The measure of flow of electricity through a given area is referred to as electric flux. It is a quantity that contributes towards analysing the situation better in electrostatic. It is closely associated with Gauss’s law and electric lines of force or electric field lines. Although flux is a general term and not limited to electromagnetics it can be defined for any vector quantity. This concept helps us understand the behaviour of the electric field, and this helps us to determine the electric field in some cases. This article aims to discuss details related to electric flux and will help students learn the basics with ease.

What is Flux?

The term flux means the effective amount of a quantity passing through a given area. It is not limited to electric or magnetic flux, but rather flux of any vector through an area can be defined as,
Flux of vector $\overrightarrow{P}$ through an area $\overrightarrow{S}$ can be given as,
Flux $=\overrightarrow{P}\cdot \overrightarrow{S}=\left|\overrightarrow{P}\right|\left|\overrightarrow{S}\right|\text{cos}\left(\theta \right)$

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In fluids, we have a flux of flow of the fluid which is given as,

$\overrightarrow{V}\cdot \overrightarrow{A}=\left|\overrightarrow{V}\right|\left|\overrightarrow{A}\right|\cos \left(\theta \right)=\left|\overrightarrow{V}\right|\left|\overrightarrow{A}\right|$

Electric Flux: Details

Electric flux can be defined as the total amount of electric field lines (amount of electric field) passing through the given area. It is denoted by $\varphi$.
Example: Flux of a uniform electric field $\overrightarrow{E}$ through the given area $\overrightarrow{S}$ is defined as,

Electric flux $=\overrightarrow{E}\cdot \overrightarrow{S}=\left|\overrightarrow{E}\right|\left|\overrightarrow{S}\right|\cos \left(\theta \right)$
For non-uniform Electric field,
$\varphi =\int \overrightarrow{E}\cdot \overrightarrow{\mathrm{d}s}$
It is a scalar quantity, and its unit is $\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-1}$ or $\mathrm{V}\mathrm{m}$ (volt metre).

Electric Field Lines

Electric field lines or electric lines of force is a hypothetical concept which we use to understand the concept of Electric field.
We have the following rules, which we use while representing the field graphically.

1. It represents the electric field in the space in both magnitude and direction.
2. It emerges from a positive charge and sinks into a negative charge.
3. It can be a straight line or a curved line.
4. It cannot be a closed curve.
   Electric field lines cannot be closed lines because they cannot emerge and sink from the same point.
5. Two electric field lines cannot intersect.
   This is because the electric field line also represents the direction of the electric field and at a particular point there can be only one direction of the electric field and if it intersects then it will mean that there is two direction thus electric field lines cannot intersect.
6. The total number of the electric field that emerges or sinks from a charge is proportional to the magnitude of the charge.

Flux Through a Circular Area

Let us consider a circular area with radius $R$, kept at a distance $d$ from a charge $q$.

Now we have to calculate the flux of the electric field of the charge through the given area. Since the electric field will be non-uniform on the circular area or surface, we need to have a differential element across which the magnitude of the electric field remains constant. For this, we have differential elements as a ring of radius $x$ and thickness $\mathrm{d}x$, as each point on the ring is at the same distance from the charge; therefore, the magnitude of the electric field remains the same.
We know that
$\varphi =\int \overrightarrow{E}\cdot \overrightarrow{\mathrm{d}s}=\int \left|\overrightarrow{E}\right|\overrightarrow{\mathrm{d}s}\cos \theta$
Here,
$\left|\overrightarrow{E}\right|=\frac{q}{4\pi {\epsilon }_0\left(d^2+x^2\right)}$
And
$\mathrm{d}s=2\pi x\mathrm{d}x$
The cosine of the angle between them will be
$\text{cos}\left(\theta \right)=\frac{d}{\sqrt{\left(d^2+x^2\right)}}$
Thus, on combining the above relations, we get,
$\varphi ={\int }_0^R\frac{q}{4\pi {\epsilon }_0}\frac{2\pi xd}{{\left(d^2+x^2\right)}^{\frac{3}{2}}}\mathrm{d}x$
$\Rightarrow \varphi =\frac{qd}{2{\epsilon }_0}\left(–\frac{1}{\sqrt{\left(d^2+x^2\right)}}+\frac{1}{d}\right)$
On simplifying we get,
$\varphi =\frac{q}{2{\epsilon }_0}\left(1–\text{cos}\left(\theta \right)\right)$
Where,
$\theta$ is the angle subtended the radius of the circular area at the location of the charge.

