Physics - Electrostatics Question with Solution | TestHub

The surface charge density of a closed surface area having charge $\mathrm{Q}$ is given by $$\sigma =\frac{\text{ Charge }}{\text{ Area }}=\frac{\mathrm{Q}}{\mathrm{A}}\text{ or }\mathrm{Q}=\sigma \mathrm{A}$$ Thus, the charges on sphere P and Q having same charge density as shown in the figure below is given by
$$\begin{array}{c}{\mathrm{Q}}_{\mathrm{P}}=\sigma \times 4\pi {\mathrm{R}}^2=4\pi \sigma {\mathrm{R}}^2\\ \text{ and }{\mathrm{Q}}_{\mathrm{Q}}=\sigma \times 4\pi (2\mathrm{R}{)}^2=16\pi \sigma {\mathrm{R}}^2\end{array}$$ when they are brought in contact with each other, the total charge will be $$\begin{array}{rl}{\mathrm{Q}}_{\mathrm{t}}& ={\mathrm{Q}}_{\mathrm{P}}+{\mathrm{Q}}_{\mathrm{Q}}\\ =& 4\pi \sigma {\mathrm{R}}^2+16\pi \sigma {\mathrm{R}}^2\text{ [From Eq. (i) and (ii)] }\\ & =20\pi \sigma {\mathrm{R}}^2\end{array}$$ In connection of two charged conducting bodies, the potential will become same on both, i.e. $$\begin{array}{r}\frac{{\mathrm{Q}}_{\mathrm{P}}}{4\pi {\in }_0\mathrm{R}}=\frac{{\mathrm{Q}}_{\mathrm{Q}}}{4\pi {ϵ}_{0}2\mathrm{R}}\\ \Rightarrow \frac{{\mathrm{Q}}_{\mathrm{P}}}{\mathrm{R}}=\frac{{\mathrm{Q}}_{\mathrm{Q}}}{2\mathrm{R}}\Rightarrow \frac{{\mathrm{Q}}_{\mathrm{P}}}{{\mathrm{Q}}_{\mathrm{Q}}}=\frac{1}{2}\end{array}$$ So, the charges on the sphere $\mathrm{P}$ and $\mathrm{Q}$ after separation will be distributed as $$\Rightarrow Q_P^{\prime }=\frac{1}{3}{Q}_{\mathrm{t}}\text{ and }{\mathrm{Q}}_{\mathrm{Q}}^{\prime }=\frac{2}{3}{\mathrm{Q}}_{\mathrm{t}}$$ After separation, the new surface charge densities on $\mathrm{P}$ and $\mathrm{Q}$ will be $$\begin{array}{rl}{\sigma }_{\mathrm{P}}=\frac{{\mathrm{Q}}_{\mathrm{P}}^{\prime }}{\text{ Area }}& =\frac{1}{3}\frac{{\mathrm{Q}}_{\mathrm{t}}}{\text{ Area }}=\frac{1}{3}\frac{20\pi \sigma {\mathrm{R}}^2}{4\pi {\mathrm{R}}^2}=\frac{5}{3}\sigma \\ \text{ and }{\sigma }_{\mathrm{Q}}=\frac{{\mathrm{Q}}_{\mathrm{Q}}^{\prime }}{\text{ Area }}& =\frac{2}{3}\frac{{\mathrm{Q}}_{\mathrm{t}}}{\text{ Area }}=\frac{2}{3}\times \frac{20\pi \sigma {\mathrm{R}}^2}{4\pi (2\mathrm{R}{)}^2}\\ & =\frac{2}{3}\times \frac{5}{4}\sigma =\frac{5}{6}\sigma \end{array}$$