Two charged spherical conductors of radius R₁ and R₂ 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. Equipotential Condition:
When two charged spherical conductors of radii $R_1$ and $R_2$ are connected by a conducting wire, electric charge flows between them until both spheres reach the same electric potential ($V_1=V_2=V$).

2. Electric Potential of a Spherical Conductor:
The potential $V$ on the surface of a sphere of radius $R$ carrying a charge $Q$ is given by:

$V=\frac{1}{4\pi {\epsilon }_0}·\frac{Q}{R}$

Since the potentials are equal:

$\frac{Q_1}{R_1}=\frac{Q_2}{R_2}\Rightarrow \frac{Q_1}{Q_2}=\frac{R_1}{R_2}$

3. Surface Charge Density ($\sigma$):
Surface charge density is defined as charge per unit surface area ($\sigma =\frac{Q}{4\pi R^2}$).
Taking the ratio of the surface charge densities of the two spheres:

$\frac{{\sigma }_1}{{\sigma }_2}=\frac{Q_1/\left(4\pi R_1^2\right)}{Q_2/\left(4\pi R_2^2\right)}=\left(\frac{Q_1}{Q_2}\right)·{\left(\frac{R_2}{R_1}\right)}^2$

4. Substituting the ratio of charges:
Substitute $\frac{Q_1}{Q_2}=\frac{R_1}{R_2}$ into the expression:

$\frac{{\sigma }_1}{{\sigma }_2}=\left(\frac{R_1}{R_2}\right)·{\left(\frac{R_2}{R_1}\right)}^2=\frac{R_2}{R_1}$

Therefore, the ratio of the surface charge densities of the two spheres is R₂/R₁.