**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.”

[caption id="attachment_1892" align="aligncenter" width="404"] **FIg : 01**[/caption]

[caption id="attachment_1893" align="aligncenter" width="418"] **Fig : 02**[/caption]

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 ^{2 }C ^{-1}

**Dimension of ϕ :**

[ ϕ ] = [ M L ^{2} T ^{-3} A ^{-1} ]

**Gauss Theorem :**

**Statement :** ” The total electric flux through any closed surface area is equal to 1 / ε_{0} times the net charge enclosed by the surface. ”

**Mathematical Expression : **

ϕ = q / ε_{0}

**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 / ε_{0}

The net electric flux at the point P is given by

ϕ = ϕ_{1} + ϕ_{2} + ϕ_{3}

q / ε_{0} = ∮ **E _{1} . **d

**A**+ ∮

_{1}**E**d

_{2}.**A**+ ∮

_{2}**E**.d

**A**

_{3}As E_{1} and E_{2} are perpendicular to the surface area dA_{1} and dA_{2} respectively.

∮ **E _{1} . **d

**A**= 0 and ∮

_{1}**E**d

_{2}.**A**= 0

_{2}where E_{1}, E_{2}, and E are the electric field intensities at the top surface, bottom surface and at point P respectively.

Therefore,

q / ε_{0} = ∮ **E . **d**A _{3}**

q / ε_{0} = ∮ EdA_{3}

q / ε_{0} = E ∮ dA_{3}

q / ε_{0 }= E * ( A )

q / ε_{0} = E ( 2πrl )

E = ( 1 / 4πε_{0} ) * ( ( 2 / r ) * ( q / l )

**E = 2k****λ /r**

where λ = q / l and it is known as linear charged density and k = 1 / 4πε_{0} , ε_{0} 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 / ε_{0}

The net electric flux at the point P is given by

ϕ = ϕ_{1} + ϕ_{2} + ϕ_{3}

q / ε_{0} = ∮ **E _{1} . **d

**A**+ ∮

_{1}**E**d

_{2}.**A**+ ∮

_{2}**E**.d

**A**

_{3}As E_{1} and E_{2} are perpendicular to the surface area dA_{1} and dA_{2} respectively.

∮ **E _{1} . **d

**A**= 0 and ∮

_{1}**E**d

_{2}.**A**= 0

_{2}where E_{1}, E_{2}, and E are the electric field intensities at the top surface, bottom surface and at point P respectively.

Therefore,

q / ε_{0} = ∮ **E . **d**A _{3}**

q / ε_{0} = ∮ EdA_{3}

q / ε_{0} = E ∮ dA_{3}

q / ε_{0 }= E * ( A )

q / ε_{0} = E ( 2πrl )

E = ( 1 / 4πε_{0} ) * ( ( 2 / r ) * ( q / l )

**E = 2k****λ /r**

where λ = q / l and it is known as linear charged density and k = 1 / 4πε_{0} , ε_{0} 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 = 2k****λ / 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 / ε_{0}

Following the figure, the net electric flux is given by

ϕ = ϕ_{1} + ϕ_{2}

q / ε_{0} = ∮_{A1} **E . **d**A** + ∮_{A2} **E .**d**A**

where A_{1} and A_{2} are the surface areas at the two ends of the Gaussian surface and both are equal, that is A_{1} = A_{2} = A

q / ε_{0} =∮_{A} E** **dAcos0 + ∮_{A} E** **dAcos0

q / ε_{0} = 2E ∮_{A }dA ( cos 0 = 1 )

q / ε_{0} = 2EA

E = ( q / A ) / 2ε_{0}

**E = σ / 2ε _{0}**

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 / ε_{0}

The net electric flux at point P is given by

ϕ = ∮_{A} **E . **d**A**

q / ε_{0} = ∮_{A }E dAcos0

q /ε_{0} = E ∮_{A }A

q / ε_{0} = E * 4πr^{2}

E = ( 1 / 4π ε_{0 }) * ( q / r^{2} )

**E = kq / r ^{2}**

where k = 1 / 4π ε_{0}

ε_{0 }is the absolute permitivity of free space.

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

R = r

**E = kq / R ^{2}**

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 / ε_{0}

The net electric flux at point P is given by

ϕ = ∮_{A} **E . **d**A**

q / ε_{0} = ∮_{A }E dAcos0

q /ε_{0} = E ∮_{A }A

q / ε_{0} = E * 4πr^{2}

E = ( 1 / 4π ε_{0 }) * ( q / r^{2} )

**E = kq / r ^{2}**

where k = 1 / 4π ε_{0}

ε_{0 }is the absolute permitivity of free space.

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

R = r

**E = kq / R ^{2}**

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 = 4πr^{3} / 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 * ( 4πr^{3} / 3 ) / ( 4πR^{3} / 3 )

q’ = q (r / R )^{3}

Now, the electric intensity at point P is given by

E = Kq’ / r^{2}

E = K (r / R )^{3 }/ r^{2}

E = Kqr / R^{3}

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