JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Electrostatics

The central charge $Q$ sets up a radial field in region I ($r<a$), $E=\frac{kQ}{r^2}\ne 0$. Within the metal of the shell, region II ($a<r<b$), the electrostatic field must vanish, so $E=0$ there, the inner surface drawing an induced charge $-Q$ and the outer surface $+Q$. Outside the shell, region III ($r>b$), the total enclosed charge is again $Q$, giving $E=\frac{kQ}{r^2}\ne 0$. The field is therefore non-zero in region I, zero in region II, and non-zero in region III.