B^2} \]

6. **Substituting \( Q_A \) in Terms of \( Q_B \):**
- Substitute \( Q_A = \frac{1}{2} Q_B \) into the expression for \( E_A \):
\[ E_A = \frac{K \left(\frac{1}{2} Q_B\right)}{R_A^2} = \frac{K Q_B}{2 R_A^2} \]

7. **Finding the Ratio of Electric Fields:**
- Now, we can find the ratio \( \frac{E_A}{E_B} \):
\[ \frac{E_A}{E_B} = \frac{\frac{K Q_B}{2 R_A^2}}{\frac{K Q_B}{R_B^2}} = \frac{R_B^2}{2 R_A^2} \]

8. **Substituting the Values of Radii:**
- Substitute \( R_A = 0.001 \, \text{m} \) and \( R_B = 0.002 \, \text{m} \):
\[ \frac{E_A}{E_B} = \frac{(0.002)^2}{2 (0.001)^2} = \frac{0.000004}{2 \times 0.000001} = \frac{0.000004}{0.000002} = 2 \]

### Final Result:
The ratio of the magnitudes of the electric fields at the surfaces of spheres A and B is:
\[ \frac{E_A}{E_B} = 2 \]