Gauss Law

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Gauss theorem relates the flow of the net electric field lines (flux Φ)  to the charges enclosed within the surface. 

Let’s write the Gauss law statement:

The net electric flux ΦNET  through a closed surface (any 3-D closed surface) is 1/ε0 times the net charge enclosed by the surface. Mathematically, we express it as:

                           ΦNET  =  ∮ E\[^{→}\] . dS\[^{→}\] = qinside0

Let’s consider a small surface whose area vector is dS\[^{→}\] is perpendicular to it. 

Let’s assume an electric field E\[^{→}\] is existing on this surface, as shown below:

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Here, Ө is the angle between E\[^{→}\] & dS\[^{→}\] .

A small flux dΦ generates in this area whose value  = EdS Cos Ө 

                                   dΦ = E\[^{→}\] . dS\[^{→}\]

The net flux dΦ = ∫ E\[^{→}\] . dS\[^{→}\] , if E\[^{→}\] is constant, then

                  ΦNET = E\[^{→}\] ∫dS\[^{→}\] =  E\[^{→}\] . S\[^{→}\] …(1)

Eq (1) is the original formula for flux. However, Gauss eliminated the integral form and provided us with a simple expression which is the Gauss law formula:


ΦCLOSED = qinside0     


∴ ΦCLOSED  = ∮CLOSED E\[^{→}\] . dS\[^{→}\] = qinside0         

Gauss Law Derivation

Let’s consider a charge q placed at point O inside the sphere. Now, we have to determine the amount of flux coming out of it.

According to Gauss law, the flux coming out will be qinside0.

Let us validate the above analogy by using mathematics:

Let’s say, we place a charge q at point O inside the closed sphere. 

Take a point P on the surface and consider a small area ΔS around this point on the surface.     

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This area vector is always perpendicular to the surface, so it passes through the centre, as shown in Fig.1.

The electric field E\[^{→}\] is away from the charge represented by a red arrow as shown below.

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We can see the vectors E\[^{→}\] & dS\[^{→}\] are in the same direction. This means the value of Ө = 0°.

Here, if we find the small flux, i.e., dΦ, it will be:

                                    dΦ = EdS Cos 0° = EdS

∴ EdS remains the same everywhere on the surface, i.e., E = q/4πε0r² remains the same at any point on this surface.

So, the net electric flux will be:

                      Φ = ∮Closed surface  dΦ = ∮Sphere E . dS = E ∮ dS, and

                     ∮Sphere dS = S = 4πr²

Now, putting the value of E and S in the above equation, we get,

                      Φ = q/4πε0r². 4πr² = q/ε

For a closed surface, we can write the equation as:

                       ΦCLOSED  =  qinside0

So, the total flux through this closed surface is equal to the net charge inside the sphere divided by ε0. This is the Gauss law.

Note: According to Gauss’s law, the net flux remains unaffected by the size and shape of the closed surface.

Application of Gauss law       

1. Charged Conductor

We know that an electric conductor has many free electrons and when placed in an electric field. These electrons redistribute themselves to make the electric field zero at all points inside the conductor.

Consider a charged conductor placed inside an electric field. Draw a surface S by joining the internal points of this conductor, as shown below:

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As the electric field inside a conductor is zero. So  ∮ E\[^{→}\] . dS\[^{→}\] is also zero. But from Gauss law, this equals the net charge inside this conductor divided by ε0. Hence, qENCLOSED is zero. This means the flux generates on the surface of this conductor, and inside it’s always zero.

2. Charge Placed Inside the Cavity

If the charge q is placed within the cavity, as shown below:

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The net flux is zero.

3. Charge Placed Outside the Conductor

Let’s understand mathematically how flux is zero when any charge is placed outside the conductor.

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We can see the electric field EA is inward, so it is given by E =  ( - ) q/4πε0r², and Eq is outside the 3-D closed surface. So it is given E = q/4πε0

We know that   ΦNET  =  ∮  E\[^{→}\] . dS\[^{→}\]

For the first point A, Φ1  = EdS Cos 180° = - EdS, and

For the second point B, Φ2  = Eds Cos 0° = EdS

ΦTotal  = Φ1 + Φ2  = - EdS + EdS = 0

This means ΦNET  is zero.

4. Charges Inside and Outside the Surface

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Here, the net flux will be:

ΦNET  = (q1 - q2 - q5)/ε0=  ∮ E\[^{→}\] . dS\[^{→}\]

However, no flux will be generated because of q3 and q4.

Suppose we consider an electric field at any point on the surface. We are not sure whether the electric field is due to q1, q2, or q3 because ∮ E\[^{→}\] is the net electric field due to all the charges, be it inside or outside the surface.

Here, we need to find the electric field due to all the charges present inside and outside the surface and then integrating them ∮RESULTANT E\[^{→}\] to find the resultant electric field.

This means the electric field is the resultant electric field due to inside and outside charges. However, the electric flux is due to inside charges only.

FAQ (Frequently Asked Questions)

Q1: What Properties Must a Useful Gaussian Surface Have?

Ans: The properties that a useful Gaussian surface must have are:

  1. The surface should be symmetric about the charge distribution.

  2. The electric field must be symmetric (equal/constant/same) at all points of the Gaussian surface.  

  3. Ө between E & dS must be invariant at all points on the surface.

  4. The Gaussian surface must not pass via a point charge.

Q2: What Do You Mean By the Gauss Theorem?

Ans: The Gauss theorem states that the total electric flux across any closed surface in an electric field is 4π times the electric charge enclosed by it.

Q3: Determine the Electric Flux for a Gaussian Surface Containing 200 Million Electrons.

Ans: We are given with a number of electrons (n) = 200 million or 2 x 10⁸

We also know that q = ne

So, q = 2 x  10⁸ x 1.6 x  10⁻¹⁹C = 3.2 x  10⁻¹¹C

Using Gauss law,

Φ  = q /ε₀ = 3.2 x 10⁻¹¹/8.85 x 10⁻¹² = 3.6 Nm²/C          

Q4: Write One Application for the Gauss law.

(charge per unit length) λ as a Gaussian surface.

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From Gauss law,

E . dS = qinside /ε₀  = λl/ε₀, and  E = λl/2πε₀r 

E * 2πrl = λl/ε₀  (∵ dS→ = 2πrl)