Two charged spherical conductors of radii R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres ${\mathrm{\sigma }}_1/{\mathrm{\sigma }}_2$ is:

Potential at the surface of spherical conductor of radius R carrying charge Q, ${\mathrm{V}}_{\mathrm{S}}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}$
Let Q₁ and Q₂ are the charges on two spherical conductors of radii R₁ and R₂ respectively. When these two charged spherical conductors connected by a wire, the potential at their surfaces becomes equal.
$\therefore {\mathrm{V}}_{{\mathrm{S}}_1}={\mathrm{V}}_{{\mathrm{S}}_2}$
$\frac{{\mathrm{Q}}_1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}_1}=\frac{{\mathrm{Q}}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}_2}\Rightarrow \frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} ....\left(\mathrm{i}\right)$
Surface charge density, $\mathrm{\sigma }=\frac{\text{ Charge }}{\text{ Area }}$
$\therefore \frac{{\mathrm{\sigma }}_1}{{\mathrm{\sigma }}_2}=\frac{\left(\frac{{\mathrm{Q}}_1}{4{\mathrm{\pi R}}_1^2}\right)}{\left(\frac{{\mathrm{Q}}_2}{4{\mathrm{\pi R}}_2^2}\right)}=\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}\times \frac{{\mathrm{R}}_2^2}{{\mathrm{R}}_1^2}=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\times \frac{{\mathrm{R}}_2^2}{{\mathrm{R}}_1^2}=\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}$