Question

A large plane sheet of charge having a surface charge density of $5.0 \times {10^{ - 16}}C{\text{ }}{m^{ - 2}}$ lies in the X-Y plane. Find the electric flux through a circular area of radius $0.1m$, if the normal to the circular area makes an angle of $60^\circ$ with the z-axis

Answer 1

Hint In a plane sheet charge, we have to find the electric flux through a circular area. The radius of the circular potion is given. The charge density of the sheet is given. We should know the electric field due to the plane sheet derivation to find the electric field. Because electric flux is the product of area and electric field vectors.

Complete step by step answer
Electric flux is defined as the net electric field, crossing through the given area.
$\Phi = \vec E.\vec A$
$\Phi = E.A\cos \theta {\text{ }} \to {\text{1}}$
$\Phi$ is the surface charge density
E is the electric field
A is the area
$\theta$ is the angle between the plane sheet and the normal to the plane sheet
Given,
Surface charge density of the plane sheet, $\sigma = 5.0 \times {10^{ - 16}}C{\text{ }}{m^{ - 2}}$
Electric field due to the plane sheet is given by the equation:
$\Rightarrow E = \dfrac{\sigma }{{2{\varepsilon _0}}}$
$\sigma$ is the surface charge density
${\varepsilon _0}$ is the permittivity of free space
The permittivity of free space (vacuum) ${\varepsilon _0}$ is a constant value.
The value of ${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$
$\Rightarrow E = \dfrac{{5 \times {{10}^{ - 16}}}}{{2 \times 8.85 \times {{10}^{ - 12}}}}$
$\Rightarrow E = 0.28248587 \times {10^{ - 4}}N{C^{ - 1}}$
We have to find the electric flux through a circular area of radius, $r = 0.1m$
Area of circle is
$\Rightarrow A = \pi {r^2}$
Area of the circular potion,
$\Rightarrow A = \pi {r^2}$
$\Rightarrow A = 3.14 \times {\left( {0.1} \right)^2}$
$\Rightarrow A = 0.0314{m^2}$
Now, the electric flux through a circular area is
$\Rightarrow \Phi = E.A\cos \theta$
The normal to the circular area makes an angle of $60^\circ$ with the z-axis
$\Rightarrow \Phi = 0.28248587 \times {10^{ - 4}} \times 0.0314\cos 60^\circ$
$\Rightarrow \Phi = 0.4435 \times {10^{ - 6}}N{m^2}{C^{ - 1}}$

Note The measure of charges accumulated over a surface is known as surface charge density. Let us consider a surface of area A and the net charge on the surface is q, then the surface charge density is given by
$\Rightarrow \sigma = \dfrac{q}{A}$
$\sigma$ is the surface charge density
A is the area of the surface
q is the net charge accumulated on the surface