How to find the electric field using Gauss law?

According to Gauss’s Law, the flux of an electric field through a closed surface equals the charge enclosed divided by a constant. It can be demonstrated that the flux will always be equal to the charge enclosed, regardless of the shape of the closed surface. This law is utilised to find electric fields when charge distribution is given. This article also tells you how to find electric fields using Gauss law. 

What is Gauss Law?

In electronics, Gauss Law for electric fields states that the electric flux across any closed surface is inversely proportional to the net electric charge enclosed by the surface. For example, a unit charge q is placed inside a cube of side a. Now, as per the Gauss Law, the flux through each surface is 

ϕE = Q/ E0

Coulomb’s law calculates the electric field, but the distribution of the electric field is studied through Gauss law. Any charges situated outside the surface do not contribute to the flow of electricity. 

Applications Of Gauss Law

Gauss’ Law can solve complex electrostatic problems with unique symmetries like cylindrical, spherical or planar symmetry. Gauss law is also helpful in calculating the electrical field, which is a complex concept and requires difficult integration. We can use the Gauss law to evaluate the electrical field straightforwardly.

Electric field Using Gauss Law due to infinite wire

You have to take an infinitely long wire with a linear charge density. We use a cylindrical Gaussian surface to compute the electric field. Because the electric field E is radial, the flux through the surface will be zero.

The flux is zero because the electric field and the area vector are perpendicular to each other. Therefore, we can say that the magnitude of the electric field is constant because it is perpendicular to every point on the curved surface.

The area of the surface of the curved cylinder is 2πrl. The electric flux flowing through a curve is equal to E× 2πrl. 

Then as per Gauss law,

ϕ = q ⁄ ε0

E × (2 π r l) = λ l ⁄ ε0

E = λ ⁄ 2 π ε0r

Electric field using Gauss Law due to infinite plane sheet

Take an infinite plane sheet with the cross-sectional area A and the surface charge density σ. Then the position of the sheet is as follows.

An infinite charge sheet produces an electric field that is perpendicular to the plane of the sheet. Take into account a cylindrical Gaussian surface with an axis parallel to the plane of the sheet. The electric field E is calculated using Gauss’ Law as follows:

ϕ = q ⁄ ε0

In a continuous charge distribution, charge q equals the charge density (σ) times the area (A). When discussing net electric flux, we will only consider electric flow from the two ends of the fictitious Gaussian surface. The curved surface area and electric field are perpendiculars, resulting in zero electric flux. As a result of this, the net electric flux is as follows:

ϕ = E A − (− E A)

ϕ = 2 E A

So,

2 E A = σ A ⁄ ε0

E = σ ⁄ 2 ε0

Electric field using Gauss law due to infinite shell

Consider a thin spherical shell with radius “R” and surface charge density of σ. Spherical symmetry characterises the shell. The electric field generated by the spherical shell can be determined in two main ways:

• Outside of the spherical shell
• Within the spherical shell

• Consider the field outside the spherical shell.

To obtain the electric field, locate a point P outside the spherical shell at a radial distance from the centre of the spherical shell. We use a Gaussian spherical surface with radius r and centre O because of symmetry. All points on the surface are equally spaced “r” from the sphere’s centre, so the Gaussian surface will transmit through P and encounter a constant electric field E all around. As a result, the total electric flux is as follows:

ϕ = q ⁄ ε0 = E × 4 π r2

The charge in the spherical surface is q = σ × 4 π R2 

Then, 

E × 4 π r2 = σ × 4 π R2 ⁄ ε0

E = σ R2 ⁄ ε0r2

You can also write an electric field in terms of a charge.

E = k q ⁄ r2

• Electric field using Gauss law inside the spherical shell

Let us examine a point P within the spherical shell and see how the electric field exists. We can use symmetry to generate a spherical Gaussian surface with a radius of r that passes through P, is centred at O and has a radius of P. Now, according to Gauss’s Law,

ϕ = q ⁄ ε0 = E × 4 π r2

Because the surface charge density is distributed outside the surface, no charge is contained within the shell. As a result, the electric field calculated from the above formula is also zero, i.e.

E = 0 since q = 0.

Conclusion 

This article explains the electric field using Gauss law and the electric field using Gauss law examples. Gauss law defines the amount of flux inside a closed space due to a charged particle. This law can also be used to find electric fields on many surfaces. This law applies to any closed surface. The only requirement for Gauss law to be useful is that the distributed charge should be symmetric.