Obtain the expression of electric field by ......

$(i)$ infinite size and with uniform charge distribution.

$(ii)$ thin spherical shell with uniform charge distribution at a point outside it.

$(iii)$ thin spherical shell with uniform charge distribution at a point inside it.

$(i)$ 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 parallel piped 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 $-x$-direction while the unit vector normal to surface $2$ is in the $+x$-direction.

Therefore, flux $\vec{E} \cdot \overrightarrow{\Delta S}$ through both the surfaces are equal and add up.

Therefore the net flux through the Gaussian surface is $2EA.$

The charge enclosed by the closed surface is $\sigma \mathrm{A}$.

Therefore by Gauss's law,

$2 \mathrm{EA}=\frac{\sigma \mathrm{A}}{\epsilon_{0}}$

$\therefore \mathrm{E}=\frac{\sigma}{2 \epsilon_{0}}$

$\therefore \overrightarrow{\mathrm{E}}=\frac{\sigma}{2 \epsilon_{0}} \cdot \hat{n} \quad \ldots(1)$

where $\hat{n}$ is a unit vector normal to the plane and going away from it.

$\mathrm{E}$ is directed away from the plate if $\sigma$ is positive and toward the plate if $\sigma$ is negative.

For a finite large planar sheet equation $(1)$ is approximately true in the middle regions of the planar sheet, away from the ends.

$(2)$ Let $\sigma$ be the uniform surface charge density of a thin spherical shell of radius $\mathrm{R}$.

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 $r$ and with centre $o$ passing through $P$.

The electric field at each point of the Gaussian surface has the same magnitude $\mathrm{E}$ 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 $\Delta \mathrm{S}$, the flux through the Gaussian surface is $\mathrm{E} \times 4 \pi r^{2}$.

The charge enclosed is $\sigma \times 4 \pi \mathrm{R}^{2}$.

By Gauss's law,

$\mathrm{E} \times 4 \pi r^{2}=\frac{\sigma}{\varepsilon_{0}} 4 \pi \mathrm{R}^{2}$

$\therefore\mathrm{E}=\frac{\sigma \mathrm{R}^{2}}{\varepsilon_{0} r^{2}}=\frac{q}{4 \pi \varepsilon_{0} r^{2}}=\frac{k q}{r^{2}}$

$\therefore\overrightarrow{\mathrm{E}}=\frac{\sigma \mathrm{R}^{2}}{4 \pi \varepsilon_{0}r^{2}} \hat{r}=\frac{k q}{r^{2}} \hat{r}$

$(iii)$ As shown in figure, surface charge density on spherical shell of radius $\mathrm{R}$ 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 $\mathrm{E} \times 4 \pi r^{2}$. In this case, the Gaussian surface encloses no charge.

Gauss's law gives,

$\mathrm{E} \times 4 \pi r^{2}=0 \quad\left(\because \frac{q}{\varepsilon_{0}}=0\right.$ as $\left.q=0\right)$

$\therefore \mathrm{E}=0 \quad(r < \mathrm{R})$

Thus, the field due to a uniformly charged thin shell is zero at all points inside the shell.