Correct Option: (B) $\frac{r_2}{r_1}$

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

When two charged metallic spheres are connected by a conducting wire, charge flows until they reach the same electric potential ($V_A = V_B$). The electric potential at the surface of a sphere of radius $r$ with charge $q$ is given by:

$$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$

The electric field at the surface of a charged sphere is given by:

$$E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$$

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

Since the spheres are at the same potential ($V_A = V_B$):

$$\frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1} = \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2} \implies \frac{q_1}{q_2} = \frac{r_1}{r_2}$$

Now, express the ratio of the electric fields $E_A$ and $E_B$:

$$\frac{E_A}{E_B} = \frac{\frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1^2}}{\frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2^2}} = \frac{q_1}{q_2} \cdot \frac{r_2^2}{r_1^2}$$

Substitute the ratio $\frac{q_1}{q_2} = \frac{r_1}{r_2}$ into the equation:

$$\frac{E_A}{E_B} = \left( \frac{r_1}{r_2} \right) \cdot \left( \frac{r_2^2}{r_1^2} \right) = \frac{r_2}{r_1}$$

Step 3 (Final Concluded Answer): [0.25 Mark]

The ratio of the electric fields at the surfaces of the spheres is:

$$\frac{E_A}{E_B} = \frac{r_2}{r_1}$$

Therefore, the correct option is (B).

💡 EL10 Concept Insight: Think of it this way: smaller spheres have a more "curved" surface, which packs the electric field lines much tighter together, making the field stronger at the surface compared to a larger, flatter sphere at the same voltage.