# CLASS 12 PHYSICS, MODULE – 01, ELECTROSTATIC,

### CHAPTER – 03, GAUSS THEOREM

In the previous chapters, we have learnt the concept of electric charge, electric field, electric force, electric field intensity and the electric lines of force. Let us discuss the electric flux.

# Electric Flux

The word ” flux “ comes from the latin word ” fluxus “ which means to flow. Any thing that flow from one point to another point, may be termed as the flux. Here, the electric flux means ” Flow of electric lines of force from one point to another point.” FIg : 01 Fig : 02

In fig : 02, we can see the definition of electric flux whereas  in fig : 03, we can see the meaning of the electric flux.

Definition : The total electric lines of force passing through a particular surface area perpendicularly, are called as the electric lines of force.

It represents the net number of electric lines of force passing through the particular surface perpendicularly.

In fig : 02, we can observe that

Number of electric lines of force coming to the surface ( C ) = 3

Number of electric lines of force going outside  to the surface ( G ) = 4

Net number of electric lines of force = G – C = 4 – 3 = 1

Hence, this net number of electric lines of force represents the electric flux that is electric flux = 1 unit.

Properties of electric flux :

1 ) It represents the net number of electric lines of force passing through the particular surface.

2 ) It is independent of medium.

3 ) It is scalar quantity.

4 ) It depends on the amount of charge from which it originates.

5 ) It depends on the angle between the surface area and the electric field.

6 ) It becomes zero for the surface area perpendicular to the electric field.

7 ) It will be negative for the direction of the surface area opposite to the electric field.

Mathematical Expression :

Let us consider that

E = Electric field

A = Area of surface through which the electric lines of force

θ = angle between the surface area and the electric lines of force.

Then the electric flux is given by

ϕ = E . A

ϕ = E A cosθ

Special case :

1 ) If θ = 0 degree, then  ϕ = EA

2 ) If θ = 90 degree, then ϕ = 0

3 ) If θ = 180 degree, then ϕ = -EA

Units Of ϕ :

Its SI unit is V – m or N m -1

Dimension of ϕ :

[ ϕ ] = [ M L 2-3-1  ]

# Gauss Theorem :

Statement : ” The total electric flux through any closed surface area is equal to 1 / ε0 times the net charge enclosed by the surface. ”

Mathematical Expression :

ϕ = q / ε0

Gaussian Surface :  This is the imaginary surface which encloses the charge symmetrically.

Application Of Gauss Theorem :

It may be used to calculate the value of electric intensity due to

1 ) infinite line of charge

2 ) infinitely long cylinder

3 ) infinite plane sheet of the charge

4 ) uniformly charged thin spherical shell

5 ) uniformly charged non conducting solid sphere

#### 1 ) Electric Intensity due to infinite line of charge : Let +q is distributed over a line of length ‘ l ‘. Let there is a point P at which electric intensity is required to calculate and it is at a distance of ‘ r ‘ from the line. A Gaussian surface which is cylinder of length ‘ l ‘ and base radius ‘ r ‘ is drawn.

From Gaussian Theorem the electric flux is given by

ϕ = q /  ε0

The net electric flux at the point P is given by

ϕ = ϕ1 + ϕ2 + ϕ3

q / ε0E1dA1 + ∮ E2 .dA2 + ∮ E .dA3

As E1 and E2 are perpendicular to the surface area dA1 and dA2 respectively.

E1dA1 = 0 and  ∮ E2dA2 = 0

where E1, E2, and E are the electric field intensities at the top surface, bottom surface and at point P respectively.

Therefore,

q / ε0 =  E . dA3

q / ε0 =  ∮ EdA3

q /  ε0 =  E ∮ dA3

q /  ε= E * ( A )

q /  ε0 = E ( 2πrl )

E = ( 1 / 4πε0 ) * ( ( 2 / r ) * ( q / l )

E =  2kλ /r

where λ = q / l and it is known as linear charged density and k =  1 / 4πε0 ,  ε0 is the absolute permitivity of free space.

#### 2 ) Electric Intensity due to infinitely long cylinder :

##### Case 1 : ( When the point at which the electric intensity is calculated is situated outside the cylinder ) Let there is infinitely long cylinder having radius R. Let there is point P outside the cylinder at which the electric intensity is to be calculated. The distance between the point P and the axis of the cylinder is ‘ r ‘.

Let us draw the Gaussian surface of radius ‘ r ‘ and length ‘ l ‘.

From Gaussian Theorem the electric flux is given by

ϕ = q /  ε0

The net electric flux at the point P is given by

ϕ = ϕ1 + ϕ2 + ϕ3

q / ε0E1dA1 + ∮ E2 .dA2 + ∮ E .dA3

As E1 and E2 are perpendicular to the surface area dA1 and dA2 respectively.

