To solve this problem, we need to understand the concept of electric potential for charged conducting spheres.$\\$The electric potential $V$ at the surface of a charged conducting sphere is given by:$\\ V = \frac{Q}{4 \pi \varepsilon_0 R} \\$where $Q$ is the charge on the sphere, $\varepsilon_0$ is the permittivity of free space, and $R$ is the radius of the sphere.$\\$Given that both spheres have equal charges, $Q_1 = Q_2 = Q. \\$Let $V_1$ and $V_2$ be the potentials on the spheres with radii $R_1$ and $R_2$ respectively.$\\$Then, the potentials are:$\\ V_1 = \frac{Q}{4 \pi \varepsilon_0 R_1} \\ V_2 = \frac{Q}{4 \pi \varepsilon_0 R_2} \\$Since $R_1 >> R_2,$ it follows that $\frac{1}{R_1} << \frac{1}{R_2}. \\$Therefore, $V_1 < V_2. \\$This means the potential is more on the smaller sphere.$\\$Thus, the correct option is: More on smaller sphere (Option 2).