Gauss law states that the enclosed electric charge is directly proportional to the total flux of an electric field. The electric flux is the electric field that passes through a given area multiplied by the area of the surface on a plane perpendicular to the field. Gauss law is used for any closed surfaceAs per Gauss’s law, the complete motion connected with a Gaussian surface is 1/ε0 times the charge encased by a Gaussian surface.

## Gauss’s Law Formula

According to this law, the total flux enclosed in a closed surface is proportional to the absolute charge enclosed by the surface.

Φ is the total flux, the total electric charge Q total charge in a closed surface or a given surface, and ε0is the electric constant, which can be portrayed as follows.

Q=ε0

In this manner, Gauss’s law formula can be expressed as shown below

E=Q/ε0

Derivation

Considering total flux through a sphere of radius, r encloses a point charge q at its centre. Divide the sphere into small area elements.

The flux through an area element ΔS is

Δɸ=E.ΔS=q4πε0r2r.ΔS

We used Coulomb’s law for the electric field due to a single charge q. Now, since the normal sphere at every point is along the radius vector at that point, the area element ΔS and rˆ have the same direction. Therefore,

Δɸ=q4π0r2ΔS

Since the magnitude of a unit vector is 1, the total flux through the sphere is obtained by adding up flux through all the different elements:

ɸ= 𝝨q4π0r2ΔS

Since each area element of the sphere are at the same distance r from the charge,

ɸ= 𝝨q4π0r2ΔS=q4π0r2S

Now S, the total area of the sphere, equal 4πr2

ɸ= q4π0r2×4πr2=q0

## Importance of Gauss’s law

It is used to solve complex electrostatic problems, including unique shapes, for example, barrel-shaped, circular, or planar balance.

This can be exceptionally valuable to work out field force because of an endless, lengthy, consistently charged wire.

Assuming the charge distribution needs symmetry of use, in those cases, we can utilise this law to ascertain point charge fields of the particular charge components which are available in the object.

This law can be used to improve the evaluation of the electric field critically without any problem.

Where solving the electric field is difficult, then, at that point, this law is utilised in basic form.

## Problems illustrating Gauss’s law

Example 1: Determine the electric flux for a closed surface that contains 200 million electrons.

Solution:

ɸ=q/ε0

ɸ=200 × 106(1.6 ×10-19)/8.85×10-12

ɸ=3.764 Nm2/C

Example 2: A uniformly charged solid spherical insulator had a radius of 0.25 m, and the total charge in the volume is 3.5 pC. Find the e-field at a position of 0.15 m from the sphere’s centre.

Solution:

E=[q/4πε0R3]r

E=[3.5×10-12/4πε0(0.25)3](0.15)

E=0.116N/C

Example 3: An enclosed Gaussian surface in the 3D space where the electric flux is measured. The Gaussian surface is spherical, enclosed with 50 electrons, and has a radius of 0.8 metres.

Find the electric flux having a distance of 0.8 metres to the field measured from the centre of the surface.

Calculate the electric charge that passes through the surface.

Explain the relation that exists between the enclosed charge and the electric charge.

Solution:

ɸ= Q/ε0

ɸ=[50(1.60×10-19)/8.85×10-12]

E=0.113/4π(0.8)2

### Conclusion

Gauss’s law is applied to any closed surface. It is an important tool since it allows the assessment of the amount of enclosed charge by mapping the field on a surface outside the charge distribution. For calculations of sufficient shapes, it improves the analysis of the electric field. It is one of Maxwell’s four equations, which shapes the basis of classical electrodynamics. Gauss’s law can be utilised to determine Coulomb’s law and the other way around.