E1dA1 = 0 and  ∮ E2dA2 = 0

where E1, E2, and E are the electric field intensities at the top surface, bottom surface and at point P respectively.

Therefore,

q / ε0 =  E . dA3

q / ε0 =  ∮ EdA3

q /  ε0 =  E ∮ dA3

q /  ε= E * ( A )

q /  ε0 = E ( 2πrl )

E = ( 1 / 4πε0 ) * ( ( 2 / r ) * ( q / l )

E =  2kλ /r

where λ = q / l and it is known as linear charged density and k =  1 / 4πε0 ,  ε0 is the absolute permitivity of free space.

##### Case 2 : ( When the point P lies on the cylinder )

In this case R = r that is the Gaussian surface is drawn on the cylinder.

E =  2kλ / R = Constant

##### Case 3 : ( When P lies inside the cylinder ) In this case, r < R that is the Gaussian surface is drawn inside the surface.

There are no electric lines of force passing through the cylinder, therefore the electric intensity inside the cylinder is zero

E = 0

#### 3 ) Electric Intensity due to infinite sheet of the charge : Let us consider that there is a sheet having positive charge +q. The Gaussian surface ( cylindrical shape ) is drawn through the sheet.

From Gaussian Theorem the electric flux is given by

ϕ = q /  ε0

Following the figure, the net electric flux is given by

ϕ = ϕ1 + ϕ2

q / ε0A1 E . dA + ∮A2 E .dA

where A1 and A2 are the surface areas at the two ends of the Gaussian surface and both are equal, that is A1 = A2 = A

q / ε0 =A E dAcos0 + ∮A E dAcos0

q / ε0 = 2E dA ( cos 0 = 1 )

q / ε0 = 2EA

E = ( q / A ) / 2ε0

E = σ / 2ε0

σ is the Surface charge density which is equal to q / A

Direction of E :

i ) If the sheet has positive charge then the electric field will be away from the sheet.

ii ) If the sheet has negative charge then the electric field will be towards the sheet.

Note : The value of electric field intensity does not depend on the distance from sheet.

#### 4 ) Electric Intensity due to uniformly charged thin spherical shell :

##### Case 1 : ( When the point P at which the electric intensity is to be calculated lies outside the spherical shell ) Let there is a spherical shell of radius ‘ R ‘ over which +q charge is spread. Let there is point P at which the electric intensity at a distance of ‘ r ‘ from the shell , is required to calculate and this point is outside the shell. Let us draw a Gaussian surface which is spherical shape of radius ‘ r ‘.

From Gaussian Theorem the electric flux is given by

ϕ = q /  ε0

The net electric flux at point P is given by

ϕ = A E . dA

q / ε0 = ∮E dAcos0

q /ε0 = E A

q / ε0 = E * 4πr2

E = ( 1 / 4π ε) * ( q / r2 )

E = kq / r2

where k = 1 / 4π ε0

εis the absolute permitivity of free space.

##### Case 2 : ( When the point on the spherical shell )

R = r

E = kq / R2

This is a constant value.

##### Case 3 : ( The point at which the electric intensity is required to calculate, is inside the shell )

E = 0

That is the electric intensity is zero inside the sphere as there is no electric lines of force.

#### 5 ) Electric Intensity due to uniformly charged non conducting solid sphere :

##### Case 1 : ( When the point of observation lies outside the sphere ) Let there is a nonconducting solid sphere of radius ‘ R ‘ over which +q charge is spread. Let there is point P at which the electric intensity at a distance of ‘ r ‘ from the sphere , is required to calculate and this point is outside the shell. Let us draw a Gaussian surface which is spherical shape of radius ‘ r ‘.

From Gaussian Theorem the electric flux is given by

ϕ = q /  ε0

The net electric flux at point P is given by

ϕ = A E . dA

q / ε0 = ∮E dAcos0

q /ε0 = E A

q / ε0 = E * 4πr2

E = ( 1 / 4π ε) * ( q / r2 )

E = kq / r2

where k = 1 / 4π ε0

εis the absolute permitivity of free space.

##### Case 2 :( When the point P lies on the surface of the solid sphere )

R = r

E = kq / R2

which is a constant value.

##### Case 3 :( When the point P lies inside the surface of the solid sphere )

Let the Gaussian surface which is sphere of radius r is drawn.

q’ = amount of charge enclosed by the Gaussian surface

v’ = volume of the Gaussian surface = 4πr3 / 3

ρ = charge per unit volume

It is constant both for the solid sphere and the Gaussian surface.

Hence

q’ / v’ = q / v

q’ = qv’ / v

q’ = q * ( 4πr3 / 3 ) / ( 4πR3 / 3 )

q’ = q (r / R )3

Now, the electric intensity at point P is given by

E = Kq’ / r2

E = K (r / R )/ r2

E = Kqr / R3

that is E is directly proportional to the distance of the point of observation.