Question

Given a uniform electric field $\vec E = 5 \times {10^3} \hat i N/C$ , Find the flux of this field through a square of $10cm$ on a side whose plane is parallel to the $y - z$ plane. What could be the flux through the same square if the plane makes $30^\circ$ angle with the $x -$axis?

Answer 1

We can easily solve this question using the Gauss Law. Gauss law will give us the flux passing through an area with a given electric field. We will have to find the surface area and then the angle between the electric field and the surface area. The product of the electric field and the perpendicular surface area will give us the answer.

Formulae used:
$\Rightarrow \phi = Eds\cos \theta$
Here $\phi$ is the amount of flux entering the surface, $E$ is the electric field through the surface area and $\theta$ is the angle between electric field and the perpendicular to the surface.

Complete step by step answer:
In the question the electric field is given as
$\vec E = 5 \times {10^3}\hat i N/C$
And the side of the square is given as
$\Rightarrow a = 10cm = \dfrac{{10}}{{100}}m = 0.1m$
So the surface area of the square will be
$\Rightarrow S = a \times a = 0.1 \times 0.1 = 0.01{m^2}$

It’s given in the question that the square makes an angle of $30^\circ$ angle with the $x -$axis. So the angle between the electric field and the perpendicular to the surface will be
$\Rightarrow \theta = 90^\circ - 30^\circ = 60^\circ$

Therefore the flux entering the square of side $10cm$ with uniform electric field of $\vec E = 5 \times {10^3}\hat i N/C$ and angle $60^\circ$ the between the electric field and the perpendicular to the surface will be,
$\Rightarrow \phi = Eds\cos \theta$
$\Rightarrow \phi = 5 \times {10^3} \times 0.01 \times \cos 60^\circ = 25N{m^2}{C^{ - 1}}$

So the answer will be $25N{m^2}{C^{ - 1}}$.

Additional information:
Gauss Law is very important in terms of physics. This law can be used to find the electric field due to an infinite long straight charged wire, electric field due to an infinite long straight charged plane sheet, electric field between two parallel charged sheets and electric field on a uniformly charged sphere. These calculations help in setting up of the various electrical instruments and also helps us to understand the concept of electrostatic shielding.

Note: While solving this question, we should be very careful about the angle. Always the angle between the electric field and the perpendicular between the surface area should be taken as we always take the vector of surface area perpendicular to the surface. Also we should be careful for the units.