JEE Main · 2008 · 27 Apr 2008 · Offline · Electrostatics

By Gauss's law a uniformly charged spherical shell encloses no charge for points inside it, so the field is exactly zero throughout $r<R$. At the surface the full charge is enclosed and the field jumps to $\frac{Q}{4\pi {\epsilon }_0{R}^2}$, and for $r>R$ the shell behaves like a point charge at the centre, the field falling as $\frac{Q}{4\pi {\epsilon }_0{r}^2}$. The graph that stays at zero up to $r=R$, rises abruptly there, and then decays as $1/r^2$ is the one that matches.