Plane Angle and Solid Angle

As the plane angle is defined using a circle, the solid angle is defined with the help of the sphere.

Plane angle	Solid angle
$\theta =\frac{l}{R}$	$\omega =\frac{S}{R^2}$
$\theta \ne \frac{l}{R}$	$\omega \ne \frac{S}{R^2}$
$\text{d}\theta =\frac{\text{d}l}{R}$	$\text{d}\omega =\frac{\text{d}S}{R^2}$
$\text{d}\theta \ne \frac{\text{d}l}{r}$ $\text{d}\theta =\frac{\text{d}l\cos \left(\alpha \right)}{r}$	$\text{d}\omega \ne \frac{\text{d}S}{r^2}$ $\text{d}\omega =\frac{\text{d}s\cos \left(\alpha \right)}{R^2}$
$\theta =2\pi$	$\omega =4\pi$
$\theta =0$	$\omega =0$

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The relation between the solid angle and the plane angle is given by,

$\omega =2\pi \left(1–\text{cos}\left(\theta \right)\right)$

Electric Flux Through a Closed Surface due to Internal Charge

Let us consider a charge inside of a closed surface.

We know that the flux through the differential element will be given by,
$\varphi =\int \overrightarrow{E}.\overrightarrow{\text{d}s}$
$\Rightarrow \varphi =\int \left|\overrightarrow{E}\right|.\left|\overrightarrow{\text{d}s}\right|\cos \left(\alpha \right)$
$\Rightarrow \varphi =\frac{q}{4\pi {\epsilon }_0}\int \frac{\text{d}s\cdot \cos \left(\alpha \right)}{r^2}$
We know that,
$\text{d}\omega =\frac{\text{d}s\cdot \cos \left(\alpha \right)}{r^2}$
Thus, the equation becomes,
$\Rightarrow \varphi =\frac{q}{4\pi {\epsilon }_0}\int \text{d}\omega$
We know that for a closed surface, the total solid angle projected at the centre will be $4\pi .$
$\Rightarrow \varphi =\frac{q}{4\pi {\epsilon }_0}\cdot 4\pi$
$\Rightarrow \varphi =\frac{q}{{\epsilon }_0}$
If the charge lies outside the surface, flux entering the surface will be equal to flux leaving the surface, giving net flux through a closed surface zero.

Gauss’s Law

According to Gauss’s law, the total electric flux through a closed Gaussian surface is equal to the total charge enclosed by the surface divided by the ${\epsilon }_0.$
${\varphi }_{\text{closed}\text{surface}}=\frac{q_{\text{enclosed}}}{{\epsilon }_0}$
Example: Let us consider a system of charge $q_1,q_2,Q_1,Q_2.$
Let the electric field due to these charges at the differential area element be, $\overrightarrow{E_1},\overrightarrow{E_2},\overrightarrow{E^{\prime }{}_1},\overrightarrow{E^{\prime }{}_2}$
$\varphi =\oint \overrightarrow{E}\cdot \overrightarrow{\text{d}s}$
$\Rightarrow \varphi =\oint \left(\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E^{\prime }{}_1}+\overrightarrow{E^{\prime }{}_2}\right)\cdot \overrightarrow{\text{d}s}$
$\Rightarrow \varphi =\oint \overrightarrow{E_1}\cdot \overrightarrow{\text{d}s}+\oint \overrightarrow{E_2}\cdot \overrightarrow{\text{d}s}+\oint \overrightarrow{E^{\prime }{}_1}\cdot \overrightarrow{\text{d}s}+\oint \overrightarrow{E^{\prime }{}_2}\cdot \overrightarrow{\text{d}s}$
$\Rightarrow \varphi =\frac{q_1}{{\epsilon }_0}+\frac{q_2}{{\epsilon }_0}+0+0=\frac{q_1+q_2}{{\epsilon }_0}$

Sample Problems

Q.1. Find the total electric flux through the curved surface of the cylinder shown in the figure.

