Physics - Electrostatics Question with Solution | TestHub

Induction puts $-q_a$ and $-q_b$ uniformly on the two cavity walls and $q_a+q_b$ uniformly on the outer surface (the metal shields the cavities).

(A) Outside ($r>R$): the field is that of $q_a+q_b$ at the centre: $\frac{K(q_a+q_b)}{r^2}$ - correct.

(B) Inside cavity b: $V(r)=\frac{K{q}_b}{r}-\frac{K{q}_b}{b}+\frac{K(q_a+q_b)}{R}$ - correct.

(C) ${\sigma }_{outer}=\frac{q_a+q_b}{4\pi R^2}$ - correct.

(D) Each charge sits at its cavity's centre, so the induced wall charge is uniform and exerts no force on it: net force zero - (D) wrong.