Question

Why the gauss theorem for infinite plane sheet is independent of distance?

Answer 1

Let us first consider the electric field on the axis of a uniformly charged ring - you will see why this is relevant soon. This situation is shown in the diagram

We know that the net electric field will point towards the centre of the ring, since all vertical components will cancel out due to the symmetry of the shape. This means we need only consider the components in the direction of the ring. First we will consider the force on particle P due to the red element highlighted. By Coulombs law we know that the contribution to the field will be

Since all the terms are constant this means that the total electric field due to the ring will be

Now we will consider the problem of the infinite sheet. We think of the sheet as being composed of an infinite number of rings. We will let the charge per unit area equal sigma

This is a very interesting result as the distance from the plane does not affect the electric field produced by it. 
Let us first consider the electric field on the axis of a uniformly charged ring - you will see why this is relevant soon. This situation is shown in the diagram

We know that the net electric field will point towards the centre of the ring, since all vertical components will cancel out due to the symmetry of the shape. This means we need only consider the components in the direction of the ring. First we will consider the force on particle P due to the red element highlighted. By Coulombs law we know that the contribution to the field will be

Since all the terms are constant this means that the total electric field due to the ring will be

Now we will consider the problem of the infinite sheet. We think of the sheet as being composed of an infinite number of rings. We will let the charge per unit area equal sigma

This is a very interesting result as the distance from the plane does not affect the electric field produced by it. 

Let us first consider the electric field on the axis of a uniformly charged ring - you will see why this is relevant soon. This situation is shown in the diagram:

We know that the net electric field will point towards the centre of the ring, since all vertical components will cancel out due to the symmetry of the shape. This means we need only consider the components in the direction of the ring. First we will consider the force on particle P due to the red element highlighted. By Coulombs law we know that the contribution to the field will be

Since all the terms are constant this means that the total electric field due to the ring will be

Now we will consider the problem of the infinite sheet. We think of the sheet as being composed of an infinite number of rings. We will let the charge per unit area equal sigma

This is a very interesting result as the distance from the plane does not affect the electric field produced by it. 

Using Gauss's law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it. ... b) Gauss's theorem states that the total electric flux through a closed surface is equal to times the net charge enclosed by the surface.

Answer 2

Answer:

Explanation:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

∮E⃗ .d⃗ s=1∈0q .

According to Gauss Law,

Φ = → E.d → A

Φ = Φcurved + Φtop + Φbottom

Φ = → E . d → A = ∫E . dA cos 0 + ∫E . dA cos 90° + ∫E . dA cos 90°

Φ = ∫E . dA × 1

Due to radial symmetry, the curved surface is equidistant from the line of charge and the electric field in the surface has a constant magnitude throughout.

Φ = ∫E . dA = E ∫dA = E . 2πrl

The net charge enclosed by the surface is:

qnet = λ.l

Using Gauss theorem,

Φ = E × 2πrl = qnet/ε0 = λl/ε0

E × 2πrl = λl/ε0

E = λ/2πrε0