Physics - Electrostatics Question with Solution | TestHub

Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same. $$\begin{array}{ll}\therefore & V_A=V_B\\ \text{ or }& \frac{Q_A}{R_A}=\frac{Q_B}{R_B}\left(\because V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}\right)\end{array}$$ Since, $R_A>R_B$ $$\therefore Q_A>Q_B$$ $\therefore$ Option (b) is correct. Potential is also equal to, $$\begin{array}{rlr}V& =\frac{\sigma R}{{\epsilon }_0}& \\ V_A& =V_B& \\ \therefore {\sigma }_A{R}_A& ={\sigma }_B{R}_B& \\ \text{ or }{\sigma }_B& \frac{{\sigma }_A}{{\sigma }_B}& =\frac{R_B}{R_A}\end{array}$$ $\therefore$ Option (c) is correct. Electric field on surface, $$E=\frac{\sigma }{E_0}\text{ or }E\propto \sigma$$ or ${\sigma }_A<{\sigma }_B$ Since, ${\sigma }_A<{\sigma }_B$ $\therefore E_A<E_B$ $\therefore$ Option (d) is also correct. $\therefore$ Correct options are (a), (b), (c) and (d). Analysis of Question (i) Question is simple. (ii) $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R^2}=\frac{\sigma }{{\epsilon }_0}$ (On surface) As, $\sigma =$ surface charge density $$=\frac{Q}{4\pi R^2}$$ (iii) $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}=\frac{\sigma R}{{\epsilon }_0}$