Physics - Electrostatics Question with Solution | TestHub

Charge Q at the cavity centre induces $-Q$ uniformly on the cavity wall and $+Q$ on the outer surface.

Field of this induced $-Q$ at the conductor centre is $\frac{kQ}{(R/2{)}^2}=\frac{Q}{\pi {ϵ}_0{R}^2}$, not $\frac{Q}{2\pi {ϵ}_0{R}^2}$, and the outside field at $2R$ is $\frac{kQ}{4R^2}$ - (A), (B) false.

The cavity charges and the conductor's surface charges form an isolated action-reaction pair, so their mutual net force is zero - (C) correct; the induced wall charge is uniform, so forces on it cancel by symmetry - (D) correct.