(N/A) For a thin spherical shell of radius $R$ carrying a total charge $Q$:
$1$. Inside the shell $(r < R)$: According to Gauss's Law,the electric field $E$ inside a charged spherical shell is zero because there is no enclosed charge $(q_{enclosed} = 0)$. Thus,$E = 0$.
$2$. Outside the shell $(r \geq R)$: The shell behaves as a point charge concentrated at its centre. The electric field at a distance $r$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$. This shows that $E \propto \frac{1}{r^2}$.
$3$. The graph shows $E$ on the y-axis and distance $r$ on the x-axis. For $r < R$,the graph lies on the x-axis $(E=0)$. At $r = R$,there is a sharp increase in the electric field. For $r > R$,the field decreases following the inverse square law.