# Gauss’s Law

Science is a fascinating subject that is full of fascinating information. The more one delves into the principles of science and its associated disciplines, the more knowledge and information there is to be gained. The Gauss Law, which analyses electric charge, a surface, and the issue of electric flux, is one such research topic. Let us learn more about the law and how it functions so that we may comprehend the equation of the law.

### What is Gauss Law?

The total electric flux out of a closed surface is equal to the charge contained divided by the permittivity, according to Gauss Law. The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane perpendicular to the field.

The total flux associated with a closed surface equals 1 ⁄ ε

_{0}times the charge encompassed by the closed surface, according to the Gauss law.

∮ E ds = q ⁄ ε_{0}

For example, a point charge ‘q’ is put within a cube with edge ‘a’. The flux across each face of the cube is now q ⁄ 6ε_{0}, according to Gauss law.

The electric field is the most fundamental concept in understanding electricity. In general, the electric field of a surface is computed using Coulomb’s equation, however, understanding the idea of Gauss law is required to calculate the electric field distribution in a closed surface. It describes how an electric charge is enclosed in a closed surface or how an electric charge is present in a closed surface that is enclosed.

### 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. As a consequence, the total electric charge Q contained by the surface is: if ε_{0} is electric constant and ϕ is total flux.

Q = ϕ ε_{0}The formula of Gauss law is given by:

ϕ = Q ⁄ ε_{0}where,

- ε
_{0}is electric constant,- Q is total charge within a given surface, and
- ϕ is flux enclosed by surface.

### The Gauss Theorem

The net charge in the volume contained by a closed surface is exactly proportional to the net flux through the closed surface.

**ϕ = E dA = q _{net} ⁄ ε_{0}**

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 no charges are contained by a surface. The number of electric field lines entering the surface equals the number of field lines exiting the surface.

**The Gauss theorem statement also gives an important corollary:**

The electric flux from any closed surface is only due to the sources and sinks of electric fields enclosed by the surface. The electric flux is unaffected by any charges outside the surface. Furthermore, only electric charges may operate as electric field sources or sinks. Changing magnetic fields, for example, cannot act as electric field sources or sinks.

Because it encloses a net charge, the net flow for the surface on the left is non-zero. Because the right-hand surface does not contain any charge, the net flow is zero. The Gauss law is nothing more than a repetition of Coulomb’s law. Coulomb’s law is readily obtained by applying the Gauss theorem to a point charge surrounded by a sphere.

### Applications of Gauss Law

- In a medium with a dielectric constant of K, the strength of the electric field near a plane charged conductor E = σ ⁄ K ε
_{0}. E_{air}= σ ⁄ ε_{0}when the dielectric medium is air. - At a distance of ‘r’ in the case of an infinite charge line, E = (1 ⁄ 4 × π r ε
_{0}) (2π ⁄ r) = λ ⁄ 2π r ε_{0}, where λ is linear charge density. - In a condenser, the field between two parallel plates is E = σ ⁄ ε
_{0}, where σ is the surface charge density. - The electric field strength near a plane sheet of charge is E = σ ⁄ 2K ε0, where σ is the surface charge density.

### Sample Problems

**Problem 1: In the x-direction, there is a homogeneous electric field of size E = 50 N⁄C. Calculate the flux of this field across a plane square area with an edge of 5 cm in the y-z plane using the Gauss theorem. Assume that the normal is positive along the positive x-axis.**

**Solution:**

Given:

Electric field, E = 50 N⁄C

Edge length of square, a = 5 cm = 0.05 m

The flux of the field across a plane square, ϕ = ∫ E cosθ ds

As the normal to the area points along the electric field, θ = 0.

Also, E is uniform so, Φ = E ΔS = (50 N⁄C) (0.05 m)

^{2}= 0.125 N m^{2 }C^{-1}.Hence, the flux of the given field is

0.125 N m.^{2 }C^{-1}

**Problem 2: How do we choose an appropriate Gaussian Surface for different cases?**

**Solution:**

In order to select an acceptable Gaussian Surface, we must consider the fact that the charge-to-dielectric constant ratio is supplied by a (two-dimensional) surface integral over the charge distribution’s electric field symmetry.

We’ll need to know about three potential scenarios.

- When the charge distribution is spherically symmetric, it is called spherical.
- When the charge distribution is cylindrically symmetric, it is called cylindrical.
- When the charge distribution exhibits translational symmetry along a plane, it is called a pillbox.
Depending on where we want to compute the field, we may determine the size of the surface. The Gauss theorem is useful for determining the direction of a field when there is symmetry, as it informs us how the field is directed.

**Problem 3: How to find the electric field using Gauss law?**

**Solution:**

Normally, the Gauss law is employed to calculate the electric field of symmetric charge distributions. When using this law to solve the problem of the electric field, there are numerous processes required. The following are the details:

- First, we must determine the charge distribution’s spatial symmetry.
- The next step is to select a proper Gaussian surface that has the same symmetry as the charge distribution. Its ramifications must also be determined.
- Calculate the flux across the surface by evaluating the integral ϕ
_{s}E over the Gaussian surface.- Calculate the amount of charge contained within the Gaussian surface.
- Calculate the charge distribution’s electric field.
However, in order to determine the electric field, pupils must remember the three forms of symmetry. The following are the several forms of symmetry:

- Symmetry on a sphere
- Symmetry in a cylindrical shape
- Symmetry on a plane

**Problem 4: There are three charges q1, q2, and q3 having charges 4 C, 7 C and 2 C enclosed in a surface. Find the total flux enclosed by the surface.**

**Solution:**

Total charge Q,

Q = q

_{1}+ q_{2}+ q_{3}= 4 C + 7 C + 2 C

= 13 C

The total flux, ϕ = Q ⁄ ε

_{0}ϕ = 13 C ⁄ (8.854×10

^{−12}F ⁄ m)ϕ = 1.468 N m

^{2}C^{-1}Therefore, the total flux enclosed by the surface is 1.584 N m

^{2}C^{-1}.

**Problem 5: What is the differential form of the Gauss theorem?**

**Solution:**

The electric field is related to the charge distribution at a certain location in space by the differential version of Gauss law. To clarify, according to the law, the electric field’s divergence (E) is equal to the volume charge density (ρ) at a given position. It’s written like this:

ΔE = ρ ⁄ ε

_{0}Here, ε

_{0}is the permittivity of free space.

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