NEET UG 2019 — Physics Electromagnetism

The surface charge density of a closed surface area having charge $\mathrm{Q}$Q is given by $$\sigma=\frac{\text { Charge }}{\text { Area }}=\frac{\mathrm{Q}}{\mathrm{A}} \text { or } \mathrm{Q}=\sigma \mathrm{A}$$σ= Area  Charge ​=AQ​ or Q=σA Thus, the charges on sphere P and Q having same charge density as shown in the figure below is given by

$$\begin{gathered} \mathrm{Q}_{\mathrm{P}}=\sigma \times 4 \pi \mathrm{R}^2=4 \pi \sigma \mathrm{R}^2 \\ \text { and } \mathrm{Q}_{\mathrm{Q}}=\sigma \times 4 \pi(2 \mathrm{R})^2=16 \pi \sigma \mathrm{R}^2 \end{gathered}$$QP​=σ×4πR2=4πσR2 and QQ​=σ×4π(2R)2=16πσR2​ when they are brought in contact with each other, the total charge will be $$\begin{aligned} \mathrm{Q}_{\mathrm{t}} & =\mathrm{Q}_{\mathrm{P}}+\mathrm{Q}_{\mathrm{Q}} \\ = & 4 \pi \sigma \mathrm{R}^2+16 \pi \sigma \mathrm{R}^2 \text { [From Eq. (i) and (ii)] } \\ & =20 \pi \sigma \mathrm{R}^2 \end{aligned}$$Qt​=​=QP​+QQ​4πσR2+16πσR2 [From Eq. (i) and (ii)] =20πσR2​ In connection of two charged conducting bodies, the potential will become same on both, i.e. $$\begin{aligned} \frac{\mathrm{Q}_{\mathrm{P}}}{4 \pi \in_0 \mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{Q}}}{4 \pi \epsilon_0 2 \mathrm{R}} \\ \Rightarrow \quad \frac{\mathrm{Q}_{\mathrm{P}}}{\mathrm{R}}=\frac{\mathrm{Q}_{\mathrm{Q}}}{2 \mathrm{R}} \Rightarrow \frac{\mathrm{Q}_{\mathrm{P}}}{\mathrm{Q}_{\mathrm{Q}}}=\frac{1}{2} \end{aligned}$$4π∈0​RQP​​=4πϵ0​2RQQ​​⇒RQP​​=2RQQ​​⇒QQ​QP​​=21​​ So, the charges on the sphere $\mathrm{P}$P and $\mathrm{Q}$Q after separation will be distributed as $$\Rightarrow \quad Q_P^{\prime}=\frac{1}{3} Q_{\mathrm{t}} \text { and } \mathrm{Q}_{\mathrm{Q}}^{\prime}=\frac{2}{3} \mathrm{Q}_{\mathrm{t}}$$⇒QP′​=31​Qt​ and QQ′​=32​Qt​ After separation, the new surface charge densities on $\mathrm{P}$P and $\mathrm{Q}$Q will be $$\begin{aligned} \sigma_{\mathrm{P}}=\frac{\mathrm{Q}_{\mathrm{P}}^{\prime}}{\text { Area }} & =\frac{1}{3} \frac{\mathrm{Q}_{\mathrm{t}}}{\text { Area }}=\frac{1}{3} \frac{20 \pi \sigma \mathrm{R}^2}{4 \pi \mathrm{R}^2}=\frac{5}{3} \sigma \\ \text { and } \sigma_{\mathrm{Q}}=\frac{\mathrm{Q}_{\mathrm{Q}}^{\prime}}{\text { Area }} & =\frac{2}{3} \frac{\mathrm{Q}_{\mathrm{t}}}{\text { Area }}=\frac{2}{3} \times \frac{20 \pi \sigma \mathrm{R}^2}{4 \pi(2 \mathrm{R})^2} \\ & =\frac{2}{3} \times \frac{5}{4} \sigma=\frac{5}{6} \sigma \end{aligned}$$σP​= Area QP′​​ and σQ​= Area QQ′​​​=31​ Area Qt​​=31​4πR220πσR2​=35​σ=32​ Area Qt​​=32​×4π(2R)220πσR2​=32​×45​σ=65​σ​