A non-conducting sphere of radius $R$ has a positive charge which is distributed over its volume with density $\rho ={\rho }_{\circ }\left(1-\frac{x}{R}\right)$per unit volume, where a is distance from the centre. If dielectric constant of material of the sphere is $K= 1$, calculate energy stored in surrounding space and total self energy of the sphere.

strength of electric field at a point is $E$, energy stored per unit volume of di-electric at that point is $\frac{1}{2}{\epsilon }_{\circ }K{E}^2$. Hence, to calculate density of electrostatic energy is to be calculated. But for its calculation, al charge on sphere must be known. Hence, first consider a thin spherical shell inside the sphere as shown in Fig.

Let radius of the shell be $x$ and let radial thickness be $dx$.

Volume of the shell$=4{\mathrm{\pi x}}^2\mathrm{dx}$

Charge on the shell,$dq={\rho }_{\circ }\left(1-\frac{x}{R}\right)4{\mathrm{\pi x}}^2\mathrm{dx}$

Total charge on the shell is $Q=4\mathrm{\pi }{\rho }_{\circ }{\int }_0^R\left(1-\frac{x}{R}\right)x^2\mathrm{dx}.......\left(1\right)$

or,

$Q=\frac{1}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_{\circ }$

Now, consider a concentric spherical shell (in surrounding space) of radius $r (>R)$ and radial thickness

Electric field at a radial distance r from the centre of this shell, $E_S=\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{Q}{r^2}=\frac{{\rho }_{\circ }{R}^3}{12{\epsilon }_{\circ }{r}^2}$

Energy stored in the surrounding space, $d{U}_S=\left(\frac{1}{2}{\epsilon }_{\circ }{E_S}^2\right)\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right)$

$d{U}_S=\frac{{{\mathrm{\pi \rho }}_{\circ }}^2{\mathrm{R}}^6}{72{\epsilon }_{\circ }}dr$

$U_S=\frac{{{\mathrm{\pi \rho }}_{\circ }}^2{\mathrm{R}}^6}{72{\epsilon }_{\circ }}{\int }_{r=R}^{r=\infty }\frac{dr}{r^2}$

$U_S=\frac{{{\mathrm{\pi \rho }}_{\circ }}^2{\mathrm{R}}^5}{72{\epsilon }_{\circ }}$ ans.

Self energy of the sphere is total electrostatic energy stored in surrounding space and inside the sphere To calculate energy stored inside the sphere, consider a concentric spherical shell of radius $x$ thickness $dx$. Electric field at surface of this shell is due to charge on inner concentric solid sphere of Charge on this concentric sphere is
$q=\int dq=4{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{\rho }}_{\circ }{\int }_0^{\mathrm{x}}\left(1-\frac{\mathrm{x}}{\mathrm{R}}\right){\mathrm{x}}^2 \mathrm{dx}=\frac{\mathrm{\pi \rho }\circ {\mathrm{x}}^3}{3\mathrm{R}}\left(4\mathrm{R}-3\mathrm{x}\right)$

Electric field at surface of this shell, $E_i=\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{q}{x^2}$

or,   $E_i=\frac{{\rho }_{\circ }\mathrm{x}\left(4\mathrm{R}-3\mathrm{x}\right)}{12{\epsilon }_{°}R}$

Energy stored in the shell, $d{U}_i=\left(\frac{1}{2}{\epsilon }_{\circ }{E_i}^2\right)\left(4{\mathrm{\pi x}}^2\mathrm{dx}\right)$

$d{U}_i=\frac{{{\mathrm{\pi \rho }}_{\circ }}^2}{72{\epsilon }_{\circ }{R}^2}{\left(4R-3x\right)}^2{x}^4\mathrm{dx}$

Total energy inside the sphere is $U_i=\frac{{{\mathrm{\pi \rho }}_{\circ }}^2}{72{\epsilon }_{\circ }{R}^2}{\int }_{x=0}^{x=R}{\left(4R-3x\right)}^2{x}^4\mathrm{dx}$

$U_i=\frac{17{{\mathrm{\pi \rho }}_{\circ }}^2{R}^5}{2520{\epsilon }_{\circ }{R}^2}$

Self energy of the sphere$=U_S+U_i=\frac{13{{\mathrm{\pi \rho }}_{\circ }}^2{R}^5}{630{\epsilon }_{\circ }}$ans.