Class XII Gauss Theorem

 

What is Gauss Law?

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

$\oint{\vec{E}.\vec{d}s=\frac{1}{{{\in }_{0}}}q}$ .

For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.

The electric field is the basic concept to know about electricity. Generally, the electric field of the surface is calculated by applying Coulomb’s law, but to calculate the electric field distribution in a closed surface, we need to understand the concept of Gauss law. It explains the electric charge enclosed in a closed or the electric charge present in the enclosed closed surface.

Gauss Law Formula

As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;

Q = ϕ ϵ0

The Gauss law formula is expressed by;

ϕ = Q/ϵ0

Where,

Q = total charge within the given surface,

ε0 = the electric constant.

The Gauss Theorem

The net flux through a closed surface is directly proportional to the net charge in the volume enclosed by the closed surface.

Φ = → E.d → A = qnet/ε0

In simple words, the Gauss theorem relates the ‘flow’ of electric field lines (flux) to the charges within the enclosed surface. If there are no charges enclosed by a surface, then the net electric flux remains zero.

This means that the number of electric field lines entering the surface is equal to the field lines leaving the surface.

The Gauss theorem statement also gives an important corollary:

The electric flux from any closed surface is only due to the sources (positive charges) and sinks (negative charges) of electric fields enclosed by the surface. Any charges outside the surface do not contribute to the electric flux. Also, only electric charges can act as sources or sinks of electric fields. Changing magnetic fields, for example, cannot act as sources or sinks of electric fields.

Gauss Law in Magnetism

The net flux for the surface on the left is non-zero as it encloses a net charge. The net flux for the surface on the right is zero since it does not enclose any charge.

Applications of Gauss Law

1. In the case of a charged ring of radius R on its axis at a distance x from the centre of the ring. E = $\frac{1}{4\pi {{\in }_{0}}}\frac{qx}{{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}$. At the centre, x = 0 and E = 0.

2. In case of an infinite line of charge, at a distance ‘r’. E = (1/4 × πrε0) (2π/r) = λ/2πrε0. Where λ is the linear charge density.

3. The intensity of the electric field near a plane sheet of charge is E = σ/2ε0K where σ = surface charge density.

4. The intensity of the electric field near a plane charged conductor E = σ/Kε0 in a medium of dielectric constant K. If the dielectric medium is air, then Eair = σ/ε0.

5. The field between two parallel plates of a condenser is E = σ/ε0, where σ is the surface charge density.

Electric Field due to Infinite Wire – Gauss Law Application

Consider an infinitely long line of charge with the charge per unit length being λ. We can take advantage of the cylindrical symmetry of this situation. By symmetry, The electric fields all point radially away from the line of charge, there is no component parallel to the line of charge.

We can use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface.

Applications of Gauss Law – Electric Field due to Infinite Wire

As you can see in the above diagram, the electric field is perpendicular to the curved surface of the cylinder. Thus, the angle between the electric field and area vector is zero and cos θ = 1

The top and bottom surfaces of the cylinder lie parallel to the electric field. Thus the angle between area vector and the electric field is 90 degrees and cos θ = 0.

Thus, the electric flux is only due to the curved surface

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