1. Condition after connecting by a wire

When two conducting spheres are connected by a conducting wire and equilibrium is reached, their potentials become equal:

$V_A = V_B$

For spherical conductors, the potential is

$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$

So,

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

Given:

$R_A = 1\text{ mm}, \qquad R_B = 2\text{ mm}$

Hence,

$\frac{Q_A}{Q_B} = \frac{R_A}{R_B} = \frac{1}{2}$

2. Electric field at the surface of a charged sphere

The electric field just outside the surface of a spherical conductor is

$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$

Therefore,

$\frac{E_A}{E_B} = \frac{Q_A/R_A^2}{Q_B/R_B^2}$

Using

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

we get

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

Substitute the radii:

$\frac{E_A}{E_B} = \frac{2}{1}$

So the ratio is

$E_A : E_B = 2:1$

3. Check options

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

Thus, the correct option is C.