A charge q is uniformly distributed throughout the volume of a sphere of radius R and permittivity k =1. Let U₁ and U₂ be the electrostatic energy stored within and outside the sphere. Then U₁ : U₂ is equal to 1/x where ‘x’ is

Electrostatic energy density, $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2$

$\begin{array}{l}\Rightarrow \text{ Energy }=\mathrm{U}=\int \frac{1}{2}{\mathrm{ϵ}}_0{\mathrm{E}}^2\mathrm{dV}=\frac{1}{2}{\mathrm{ϵ}}_0\int {\mathrm{E}}^2\mathrm{dV}\\ {\mathrm{E}}_{\mathrm{in}}=\frac{\mathrm{\rho r}}{3{\mathrm{ϵ}}_0},{\mathrm{E}}_{\text{out }}=\frac{{\mathrm{\rho R}}^3}{3{\mathrm{ϵ}}_0{\mathrm{r}}^2}\\ {\mathrm{U}}_{\mathrm{in}}=\frac{1}{2}{\mathrm{\epsilon }}_0{\int }_0^{\mathrm{R}} {\mathrm{E}}_{\mathrm{in}}^2\mathrm{dV}=\frac{1}{2}{\mathrm{\epsilon }}_0{\int }_0^{\mathrm{R}} {\left(\frac{\mathrm{\rho r}}{3{\mathrm{ϵ}}_0}\right)}^{2}4{\mathrm{\pi r}}^2\mathrm{dr}\\ =\frac{2{\mathrm{\pi \rho }}^2{\mathrm{r}}^5}{45{\mathrm{ϵ}}_0}={\mathrm{U}}_1\\ {\mathrm{U}}_{\text{out }}=\frac{1}{2}{\mathrm{ϵ}}_0{\int }_{\mathrm{R}}^{\mathrm{\infty }} {\mathrm{E}}_{\text{out }}^2\mathrm{dV}=\frac{1}{2}{\mathrm{ϵ}}_0{\int }_{\mathrm{R}}^{\mathrm{\infty }} {\left[\frac{{\mathrm{\rho R}}^3}{3{\mathrm{ϵ}}_0{\mathrm{r}}^2}\right]}^{2}4{\mathrm{\pi r}}^2\mathrm{dr}\\ =\frac{{\mathrm{\rho }}^2{\mathrm{R}}^6}{9}\left(2\mathrm{\pi }\right){\mathrm{\epsilon }}_0{\int }_{\mathrm{R}}^{\mathrm{\infty }} 1/\mathrm{rdr}=\frac{2{\mathrm{\pi \rho }}^2{\mathrm{R}}^5}{9{\mathrm{\epsilon }}_0}={\mathrm{U}}_2\\ \frac{{\mathrm{U}}_1}{{\mathrm{U}}_2}=\left(\frac{2{\mathrm{\pi \rho }}^2{\mathrm{R}}^5}{45{\mathrm{\epsilon }}_0}\right)/\left(\frac{2{\mathrm{\pi \rho }}^2{\mathrm{R}}^5}{9{\mathrm{\epsilon }}_0}\right)=\frac{1}{5}-\frac{1}{\mathrm{x}}\Rightarrow \mathrm{x}=5\end{array}$