1. Condition for equilibrium after connection

When two conducting spheres are connected by a wire, charges flow until their potentials become equal.

So, $V_A = V_B$

For a spherical conductor, $V = \frac{kQ}{R}$

Hence, $\frac{kQ_A}{R_A} = \frac{kQ_B}{R_B}$

which gives $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$

Given: $R_A = 5\,\text{mm}, \qquad R_B = 10\,\text{mm}$

Thus, $\frac{Q_A}{Q_B} = \frac{5}{10} = \frac{1}{2}$

2. Electric field at the surface of a spherical conductor

The electric field just outside the surface is $E = \frac{kQ}{R^2}$

Therefore, $\frac{E_A}{E_B} = \frac{\frac{kQ_A}{R_A^2}}{\frac{kQ_B}{R_B^2}} = \frac{Q_A}{Q_B}\cdot \frac{R_B^2}{R_A^2}$

Substitute $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$

So, $\frac{E_A}{E_B} = \frac{R_A}{R_B}\cdot \frac{R_B^2}{R_A^2} = \frac{R_B}{R_A}$

Hence, $\frac{E_A}{E_B} = \frac{10}{5} = 2$

So the ratio is $E_A : E_B = 2 : 1$

3. Check the options

• A: $1:2$ ❌
• B: $2:1$ ✅
• C: $1:1$ ❌
• D: $1:4$ ❌

4. Final answer

The correct option is B.

$\boxed{E_A : E_B = 2:1}$

Note: The given separation $2\,\text{cm}$ is not needed here, since at equilibrium the connected conductors must have equal potential, and for this standard JEE treatment we use $V \propto Q/R$ for spherical conductors.