Physics - Electrostatics Question with Solution | TestHub

Initially, inner shell has charge $Q_1=Q$, outer shell has charge $Q_2=2Q$. The electric field for $R<r<2R$ is $E_1=kQ/r^2$. The electric field for $r>2R$ is $E_2=k(Q+2Q)/r^2=k(3Q)/r^2$.

When the inner shell is grounded, its potential becomes zero. Let its new charge be $Q^{\prime }$. The outer shell still has total charge $2Q$. The electric field for $R<r<2R$ is $E_1^{\prime }=k{Q}^{\prime }/r^2$. The electric field for $r>2R$ is $E_2^{\prime }=k(Q^{\prime }+2Q)/r^2$. The potential of the inner shell $V_R^{\prime }=-{\int }_R^{2R}{E}_1^{\prime }dr-{\int }_{2R}^{\infty }{E}_2^{\prime }dr=0$. This simplifies to $k{Q}^{\prime }/R-k{Q}^{\prime }/(2R)+k(Q^{\prime }+2Q)/(2R)=0$, which gives $k{Q}^{\prime }/R+kQ/R=0⟹Q^{\prime }=-Q$. So, option A is correct.

Initial charge on outer surface of outer shell: $Q_{outer,initial}=Q_2-(-Q_1)=2Q-(-Q)=3Q$. Final charge on outer surface of outer shell: $Q_{outer,final}=Q_2-(-Q^{\prime })=2Q-(-(-Q))=2Q-Q=Q$. Since $Q_{outer,final}=Q<3Q=Q_{outer,initial}$, the charge decreases. So, option B is incorrect. Potential of outer shell after grounding: $V_{2R}^{\prime }=k(Q^{\prime }+2Q)/(2R)=k(-Q+2Q)/(2R)=kQ/(2R)$. So, option C is correct. Magnitude of electric field in $R<r<2R$: Initially, $|E_1|=kQ/r^2$. Finally, $|E_1^{\prime }|=|k{Q}^{\prime }/r^2|=|k(-Q)/r^2|=kQ/r^2$. So, option D is correct.