# How To Find Electric Flux: What, How, Types, When, Why And Detailed Facts

The electric flux is the quantity of the flux line penetrating through the surface of the conductor. Let us discuss how to find electric flux in this article.

Electric flux can be found by calculating the total magnitude of the electric field produced by the conducting material and the density of charges present on its surface ad is a product of the electric field and the surface area of the conductor.

## How to Find Electric Flux Through a Surface?

A conductor is placed in an electric field, the charge carriers start to drift and conduct electric current.

The direction of the electric flux runs from the positive surface to the negatively charged surface. The total electric flux through a conductor relies upon the intensity of the electric field and the surface area of a conductor through which the electric flux penetrates.

Consider a small sheet placed in an electric field such that the electric flux lines are passing through the conducting sheet. Let A be the area of the conductor sheet.

By Gauss law,

Φ=E.da

Φ =EACosθ

The angle θ is formed by the electric field line with the normal surface of the charged conductor. If the normal of the surface is perpendicular to the electric field then the electric flux will be zero.

## How to Find Electric Flux Through a Cylinder?

The electric flux through a cylinder is a net flux density from all the surfaces of the cylinder.

The net electric flux can be calculated by using a Gauss law which says that the net electric flux through a conductor is a product of the charge of the conductor times 1/ε0.

Consider a cylinder of length ‘l’ and radius ‘r’ placed in an electric field region E such that electric flux penetrate through surface A of the cylinder and emerges out from the surface B of the cylinder.

The electric flux through the surface ‘A’ of the cylinder is

ΦA=EACosθ

The base of the cylinder is circular in shape hence the area of the circular base is πr2. The angle θ made by the normal of the surface with the electric flux is 0.

ΦA=E*πr2Cos0

ΦA=E*πr2 * 1

ΦA=Eπr2

Similarly, the electric flux through the surface ‘B’ where the flux line is running making an angle of 180 degrees to the normal of the surface B is

ΦB=E*πr2 Cos180

ΦB=E*πr2 (-1)

ΦB=-Eπr2

Whereas, the normal to the surface ‘C’ is perpendicular to the electric field, hence the electric flux through the cylindrical surface is

ΦC=E*πr2 Cos 90

ΦC=E*πr2 0

ΦC=0

Now, the net flux through the cylinder is

Φ =ΦABC

Φ =Eπr2 – Eπr2+0

Φ= 0

Hence, the net flux through a cylinder is found to be zero.

## How to Find Electric Flux Through a Cube?

A cube has 6 surfaces hence the electric flux through a cube will be in proportionate to the electric flux through these six surfaces of the cube.

This can be calculated by applying the gauss law, according to which the electric flux through a surface is equal to the charge times the inverse of the permeability of the medium.

Consider a cube placed in an electric field E. The length of each side of the cube is ‘l’. Let charge ‘q’ be present at the center of the cube.

By Gauss Law, electric flux through one side of the cube should be

Φ =q/ε0

Since there are 6 sides of the cube, the electric flux will be

Φ =q/6ε0

Suppose there is no charge present inside the cube, and the electric field is running perpendicular to the two surfaces of the cube and does not interfere with the remaining 4 sides of the cube as shown in the figure.

The electric flux lines are entering from the bottom surface of the cube and hence the electric flux is negative here and the field lines are leaving from the top surface of the cube, therefore, the electric flux is positive at this surface.

Electric flux at the bottom surface of the cube is

Φb=EA

Φb=-ql2/ε0

The electric flux at the top surface of the cube is

Φt=EA

Φt= ql2/ε0

Hence, the net electric flux through the cube is

Φ =Φbt

Φ =-ql2/ε0+ql2/ε0

Φ =0

Thus, the net electric flux in absence of the charged particle in the cube is zero. As per the Gauss law, the electric flux through the close objects is equal to the charge enclosed in that object and to the inverse of the permeability of the medium.

## How to Find Electric Flux Through a Sphere?

In a sphere, the electric field within the close sphere is zero as the charges are settled and running along the surface of the sphere.

The electric flux through the sphere in presence of a charge placed inside the spherical shell is equal to the product of the electric field on the surface of the sphere and the surface area of the sphere.

Consider a spherical shell of radius ‘r’ and a charge ‘q’ placed within the sphere. The electric field produced by the charge carrier on the sphere is

E=1/4ε0qπr2

The area of the sphere is A=4πr2

The electric flux is the amount of the electric field through the surface of the conductor, then we can write,

Φ =EA

Substituting the values found above in this we get

Φ =1/4ε0qπr2 (4πr2)

Φ =q/ε0

Hence, the electric flux through a sphere is independent of the radius of the sphere but depends upon the charge that it carries and the permeability of the medium.

