Question

Use gauss’s theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density $\sigma $.

Answer 1

Hint So to find the electric field, we will take a thin plane sheet and have charge density and then will draw a Gaussian cylinder and will find the electric flux through it. And then by using the gauss’ theorem we will be able to find it.

Complete Step by Step Solution So to find the electric field, first of all, let us assume a thin sheet having a positive charge with an even surface charge density $\sigma $ on each edge of the sheet. Now let suppose $P$be the point which is at a distance $a$and at that the electrical field is needed. So we’ll draw a Gaussian cylinder of cross-sectional $A$through purpose $P$
So, through the Gaussian surface electric flux produced will be
$\phi = E \times $Area of the circular top of the rounded cylinder
Here the curved surface area of the cylinder will be parallel to the electric lines of force, through the circular top of the cylinder the flux will be generated and it is due to the passes of the electric field charge.
$\phi = E \times 2A$
As we know, according to the theorem given by gauss’
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
And therefore the charge surrounded by the Gaussian surface will be,
$ \Rightarrow \phi = \sigma A$
And therefore it can also be written as,
$ \Rightarrow \phi = \dfrac{{\sigma A}}{{{\varepsilon _ \circ }}}$
From both the above equation, on equating we will get
$ \Rightarrow E \times 2A = \dfrac{{\sigma A}}{{{\varepsilon _ \circ }}}$
Now we will solve the above equation for the value of $E$
We get,
$ \Rightarrow E = \dfrac{\sigma }{{2{\varepsilon _ \circ }}}$

Note When the Gaussian surface is made to pass through a discrete charge, the electric field is not definable at the point where the charge is there. Hence, we shall not be able to calculate the surface integral of the electric field. This means that we shall not be able to use gauss’s law.