Gauss Theorem Proof

Description:

To prove Gauss Theorem, we need to prove Φ = q/ ε₀

We know that for a closed surface ∮ E→ . dA→ = Φ = q/ ε₀ ............(1)

First we will calculate LHS of equation (1) and prove that it is equal to RHS

Proof

Step 1

Consider a sphere with a point charge ‘q’ as its centre and radius as ‘r’. Since the charge is a point charge so the electric field will be radial in all directions.

Step 2

Take an infinitely small area on the surface.

So, the electric field at a distance ‘r’ over the Gaussian surface due to charge ‘q’ will be:  E_r = 1/4πε₀ q/r²

Step 3

Multiply both side of the equation with dS.

We get − ∮ E→ . dA→ = ∮ 1/4πε₀ q/r² dS

⇒ 1/4πε₀ q/r² ∮dS ..........(2)

(Here; 4πε₀ ,q, r are constants)

Now complete area of a sphere = ∮ dS = 4πr²

Putting the value in equation (2) we get −

∮ E→ . dA→ = Φ = q/ ε₀ ..........(3)

We can see that ; equation (1) = equation (3)

Since LHS = RHS, Hence Gauss theorem is proved.

The total flux according to our knowledge is ∮ E.dS

Since electric field E ∝ 1/r². That means it follows inverse square law.

Suppose electric field does not follow inverse square law, instead it follows inverse cube law as in case of a dipole.

In that case E ∝ 1/r³

So, in case of a dipole E = 1/4πε₀ q/r³

On Multiply both side of the equation with dS.

We get − ∮ E.dS = ∮ 1/4πε₀ q/r³dS

⇒ 1/4πε₀ q/r³ ∮ dS ............(2)

(Here; 4πε₀, q, r are constants)

Now complete area of a sphere = ∮ dS = 4πr²

Putting the value in equation (2) we get −

⇒ ∮ E_r.dS = q/ ε₀ ≠ Φ ............(3)

The above equation is not satisfying Gauss Theorem.

So Gauss Theorem is applicable only when electric field follows inverse square law.