## How to Find Electric Flux Through a Square?

The electric flux through a square is equivalent to the electric flux passing from one side of the cube.

The electric flux through a square traverses two faces of the square sheet, hence the area here doubles, and the electric flux through a square can be found by applying Gauss Law.

Suppose you have placed a square sheet having each side of length ‘l’ in the electric field E and electric flux lines are passing through this square sheet as shown in the below figure. Let charge q be present on the surface of the square.

The electric flux through the square is Φ =EA

EA=q/ε0

The surface charge density on the square is the charge present on the surface per unit area of the square that is given by the relation

σ=q/A

Where σ is a surface charge density

q is a charge of the square

A is an area of the square which is equal to A =l2

Hence, q=σ A

Substituting this in the above equation

EA=σ A / ε0

The electric flux is passing through both the surfaces of the square sheet and hence total surface area through which the electric flux is penetrating is equal to twice the surface area of the square. Therefore we can write

E.2A=σ A / ε0

Hence, we get,

E=σ A / 2ε0

This is the equation of the electric field passing through the square.

The electric flux is the product of the electric field and the surface area of the square. The total surface area of the square through which the flux is traversing is 2I2. Hence we get electric flux through a square as

Φ =EA =σI2/ε0

## How to Find Electric Flux Density?

The electric flux lines are imaginary lines penetrating through the conductor indicating the electric field direction.

The electric flux density is the number of electric flux traveling through per unit area of the conductor and is denoted as the product of the permittivity and the electric field in which the conductor is placed.

We know that the electric flux through a close surface of the conductor is

Φ =E.dA

On integrating we get

Φ=EA

If we consider a close surface to be a small element and spherically symmetric in shape the

Φ=1/4ε0qπr2 (4πr2)

Φ=q/ε0

The flux density is directly proportional to the electric flux passing through the conductor and inversely related to the surface area of the conductor.

D Φ/A

D=ε/A

Where ε is a proportionality constant and is called the permittivity of free space

Substituting value of Φ in this equation we get

D=qε/ε0A

because Φ =EA=q/ε0

E=q/ε0A

Hence, equation for electric flux density becomes

D=ε0

This is an equation to find the electric flux density through any conducting material.

## How to Find Net Electric Flux?

The net electric flux through any conductor is the sum of all the electric flux passing through all the surfaces of the conductor.

By calculating the surface area of the conductor and knowing the electric field of the conducting surface we can calculate the flux through each surface of the conductor and then add all the flux. The sum of all the flux through each of the surfaces of the conductor will give us the net electric flux through the conductor.

Suppose we have the rectangular slab placed in an electric field E such that the electric flux passes through two surfaces of the slab of length ‘l’, breath ‘b’, and height ‘h’.

The electric field is perpendicular to all the surfaces of the slab and hence Cosθ =Cos 900=0 and therefore the electric flux is nil through these surfaces Φ =0.

Except for two surfaces, the electric flux is entering through the rectangular slab through one surface hence the charge is negative and the electric flux through this surface is

Φ1 =EA

Φ1 =-qI2/ε0

Whereas the electric flux is leaving from another surface hence the charge is positive on this surface of the slab and therefore the electric flux is

Φ2=qI2/ε0

The electric flux through the rest of all the surfaces is zero and hence now we can find the net electric flux through the rectangular slab as the addition of these flux.

Φ=Φ12

Φ=-qI2/ε0+qI2/ε0=0

The net electric flux in the slab is zero as the flux is entering from one surface and leaving from another surface too.

## What is the electric flux through the conducting sphere of radius 4.8cm if a charge of 35 mC?

Given: r =4.8cm =0.048m

q= 35 mC

The electric flux through a sphere is given by the equation

Φ =q/ε0

Φ = 35*10-6/8.85*10-12

Φ =3.95* 106V.m

The electric flux through a sphere with a charge of 35 mC placed within the sphere is 3.95*106V.m.

## What is the electric flux density through a square sheet of length 13cm having a charge of +3C?

Given: q= +3C

L=13cm=0.13m

The electric field through a square sheet is

E=σ/ε0

E=q/2ε0A

E=q/2εr2

=3/2*8.85*10-12 * 0.132

=10.34*1012V/m

The electric flux density through the square sheet is

D=ε0E

=8.85*10-12*10.34*1012

=91.51 N/C

Scroll to Top