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:
$\frac{R_2}{R_1}$

Step-by-step Explanation:

1. Condition for Connected Conductors:
When two charged spherical conductors of radii $R_1$ and $R_2$ are connected by a conducting wire, electric charge flows between them until they reach electrostatic equilibrium. At equilibrium, the electric potential ($V$) on the surface of both spheres becomes equal:
$V_1=V_2$

2. Expressing Potential in terms of Charge:
The potential on the surface of a spherical conductor carrying a charge $Q$ and having radius $R$ is given by the formula:
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$
Equating the potentials of both spheres:
$\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{R_1}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{R_2}$
By canceling common terms, we get the relationship between their charges and radii:
$\frac{Q_1}{R_1}=\frac{Q_2}{R_2}\Rightarrow \frac{Q_1}{Q_2}=\frac{R_1}{R_2}$

3. Relating Charge to Surface Charge Density:
The surface charge density ($\sigma$) of a sphere represents the charge per unit surface area:
$\sigma =\frac{Q}{4\pi R^2}$
This allows us to write the charge $Q$ in terms of $\sigma$ and $R$:
$Q=\sigma \left(4\pi R^2\right)$

4. Finding the Ratio of Surface Charge Densities:
Taking the ratio of the surface charge densities ${\sigma }_1$ and ${\sigma }_2$:
$\frac{{\sigma }_1}{{\sigma }_2}=\frac{Q_1/\left(4\pi R_1^2\right)}{Q_2/\left(4\pi R_2^2\right)}=\left(\frac{Q_1}{Q_2}\right)·{\left(\frac{R_2}{R_1}\right)}^2$
Substitute the charge ratio $\frac{Q_1}{Q_2}=\frac{R_1}{R_2}$ derived in step 2:
$\frac{{\sigma }_1}{{\sigma }_2}=\left(\frac{R_1}{R_2}\right)·{\left(\frac{R_2}{R_1}\right)}^2=\frac{R_2}{R_1}$

Therefore, the ratio of the surface charge densities of the two spheres is inversely proportional to their radii, which is $\frac{R_2}{R_1}$.