Question 1. Two charged spherical conductors of radius R_{1} and \mathrm{R}_{2} are connected by a wire. Then the ratio of surface charge densities of the spheres \left(\sigma_{1} / \sigma_{2}\right) is :
(1) \frac{R_{1}}{R_{2}}
(2) \frac{R_{2}}{R_{1}}
(3) \sqrt{\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)}
(4) \frac{R_{1}^{2}}{R_{2}^{2}}
Answer: Option (2)
Explanation:
When two charged spherical conductors are connected by a wire, they attain the same electric potential.
The potential V of a charged conducting sphere of radius R is given by:
V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}For the two spheres after connection:
\frac{Q_1}{R_1} = \frac{Q_2}{R_2}Surface charge density \sigma is defined as charge per unit surface area:
\sigma = \frac{Q}{4\pi R^2}Thus,
\frac{\sigma_1}{\sigma_2} = \frac{Q_1 / R_1^2}{Q_2 / R_2^2}Substituting \frac{Q_1}{Q_2} = \frac{R_1}{R_2}, we get:
\frac{\sigma_1}{\sigma_2} = \frac{R_1}{R_2} \times \frac{R_2^2}{R_1^2} = \frac{R_2}{R_1}Hence, the correct answer is Option (2).