Application of Gauss’s Law

Application of Gauss’s Law

The Gauss Law, often known as Gauss’s flux theorem or Gauss’s theorem, is a law that describes the relationship between electric charge distribution and the consequent electric field. The total quantity of electric flux travelling through any closed surface is directly proportional to the enclosed electric charge, according to Gauss’ law. Coulomb’s law is used to compute the electrical field of a surface, but the Gauss law is needed to calculate the distribution of the electrical field on a closed surface. It explains whether there is an electrical charge encapsulated in the closed surface or whether there is an electrical charge present in the closed surface.

The Gauss Theorem

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

Φ = E.A = qnet/ε0  

E=electric field

A=surface area of the region

qnet=charge enclosed

The Gauss theorem, in basic terms, connects the ‘flow’ of electrical field lines (flux) to charges within the enclosed surface. The net electric flow stays zero if no charges are confined on the surface.

Applications of Gauss’s Law

Gauss’s Law can be utilised to address complex electrostatic issues with unusual symmetry, such as cylindrical, spherical, or planar. In other circumstances, calculating the electric field is also fairly difficult and requires a lot of integration. Gauss’s Law can be used to evaluate an electric field straightforwardly.

The following is how we use Gauss’s Law:

• Use a Gaussian surface so that evaluating the electric field is simple.
• Use symmetry to make problems easier to solve.
• Keep in mind that the gaussian surface does not have to match the real surface; it might be inside or outside the gaussian surface.

Electric field due to Infinite Wire

Consider a wire that is infinitely long and has a linear charge density. Due to the symmetry of the wire, let’s take a cylindrical gaussian surface for instance to compute the electric field. Because the electric field E is radial in direction, flux through the end of the cylindrical surface will be 0 because the electric field and the area vector are perpendiculars. The curved gaussian surface will be the only source of electric flux. The magnitude of the electric field will be constant because it is perpendicular to every point of the curved surface.

E.2πL=λL/ε0

So, E=λ/2πε0x

E=Electric field at a distance x

x=Distance from axis of cylinder

λ=charge per unit length

Ε0 = Vacuum permittivity

Electric field due to Infinite Plate Sheet

One of the applications of the gauss theorem in the calculation of electric fields is that the electric field produced by an infinite charge sheet is perpendicular to the sheet’s plane. Charge q will be the charge density (σ) times the area in a continuous charge distribution (A). When considering net electric flux, we will only consider electric flow from the two ends of the imaginary Gaussian surface. The electric field and the curved surface area are perpendicular to each other, resulting in zero electric flux. So, the net electric flux is

Φ = EA – (– EA)

Φ = 2EA=σA/ε0

E=σ/2ε0

1. Electric field due to Thin Spherical Shell

Let’s assume a thin spherical shell with a radius “R” and a surface charge density of σ. We can see that the shell possesses spherical symmetry just by looking at it. As a result, we can calculate the electric field owing to the spherical shell:

Φ = E × 4 πr2=Q/ε

E=Q/4πε

1. The electric field outside the spherical shell

Take a point P outside the spherical shell at a distance r from the centre of the spherical shell to find the electric field. For symmetry, we employ a gaussian spherical surface with radius r and centre O. Because all points are equally spaced “r” from the sphere’s centre, the gaussian surface will pass through P and experience a constant electric field E all around.

Inside the gaussian surface q, the enclosed charge will be σ × 4 πR2. Through the gaussian surface, the total electric flux will be

Φ = E × 4 πr2

The electric field inside the spherical shell

Take a point P inside the spherical shell to measure the electric field inside the shell. To create a spherical gaussian surface, it is possible to use symmetry that passes through P, is centred at O, and has a radius of r. Now, based on Gauss’s Law:

The net electric flux will be E × 4 π r2.

Conclusion

Hopefully, you have got a clear idea about Gauss law and its applications.

Gauss Law is a fundamental principle in learning and understanding electricity. It explains whether there is an electrical charge encapsulated in the closed surface or whether there is an electrical charge present in the closed surface. Even though the computation of electric fields is highly complicated, Gauss’ Law may address difficult electrostatic problems, including unique symmetries such as cylindrical, spherical, or planar symmetry.