a) A positive point charge (+ q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines originating from the point on to the surface of the plate.
Derive the expression for the electric field at the surface of a charged conductor. 

b)   Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities a and -a respectively.

a)  Representation of electric field. (due to a positive charge)

b)  Gauss’ Law states that the total flux through a closed surface is  $\frac{1}{{\epsilon }_0}$ times the net charge enclosed by

 ${\mathrm{\varphi }}_{\mathrm{E}}=\oint \overrightarrow{\mathrm{E}}.\mathrm{d} \overrightarrow{\mathrm{S}}=\frac{\mathrm{q}\text{'}\text{'}}{{\mathrm{\epsilon }}_0}$

      Let $\mathrm{\sigma }$ be the surface charge density (charge per unit area) of the given sheet and let P be a point at distance r from the sheet where we have to find $\overrightarrow{\mathrm{E}}$

Choosing point P’ symmetrical with P on the other side of the sheet, let us draw a Gaussian cylindrical surface cutting through the sheet as shown in the diagram. As at the cylindrical part of the Gaussian surface, $\overrightarrow{\mathrm{E}}$ and $\mathrm{d}\overrightarrow{\mathrm{S}}$ are at a right angle, the only surface having $\overrightarrow{\mathrm{E}}$ and $\mathrm{d}\overrightarrow{\mathrm{S}}$ parallel are the plane ends.

$$\begin{gathered} \therefore {\varnothing }_{\mathrm{E}}=\oint \overrightarrow{\mathrm{E}}.\mathrm{d}\overleftrightarrow{\mathrm{S}}+\oint \overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{S}} \\ = \oint \mathrm{E}.\mathrm{dS}+\oint \mathrm{E}.\mathrm{dS}=\mathrm{EA}+\mathrm{EA}=2\mathrm{EA} \end{gathered}$$

[As E is outgoing from both plane ends, the flux is positive.

This is the total flux through the Gaussian surface.

$$\begin{gathered} \mathrm{Using} \mathrm{Gauss}\text{'} \mathrm{larw}, {\varnothing }_{\mathrm{E}}=\frac{\mathrm{q}}{{\mathrm{E}}_0} \\ \therefore 2 \mathrm{EA}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}=\frac{\mathrm{\sigma A}}{{\mathrm{\epsilon }}_0} \mathbf{.}\mathbf{.}\mathbf{.}\mathbf{[}\mathbf{As}\mathbf{ }\mathbf{q}\mathbf{=}\mathbf{\sigma A} \\ \mathbf{\therefore }\mathbf{ }\mathbf{ }\mathbf{ }\mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \end{gathered}$$

This value is independent of r. Hence, the electric field intensity is same for all points near the charged sheet. This is called uniform electric field intensity.

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ \mathbf{In}\mathbf{ }\mathbf{region}\mathbf{ }\mathbf{I} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathrm{E}}_1=-\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ {\mathrm{E}}_2=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ \mathrm{Total} \mathrm{field} {\mathrm{E}}_1 \\ ={\mathrm{E}}_1+{\mathrm{E}}_2 \\ =\frac{-\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}=0 \\ \mathbf{In}\mathbf{ }\mathbf{region}\mathbf{ }\mathbf{II} \\ {\mathrm{E}}_{\mathrm{II}}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}=\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0} \\ \mathbf{In}\mathbf{ }\mathbf{region}\mathbf{ }\mathbf{III} \\ {\mathrm{E}}_1=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}, {\mathrm{E}}_2=-\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ \therefore \mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}+\left(-\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\right)=0 \end{gathered}$$