Obtain the expression for the electric field due to:
$(i)$ An infinite plane sheet with uniform charge distribution.
$(ii)$ $A$ thin spherical shell with uniform charge distribution at a point outside it.
$(iii)$ $A$ thin spherical shell with uniform charge distribution at a point inside it.

(N/A) $(i)$ Let $\sigma$ be the uniform surface charge density of an infinite plane sheet.
Take the $x$-axis normal to the given plane. By symmetry,the electric field will not depend on $y$ and $z$ coordinates,and its direction at every point must be parallel to the $x$-direction.
We can take the Gaussian surface to be a rectangular parallelepiped of cross-sectional area $A$,as shown.
Only the two faces $1$ and $2$ will contribute to the flux; electric field lines are parallel to the other faces and they do not contribute to the total flux.
The unit vector normal to surface $1$ is in the $-x$-direction,while the unit vector normal to surface $2$ is in the $+x$-direction.
Therefore,the flux $\vec{E} \cdot \overrightarrow{\Delta S}$ through both surfaces is equal and adds up.
Therefore,the net flux through the Gaussian surface is $2EA$.
The charge enclosed by the closed surface is $\sigma A$.
Therefore,by Gauss's law,
$2 EA = \frac{\sigma A}{\epsilon_{0}}$
$\therefore E = \frac{\sigma}{2 \epsilon_{0}}$
$\therefore \vec{E} = \frac{\sigma}{2 \epsilon_{0}} \hat{n} \quad \dots(1)$
where $\hat{n}$ is a unit vector normal to the plane and pointing away from it.
$E$ is directed away from the plate if $\sigma$ is positive and toward the plate if $\sigma$ is negative.
$(ii)$ Let $\sigma$ be the uniform surface charge density of a thin spherical shell of radius $R$.
Consider a point $P$ outside the shell with radius vector $\vec{r}$.
To calculate $\vec{E}$ at $P$,we take the Gaussian surface to be a sphere of radius $r$ with center $O$ passing through $P$.
The electric field at each point of the Gaussian surface has the same magnitude $E$ and is along the radius vector at each point.
Thus,$\vec{E}$ and $\overrightarrow{\Delta S}$ at every point are parallel,and the flux through each element is $E \Delta S$.
Summing over all $\Delta S$,the flux through the Gaussian surface is $E \times 4 \pi r^{2}$.
The charge enclosed is $q = \sigma \times 4 \pi R^{2}$.
By Gauss's law,
$E \times 4 \pi r^{2} = \frac{q}{\epsilon_{0}}$
$\therefore E = \frac{q}{4 \pi \epsilon_{0} r^{2}} = \frac{k q}{r^{2}}$
$\therefore \vec{E} = \frac{q}{4 \pi \epsilon_{0} r^{2}} \hat{r} = \frac{k q}{r^{2}} \hat{r}$
$(iii)$ As shown in the figure,the surface charge density on a spherical shell of radius $R$ is $\sigma$.
The point $P$ is inside the shell. The Gaussian surface is a sphere through $P$ centered at $O$ of radius $r$.
The flux through the Gaussian surface,calculated as before,is $E \times 4 \pi r^{2}$. In this case,the Gaussian surface encloses no charge $(q = 0)$.
Gauss's law gives,
$E \times 4 \pi r^{2} = 0$
$\therefore E = 0 \quad (r < R)$
Thus,the field due to a uniformly charged thin shell is zero at all points inside the shell.