Spherical symmetric charge system centered at origin.
Electric potential $\phi=\frac{Q}{4 \pi \varepsilon_0 R_0} \quad r \leq R_0$
$\phi=\frac{Q}{4 \pi \varepsilon_0 r} \quad r>R_0$
Answer & explanation
Correct answer: option 4
For $r>R_0, E=-\frac{d \phi}{d r}=\frac{Q}{4 \pi \varepsilon_0 r^2}$
Charge enclosed by the concentric spherical surface of $r=2 R_0$
$=\varepsilon_0 \phi_{C}=\varepsilon_0 E \times 4 \pi \times r^2=\varepsilon_0 \frac{Q}{4 \pi \varepsilon_0 r^2} 4 \pi r^2=Q$
For $r<R_0, E=-\frac{d \phi}{dr}=0$ (∵ c = constant)
For $r>R_0, E=-\frac{d \phi}{d t}=\frac{Q}{4 \pi \varepsilon_0 r^2}$
Since for $r<R_0, E=0$, hence charge will be only on the spherical surface of $r=R_0$.
For $r<R_0, E=0 \Rightarrow$ energy density $\frac{1}{2} \varepsilon_0 E^2=0$
(d)