Consider two conducting spheres of radii $R_1$ and $R_2$ with $(R_1 > R_2)$. If the two are at the same potential,the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

(B) Both spheres are at the same potential,so $V_1 = V_2$.
Using the formula for potential $V = \frac{kq}{R}$,we have $\frac{kq_1}{R_1} = \frac{kq_2}{R_2}$.
Since the charge $q = \sigma A = \sigma (4\pi R^2)$,we substitute this into the potential equation:
$\frac{k(\sigma_1 4\pi R_1^2)}{R_1} = \frac{k(\sigma_2 4\pi R_2^2)}{R_2}$.
Simplifying this,we get $\sigma_1 R_1 = \sigma_2 R_2$.
Therefore,the ratio of surface charge densities is $\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$.
Since $R_1 > R_2$,it follows that $\frac{R_2}{R_1} < 1$,which implies $\sigma_1 < \sigma_2$.
Thus,the charge density of the smaller sphere $(R_2)$ is greater than the charge density of the larger sphere $(R_1)$.