Use Gauss's law to find the electric field inside and outside a spherical shell of radius Rthat carries a uniform surface charge density${}^{\sigma }$Compare your answer to Prob. 2.7.

It is given that a spherical shell of radius${}^R$carries a uniform surface charge density${}^a$.The electric field inside and outside a spherical shell 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

Here ${}^{\mathbf{q}}$is the elemental surface area,localid="1654335054198" ${}^{{\mathbf{\epsilon }}_{\mathbf{0}}}$is the permittivity of free surface.

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

It is known that the spherical shell consist the surface charge only. So, the charge enclosed by the shell is 0, that is,${}^{q_{enclosed}=o}$

Apply Gauss law, on the Gaussian surface, as,

Thus the above equation states that for, the electric field is 0. In other words, the electric field inside the spherical shell is 0

Consider a Gaussian surface of radius $r$such that localid="1654335162136" $r>R$, outside the spherical shell as shown below:

It is known that the spherical shell consist the surface charge of density ${}^{\sigma }$For a Gaussian surface of radius ${}^r$thus the surface area is localid="1654334729577" ${}^{4{\mathrm{\pi R}}^2}$Thus, the total charge inside the Gaussian surface is localid="1654334774208" ${}^{4\pi R^2\sigma }$

Apply Gauss law, on the Gaussian surface, as,

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

Thus, the electric field outside the spherical shell is${}^{E=\frac{4{\mathrm{\pi R}}^2\mathrm{\sigma }}{r^2{\epsilon }_0}\hat{r}}$ The result is same as the result of problem 2.7.