Physics - Elecrostatics Question with Solution | TestHub

Charge and charge density on both isolated metallic solid spheres are show below.

Let the charges and the charge densities on both the spheres after connected by thin conducting wire be $Q_1^{\text{'}}, Q_2^{\text{'}}, \sigma \text{and} {\sigma }^{\text{'}}$ as shown below.

Once connected, charge on both the spheres will redistribute such that potentials at the surface of both spheres is the same.

$\therefore \frac{k{Q}_1^{\text{'}}}{R}=\frac{k{Q}_2^{\text{'}}}{2R}$

$\Rightarrow \frac{Q_1^{\text{'}}}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}=\frac{{\displaystyle Q_2^{\text{'}}}}{{\displaystyle 4\mathrm{\pi }{\epsilon }_0\left(2\mathrm{R}\right)}}$

$\therefore Q_2^{\text{'}}=2Q_1^{\text{'}}$

Using charge conservation, 

$Q_1^{\text{'}}+Q_2^{\text{'}}=Q_1+Q_2$

$\therefore \frac{Q_2^{\text{'}}}{2}+Q_2^{\text{'}}=20{\mathrm{\pi R}}^2\mathrm{\sigma }$

$\frac{3}{2}{\mathrm{Q}}_2^{\text{'}}=20{\mathrm{\pi R}}^2\mathrm{\sigma }$

$\therefore \frac{{\mathrm{Q}}_2^{\text{'}}}{4\mathrm{\pi }{\left(2\mathrm{R}\right)}^2}=\frac{2}{3}\times \frac{20{\mathrm{\pi R}}^2\mathrm{\sigma }}{16{\mathrm{\pi R}}^2}$

$\therefore \frac{{\displaystyle \mathrm{\sigma }\text{'}}}{\mathrm{\sigma }}=\frac{5}{6}$