Two charged spherical conductors of radius R₂ and R₁ are connected by a wire. Then the ratio of surface charge densities of the spheres (σ₁/σ₂) is :

The correct answer is R₂/R₁.

Step-by-Step Explanation:

1. Understanding Electrostatic Equilibrium:
When two charged spherical conductors of radius $R_1$ and $R_2$ are connected by a wire as described, electric charge flows between them until they reach electrostatic equilibrium. At equilibrium, the electric potential on the surface of both spheres becomes equal.

Let $Q_1$ and $Q_2$ represent the charges on the spheres at equilibrium.
Since the potentials on both spheres are equal:
$V_1=V_2$

The electrostatic potential $V$ at the surface of a sphere of radius $R$ is given by the formula:
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$

Equating the potentials of the two connected spheres:
$\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{R_1}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{R_2}$

This simplifies to find the ratio of the charges:
$\frac{Q_1}{R_1}=\frac{Q_2}{R_2}\Rightarrow \frac{Q_1}{Q_2}=\frac{R_1}{R_2}$

2. Determining the Surface Charge Density:
The surface charge density ($\sigma$) is defined as the charge per unit surface area of a sphere:
$\sigma =\frac{Q}{4\pi R^2}$

Writing the ratio of the surface charge densities of the two spheres ($\frac{{\sigma }_1}{{\sigma }_2}$):
$\frac{{\sigma }_1}{{\sigma }_2}=\frac{Q_1/\left(4\pi R_1^2\right)}{Q_2/\left(4\pi R_2^2\right)}$

Rearranging the equation yields:
$\frac{{\sigma }_1}{{\sigma }_2}=\left(\frac{Q_1}{Q_2}\right)\cdot {\left(\frac{R_2}{R_1}\right)}^2$

Substituting the relationship $\frac{Q_1}{Q_2}=\frac{R_1}{R_2}$:
$\frac{{\sigma }_1}{{\sigma }_2}=\left(\frac{R_1}{R_2}\right)\cdot {\left(\frac{R_2}{R_1}\right)}^2$

Simplifying the expression gives us the final ratio:
$\frac{{\sigma }_1}{{\sigma }_2}=\frac{R_2}{R_1}$