Ans: The flux through the two circular ends of the cylinder is given as,
${\varphi }_1=\frac{q}{2{\epsilon }_0}\left(1–\frac{3l}{\sqrt{9l^2+16R^2}}\right)$
And
${\varphi }_2=\frac{q}{2{\epsilon }_0}\left(1–\frac{l}{\sqrt{l^2+16R^2}}\right)$
We know that the total flux through a closed surface due to a charge enclosed is $\frac{q}{{\epsilon }_0}.$
Therefore, the flux through the curved surface will be equal to,
${\varphi }_{\text{cs}}={\varphi }_{\text{total}}–\left({\varphi }_1+{\varphi }_2\right)$
${\varphi }_{\text{csa}}=\frac{q}{{\epsilon }_0}–\frac{q}{2{\epsilon }_0}\left(2–\frac{3l}{\sqrt{9l^2+16R^2}}–\frac{l}{\sqrt{l^2+16R^2}}\right)$

Q.2. A charge ‘$q$’ is kept just at the corner of the cube of side ‘$a$’. Find the flux through the surface $ABCD$.

Ans: Let us consider the given cube to be a part of a larger cube of dimension ‘$2a$’, and at the centre lies the charge.

Now we know the flux through the larger cube will be $\frac{q}{{\epsilon }_0}.$
Also, from the symmetry the flux through each face of the cube will be the same and is equal to,
$\varphi =\frac{q}{6{\epsilon }_0}$
Further, if the face is divided into four equal parts, then the flux through each part will be one-fourth of the flux through the whole face.
Thus, the flux through the surface $ABCD$ will be,
$\varphi =\frac{q}{24{\epsilon }_0}.$

Summary

In the above article, we learnt that flux is a general term, and not limited to electromagnetics. It is a scalar quantity and its unit is $\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-1}$ or $\mathrm{V}\mathrm{m}$ (volt metre).
We also came to know that the electric flux signifies the number of electric field lines (amount of electric field) passing through a surface.
We derived that the electric flux through a circular area whose radius subtends an angle $\theta$ at the location of the charge ‘$q$’ is given as,
$\varphi =\frac{q}{2{\epsilon }_0}\left(1–\text{cos}\left(\theta \right)\right)$
We proved that the electric flux through the closed surface is zero due to the charge located outside of the closed surface and the electric flux through a closed surface due to a charge inside of the closed surface is $\frac{q}{{\epsilon }_0}.$
We learnt a new concept, that is solid angle and also the relation between the solid angle and the plane angle, which is given by,
$\omega =2\pi \left(1–\text{cos}\left(\theta \right)\right)$
We also studied the Gauss’s law, which states that the total electric flux through a closed Gaussian surface is equal to the total charge enclosed by the surface divided by the ${\epsilon }_0.$
${\varphi }_{\text{closed}\text{surface}}=\frac{q_{\text{enclosed}}}{{\epsilon }_0}.$

Frequently Asked Questions on Electric Flux

Frequently asked questions related to electric flux is listed as follows:

Q.1. Is electric flux a scalar or a vector quantity?
Ans: The electric flux is the dot product of the electric field and the area vector thus, it is a scalar quantity.

Q.2. What is the unit of the electric flux?
Ans: The unit of electric flux is $\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-1}$ or $\mathrm{V}\mathrm{m}$ (volt metre).

Q.3. What is electric flux?
Ans: The term flux means the effective amount of a quantity passing through a given area. Electric flux means, amount of electric field passing through a given area.
Electric flux $\varphi =\overrightarrow{E}\cdot \overrightarrow{S}.$

Q.4. What is the electric flux through a circular area due to a point charge at a distance?
Ans: Electric flux through a circular area whose radius subtends an angle $\theta$ at the location of the charge is given as,
$\varphi =\frac{q}{2{\epsilon }_0}\left(1–\text{cos}\left(\theta \right)\right).$

Q.5. What is Gauss’ law?
Ans: Gauss’ law states that the total electric flux through a closed Gaussian surface is equal to the total charge enclosed by the surface divided by the ${\epsilon }_0.$
${\varphi }_{\text{closed}\text{surface}}=\frac{q_{\text{enclosed}}}{{\epsilon }_0}.$

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