JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Electrostatics

The inner sphere is metallic, so its charge lies on its surface and the field is zero for $r<R$. Immediately outside its surface the field jumps to $\frac{kq}{R^2}$ and then falls as $\frac{kq}{r^2}$ up to the inner face of the shell. Within the conducting material of the shell, $a<r<b$, the field is zero again: charge $-q$ is induced on the inner face and $+q$ appears on the outer face so that the metal carries no field. Beyond the shell the charge enclosed is once more $q$, so the field restarts at $\frac{kq}{b^2}$ at $r=b$ and decays as $\frac{kq}{r^2}$. The graph is therefore zero from the centre to $R$, an inverse-square fall from $R$ to $a$, zero from $a$ to $b$, and an inverse-square fall beyond $b$.