teaching gauss law

Introduction to Gauss’s Law

Did you know Gauss’s law is also known as Gauss’s flux theorem in physics? The law is relating to the distribution of electric charge to the resulting electric field.

The law was first formulated by Joseph-Louis Lagrange in 1773 before Carl Friedrich Gauss modified it in 1813. They both discussed the attraction of ellipsoids, which is one of Maxwell’s four equations. It forms the basis of classical electrodynamics.

Coulomb’s law can be used to derive Gauss’s law and vice versa. Today we’ll be looking at the definition, equation, states, formula, applications, examples of gauss law.

gauss law to coulumb law


What is Gauss Law?

Gauss law is a total flux lined with a close surface is 1/ε0 times the charge enclosed by the closed surface.

For instance, if a point charge is placed inside a cube of edge ‘a’, the flux through each face of the cube is q/6ε0. This is what the gauss law said.

An electric field is known as the basic concept of electricity. It’s typically calculated by applying coulomb’s law when the surface is needed. The gauss law helps to calculate the electric field distribution in a close surface.

Gauss law explains the electric charge enclosed in a closed or electric charge present in the enclosed closed surface. So,

The Gauss Law States that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge. This law is one of four equations of Maxwell’s laws of electromagnetism.

Electric flux is known as the electric field passing through a given area multiplied by the area of the surface in a plane perpendicular to the field.

Another statement of gauss’s law states that the net flux of a given electric field through a given surface, divided by the enclosed charge should be equal to a constant.

Equation & Formula of Gauss Law

The formula Gauss law is expressed by;

ϕ = Q/ϵ0


Q = total charge within the given surface,

ε0 = the electric constant.

Below is the equation of gauss law in an integral form:

E⋅dA=Q/ε0 …… (1)


  • Eis the electric field vector
  • Q is the enclosed electric charge
  • ε0is the electric permittivity of free space
  • Ais the outward pointing normal area vector

Electric flux is defined as Φ=∫E⋅dA …. (2)

The electric field is understood as flux density. In gauss law, the net electric flux through any given closed surface is zero only if the volume bounded by that surface has a net charge.

Read: Electric charge – everything you need to know

Gauss Law Applications

  • The field between two parallel plates of a condenser is E = σ/ε0, where σ is the surface charge density.
  • The intensity of the electric field near a plane sheet of charge is E = σ/2ε0K where σ = surface charge density.
  • Intensity of the electric field near a plane charged conductor E = σ/Kε0in a medium of dielectric constant K. If the dielectric medium is air, then Eair = σ/ε0.
  • In the case of a charged ring of radius R on its axis at a distance x from the centre of the ring. E = \frac{1}{4\pi {{\in }_{0}}}\frac{qx}{{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}4π∈01(R2+x2)3/2qx. At the centre, x = 0 and E = 0.
  • In case of an infinite line of charge, at a distance ‘r’. E = (1/4 × πrε0) (2π/r) = λ/2πrε0. Where λ is the linear charge density.

Read: Electric force – things you must know

Solved Example

Question: There are three charges q1, q2, and q3 having charge 6 C, 5 C and 3 C enclosed in a surface. Find the total flux enclosed by the surface.

Answer: Total charge Q,

Q = q1 + q2 + q3
= 6 C + 5 C + 3 C
= 14 C

The total flux, ϕ = Q/ϵ0
ϕ = 14C / (8.854×10−12 F/m)
ϕ = 1.584 Nm2/C

Therefore, the total flux enclosed by the surface is 1.584 Nm2/C.

watch this video to have more understanding of Gauss law:

That’s it for this article “Gauss Law”. I hope the knowledge is attained, if so, kindly comment, share, and recommend this site to other technical students. Thanks!


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