A Brief Note On The Applications Of Gauss Theorem

Gauss law says that the total flux of an electric field in a closed surface and the enclosed electric charge will be directly proportional. There are four equations of electromagnetism, and Gauss’s law is one of them. We can define the electric flux as the product of an electric field that would pass through a given area and the surface area in a plane perpendicular to the field. This law calculates electric fields due to Uniformly charged Straight wire, uniformly charged Infinite plate sheet, and Uniformly charged thin spherical shell.

Applications of Gauss’s Law

Gauss law is an essential part of electromagnetism. It helps relate the charge distribution with the resultant electric field because of the charge. Joseph Lagrange formulated this law in 1773 and was followed by Carl Gauss in 1813. It has many applications, and some of them are:

Electric field due to Uniformly charged Straight wire

Suppose we take a uniformly charged straight wire with a linear charge density λ and Length (L). We would assume a cylindrical Gaussian surface to calculate the electric field. The electric flux across the end of the cylindrical surface would be zero because the electric field E is circular in direction. This is because the area vector and the electric field are perpendicular. It is seen that the electric field is upright to each point of the curved surface; it can be said that it has a constant magnitude.

If 2πrl gives the surface area of the cylindrical surface, then the electric flux through the curve would be

         E × 2πrl

According to Gauss’s Law

Φ = q/ ε0

E × 2πrl = λl/ ε0

E = λ / 2π ε0r

It must be remembered that if the charge of linear density is positive, then the direction of the electric field would be radially outward.

Electric field because of Uniformly charged Infinite plate sheet:

If we consider a uniformly charged infinite plate sheet that has a surface charge density σ with a cross-sectional area A. The infinite charges sheet will cause the direction of the electric field to be perpendicular to the plane of the sheet. If we take a cylindrical Gaussian surface that has its axis normal to the plane of the sheet, So by Gauss’s Law:

2EA = σ A/ ε0

E = σ / 2 ε0

The unit vector, which is expected to surface 1, is in the –x-direction, and the unit vector, which is expected to surface two, would be in the +x direction. Hence, the fluxes from both the surfaces would be equal and add up. So, the total flux through the Gaussian surface would be 2 EA. The closed surface enclosing the charge is σ A.

Electric Field because of a uniformly charged thin spherical shell:

If we assume a thin spherical shell with radius R and σ is its uniform surface charge density, At any point P, whether inside or outside, the field may depend only on r, and it should be radial.

The field outside the shell:

Suppose we consider a point P outside the shell with a radius vector r. In order to calculate E, at point P, we would take the Gaussian surface that is a sphere with radius r, and its centre is O which would pass through P. Every point on this sphere is equally relative to the configuration of the given charge. At every point of the Gaussian surface, the electric field would have the same intensity E and is radial at each point.

So, E and ΔS would be parallel at every point, and the flux through every element would be E ΔS. The flux across the Gaussian surface would be E × 4 πε0 r2, and σ × 4 π R2 is the charge enclosed in it.

Conclusion

Gauss’s law is used to derive the equations for electrical fields for several charged surfaces. We can take any object like a uniformly charged Straight wire, a uniformly charged Infinite plate sheet, or a thin spherical shell. We can define the electric flux as the product of the electric field, and the area is the surface perpendicular to the electric field. The law is fundamental as it helps us estimate the electrical charge enclosed inside a closed surface. The law is helpful for symmetrical geometries.

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