surface charge densities $$\sigma_1$$ and $$\sigma_2$$. The electric potential on the surface of a charged spherical conductor is given by:

$$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R}$$

Where:

$$V$$ is the potential,
$$Q$$ is the total charge,
$$R$$ is the radius of the sphere,
$$\epsilon_0$$ is the permittivity of free space.

The surface charge density ($$\sigma$$) is defined as the charge per unit area, so:

$$\sigma = \frac{Q}{4\pi R^2}$$

Combining these expressions, we can express the potential in terms of the surface charge density as follows:

$$V = \frac{\sigma R}{\epsilon_0}$$

Given that the two spheres are charged to the same potential ($$V_1 = V_2 = V$$), and the radii are $$R_1 = 4\, \mathrm{cm}$$ and $$R_2 = 5\, \mathrm{cm}$$, we can set up the equality:

$$\frac{\sigma_1 R_1}{\epsilon_0} = \frac{\sigma_2 R_2}{\epsilon_0}$$

Upon simplifying, $$\epsilon_0$$ cancels out on both sides, allowing us to solve for the ratio of $$\sigma_1$$ to $$\sigma_2$$ directly:

$$\sigma_1 R_1 = \sigma_2 R_2$$

Thus,

$$\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$$

Plugging in the given radii values:

$$\frac{\sigma_1}{\sigma_2} = \frac{5}{4}$$

This corresponds to the ratio of surface charge densities:

$$\sigma_1 : \sigma_2 = 5 : 4$$

Therefore, the correct answer is:

Option A $$5: 4$$

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