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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.

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.

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.

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

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.

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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:

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 \[\vec{\Delta S}\].

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 a problem to understand the flux of electric field:

Question: 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:

= q /( 4 x π x 𝛆o x r²)

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 = \[\vec{E}\] . \[\vec{\Delta S}\] = E ΔS Cos 0° = E ΔS

=> Δ Φ = q / (4 x π x 𝛆o x r² x 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:

= q / (4 x π x 𝛆o x r² x 4 x π x r²)

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

FAQ (Frequently Asked Questions)

Q1: A Uniform Electric Field Exists in Space. Find the Flux of this Field Via a Cylindrical Surface with the Axis Parallel to the field.

Ans: From Fig.2, look at the small area ΔS on the cylindrical surface.

The normal to the cylindrical area is perpendicular to the axis of the cylinder but the electric field is parallel to the axis of the cylinder and hence the equation becomes the following:

→ →

Δ Φ = E . ΔS

Since the electric field passes perpendicular to the area element of the cylinder, so the angle between E and ΔS becomes 90°. In this way, the equation f the electric flux turns out to be the following:

→ →

Δ Φ = E . ΔS = E ΔS Cos 90°= 0 (∵ Cos 90° = 0)

This is true for each small element of the cylindrical surface. The total flux of the surface is zero.

Q2: What Do You Mean by the Electric Flux in Electrical?

Answer: Let’s consider two types of electric fields for determining the electric flux in each:

1. Uniform Electric Field

If the electric field is uniform, the electric flux passing through the vector surface area S is:

Δ Φ = E . S = E S CosӨ |

Where E is the magnitude of the electric field has units of V/m, S is the surface area, and Ө Is the angle between E and the normal to S.

2. Non-uniform Electric Field

If the electric field is non-uniform, the electric flux dΦE via a small surface area dS is given by

Δ ΦE = E. ds CosӨ |

Here, we see that the electric field E is multiplied by the component of area perpendicular to the field