Electric Flux

We define the Electric Flux as the rate of flow of the electric field through a given area and it varies directly with the number of electric field lines going via a virtual surface.

Switch on the mosquito repellent, you smell the fragrance after some time. So, here, lines (electric field lines) of fragrance passing through the area of the room are the electric flux. 

On this page, you will learn what is electric flux, electric flux definition in detail.

Important Terms Used in the Concept of Electric Flux

Some of the important terms that are used in the concept of electric flux and one must understand these terms in order to understand the concept of electric flux. These terms are as follows

Electric Field: electric field is a field or space around a stable or moving charge in the form of a charged particle or between the two voltages. The other charged objects or particles in this space also experience some force exerted by this field, the intensity and type of force exerted will be dependent on the charge a particle carries.

Electric Charge: It is a physical property of matter that causes it to experience a force while in an electromagnetic field. It also provides the particle to have an electric field of itself. An electron has a charge of -1 while the proton has a charge of +1. Neutrons are neutral with charge equal to the 0 and also do not interfere with any other electric charges.

Electric displacement field: The term electric displacement field or electric induction is a vector field which is used in the Maxwell's equation, and In Physics, it is denoted by ‘D’.

What Is Electric Flux?

In terms of electromagnetism, electric flux is the measure of the electric field lines crossing the surface. Although an electric field cannot flow by itself, it is a way of describing the electric field strength at any distance from the charge creating the field. The electric field E can generate a force on an electric charge at any point in space.

Concept of Electric Flux

In the above text, we understood the electric flux definition. Now, let’s understand the concept of electric flux.

Let’s consider a hypothetical (imaginary) planar element of area ΔS and a uniform electric field exists on the plane surface.

We know that the number of field lines crossing a unit area placed normal to the electric 

field E at a point is a measure of the strength of the electric field at that point.

Now, draw a line normal to the surface and call one side of it the positive normal to the surface.

If we place a smaller planar element of an area ΔS normal to \[\vec{E}\] at this point electric field lines crossing this area are proportional to EΔS.

Points to Remember

Please note that it doesn’t mean that the number of field lines crossing this area is equal to E ΔS.

As the orientation of the area element also decides the amount of actual eclectic flux experienced, the area in the electric flux is a vector quantity.

S.I. unit of electric flux is volt metres (V m) and the dimensions of the electric flux are - 

\[N C^{-1} m^{2} or Kg m^{3} s^{-3} A^{-1}.\]

Flux of Electric Field

If we tilt the area element by an angle Ө (or we tilt/rotate \[\vec{E}\] with respect to the area element by an angle Ө, the number of electric field lines crossing the area will be smaller.

As the projection of area element normal to \[\vec{E}\] is ΔS CosӨ (or the component of   \[\vec{E}\] normal to area element is  E CosӨ).

Therefore, the number of electric field lines crossing area ΔS  is proportional to the following components in the equation:

Unit of Electric Flux

Dimensional Formula of Electric Flux

What is Flux in Electrical?

Let’s suppose that we draw a vector of the magnitude ΔS along with the positive normal. ΔS is called the area-vector. The equation for the electric flux through a given area is:

Where,

Φ =  Electric flux which is proportional to the number of field lines cutting the area element.

Ө  is the smaller angle between  \[\vec{E}\] and ΔS.

Now, let’s look at the following cases to determine the electric flux at certain angles:                                         

1. When \[\vec{E}\] is normal to the area element times the magnitude of area element, 

Ө = 0°, Φ is maximum. (∵ Cos 0° =1)

2. When  \[\vec{E}\] is along with the area element, Ө = 90°,  Φ is zero.

3. When  Ө >  90°, CosӨ is negative,i .e., Φ is negative.

Now, Let’s Solve Some Problems to Understand the Flux of Electric Field: 

Question1: Let’s Say that Charge q is Placed at the Centre of a Sphere. Taking Outward Normal as Positive. Determine the Flux of the Electric Field via the Surface of the Sphere Due to the Enclosed Charge.

Solution: Let us take a small element ΔS on the surface of the sphere 

(Figure.a). The electric field here is radially outward and has the following magnitude:

 = \[\frac{q} {( 4 \times π \times 𝛆o \times r^{2})}\]

Here,

q is the charge inside the sphere 

r is the radius of the sphere

𝛆o is the permittivity of free space

As the positive normal is also outward,  Ө = 0° and flux via this element are given by:

Δ Φ =  \[\vec{E}\].ΔS = E ΔS Cos 0° = E ΔS

=> Δ Φ =  \[\frac{q} {(4 \times π \times 𝛆o \times r^{2} \times  ΔS)}\]

Summing over all the elements of the spherical surface

Φ  = ∑  Δ Φ =   q / (4 x π x 𝛆o x r² ∑ ΔS)…(1)

The surface area of the sphere ΔS =  4 x π  x r²…..  putting this value in eq (1), we get:

=  \[\frac{q} { (4 \times π \times 𝛆o \times r^{2} \times 4 \times π \times r^{2})}\]

This is the required equation for the electric flux enclosed in the sphere. It means that the electric flux equation remains the same.

Question2: A rectangular surface with the measurement of sides 10 cm and 12 cm is placed inside a uniform electric field of 22 V m -1, in such a manner that the normal to the surface of the rectangle and the direction of the electric field, makes an angle of 60o. Simply find the flux of the electric field through the rectangular surface. 

Solution: Given 

The uniform electric field = E = 22 V m -1  and the angle formed between the area vector and the electric field vector is 60o. 

The flux of electric field passing through such a rectangular surface can be given by - 

Φ =  \[\vec{E}\].S = E ΔS cos Ө

 ΔS = 10 cm12 cm = 120 cm2 = 1.210 -2 m2 

 Φ =  221.210 -2 cos 60o. 

Φ =  26.410 -2 (½). = 0.132 N m 2 C -1