Question

state and prove Gauss's theorem​

Answer 1

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Answer 2

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.

Radial Direction

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: Er =

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

r3

dS

⇒

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 −

⇒ ∮ Er.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.