Application of Gauss’s Law

The Gauss Law is the most fundamental concept in electricity. We need to understand the application of Gauss law to determine the distribution of the electrical field on a closed surface. It describes the enclosed closed surface has an electrical charge.

Here, you will gain comprehensive knowledge about the Applications of Gauss’ Law, Gauss Law formula, The Gauss Law Theorem, Electric Field, FAQs, and more. If you want to ace the Class 12th exams with excellent marks, you must properly understand the Gauss Law concepts. 

What is Gauss Law?

According to the Gauss law in electrostatics, the total flux associated with a closed surface equals 1/ϵ0 times the charge encompassed by the closed surface. 

A point charge q, for example, is placed within a cube with edge ‘a’. According to Gauss’ law, the flux across each face of the cube is now q/ϵ0.

Gauss Law Formula

According to the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. 

Q = ϕ ϵ0

The formula of Gauss law is articulated by;

ϕ = Q/ϵ0

Where,

Q = the total charge on a given surface,

ε0 = the electric constant.

The Gauss Theorem

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

Φ =  EdA = qnet/ε0

The Gauss theorem connects the ‘flow’ of electric field lines (flux) to the charges within the enclosed surface in simple terms. The net electric flow stays 0 if a surface encloses no charges.

Note: This law can be defined as the repetition of Coulomb’s law. Coulomb’s law is simply obtained by applying the gauss theorem in the calculation of electric fields. 

Application of Gauss Theorem in Calculation of Electric Field

Gauss’s law can address complex electrostatic issues with unusual symmetry, such as cylindrical, spherical, or planar. Here you’ll learn how to state and prove gauss law and its application to straightforwardly evaluating the electrical field.

Electric Field due to Infinite Wire

Consider an infinitely long wire that has a linear charge density λ and length L. Due to the symmetry of the wire, we assume a cylindrical Gaussian surface 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.

The bent cylindrical surface will have a surface area of 2πrl. Through the curve, the electric flux will be E × 2πrl

Electric Field due to Infinite Plate Sheet

Consider an infinite plane sheet with a cross-sectional area A and a surface charge density σ.

The electric field generated by an infinite charge sheet will be perpendicular to the sheet’s plane. Consider a cylindrical Gaussian surface whose axis is perpendicular to the sheet’s plane.

Charge q will be σ A as a result of continuous charge distribution. We shall only consider electric flow from the two ends of the hypothetical Gaussian surface when discussing net electric flux. This is because the curved surface area and the electric field are perpendicular to each other, resulting in zero electric flux.

So the net electric flux will be

Φ is equal to EA – (– EA)

Φ is equal to 2EA

Note 1: If the surface charge density is positive, the electric field is directed away from the infinite sheet; if the surface charge density is negative, the electric field is directed towards the infinite sheet.

Note 2: The infinite sheet’s electric field is independent of its position.

Electric Field due to thin Spherical Shell

Consider a thin spherical shell with a radius “R” and a surface charge density. Shell has spherical symmetry, as may be seen by looking at it. The electric field produced by the spherical shell can be measured in two ways:

Electric Field Outside the Spherical Shell

We take a point P outside the spherical shell at a distance r from the centre of the spherical shell to find the electric field outside the shell. We use a Gaussian spherical surface with radius r and centre O for symmetry. As all points are equally spaced “r” from the sphere’s centre, the Gaussian surface will pass through P and experience a constant electric field all around. Then,

According to Gauss’s Law

σ × 4 πR2will be the encapsulated charge inside the Gaussian surface q. Through the Gaussian surface, the total electric flux will be

Φ= E × 4 πr2

Electric Field Inside the Spherical Shell

Take a point P inside the spherical shell to measure the electric field inside the shell. We can use symmetry to create a spherical Gaussian surface that passes through P, is centred at O, and has a radius of r.

E × 4 π r2 will be the net electric flow. However, because surface charge density is diffused outside the surface, the enclosed charge q will be zero, and there will be no charge inside the spherical shell.

Conclusion 

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

Even though the computation of electric fields is highly complicated, Gauss’ Law may address complex electrostatic problems, including unique symmetries such as cylindrical, spherical, or planar symmetry. 

With the proper understanding of the Gauss Law formula, The Gauss Law Theorem, Electric Field, and more, you’re likely to outperform your colleagues. The notes are fully illustrated with relevant facts and figures that would ultimately help students learn to state and prove gauss law and its application.