Use Gauss's law to find the electric field inside a uniformly charged solid sphere (charge density p) Compare your answer to Prob. 2.8.

It is given that a solid sphere of radius${}^R$carries a uniform volume charge density${}^P$.The electric field inside solid sphere has to be evaluated.

If there is a surface area enclosing a volume, possessing a charge${}^q$inside the volume then the electric field due to the surface or volume charge is given as

$\mathbf{\oint }\mathit{E}\mathbf{.}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

Here${}^q$is the elemental surface area,${}^{{\epsilon }_0}$is the permittivity of free surface.

Consider a Gaussian sphere of radius${}^r$such that${}^{r<R}$inside the solid sphere as shown below:

It is known that the solid sphere consist the volume charge of density${}^{\sigma }$.For a Gaussian sphere of radius${}^r$, thus the volume is${}^{\frac{4}{3}{\mathrm{\pi r}}^3}$. Thus, the total charge inside the Gaussian sphere is${}^{\frac{4}{3}{\mathrm{\pi r}}^3\mathrm{p}}$.

Apply Gauss law, on the Gaussian surface, as,

$$\begin{gathered} \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ =\frac{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}^{3}p}{{\epsilon }_0} \\ =\frac{4{\mathrm{\pi r}}^{3}p}{3r^2{\epsilon }_0}\hat{r} \\ =\frac{4\mathrm{\pi r}p}{3{\epsilon }_0}\hat{r} \end{gathered}$$

Thus, the electric field outside the spherical shell is$E=\frac{4\mathrm{\pi r}p}{3\epsilon }\hat{r}$.The result is same as the result of problem 2.8.