Gauss’s Law states that the total electric flux out of a closed surface equals the charge contained inside the surface divided by the absolute permittivity. The electric flux in an area is defined as the electric field multiplied by the surface area projected in a plane perpendicular to the field. Now that we’ve established what Gauss law is, let’s look at how it’s used. Application of Gauss Law is important for Class 12 students.

In this article, our main focus is on the Application of Gauss Law with a brief discussion of Gauss Law.

Table of Content

## What is Gauss Law?

According to the Gauss law, the total flux linked with a closed surface is 1/ε_{0} times the charge enclosed by the closed surface.

** Example: **When a point charge

**is placed inside a cube of edge ‘a’. Then, the flux through each face of the cube is**

**q**

**q / 6ε****.**

_{0}## Applications of Gauss Law

We’ll look at a few of the applications of Gauss law in this article. To begin with, we know that in some situations, calculating the electric field is fairly difficult and requires a lot of integration. Gauss’s Law is used to make calculating the electric field easier.

Few applications of Gauss law involve finding:

- Electric field due to infinite straight wire.
- Electric field due to infinite plate sheet.
- Electric field due to thin spherical shell.

## Electric Field due to Infinite Wire

Consider a wire that is infinitely long and has a linear charge density λ. To compute the electric field, we utilize a cylindrical Gaussian surface. The flux through the end of the surface will be 0 since the electric field E is radial. Because the electric field and the area vector are perpendicular to each other, this is the case. We may argue that the electric field’s magnitude will be constant since it is perpendicular to every point on the curved surface.

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

According to Gauss’s Law:

**ϕ = q ⁄ ε**_{0}

**E × (2πrl) = λl ⁄ ε**_{0}

Hence, Electric Field due to Infinite Wire is given as

E = λ ⁄2πε_{0}r

It’s important to note that if the linear charge density is positive, the electric field is radially outward. If the linear charge density is negative, however, it will be radially inward.

## Electric Field due to Infinite Plane Sheet

Consider an infinite plane sheet with a cross-sectional area A and a surface charge density σ. The infinite plane sheet is in the following position:

The electric field generated by an infinite charge sheet is perpendicular to the sheet’s plane. Consider a cylindrical Gaussian surface whose axis is perpendicular to the sheet’s plane. Gauss’s Law allows us to calculate the electric field E as follows:

ϕ = q ⁄ ε_{0}

Charge q will be the charge density (σ) times the area (A) in continuous charge distribution. We shall only consider electric flow from the two ends of the imagined Gaussian surface when discussing net electric flux. We may explain it by the fact that the curved surface area and the electric field are perpendicular to each other, resulting in zero electric flux. As a result, the net electric flux:

**ϕ = E A − (− E A)**

**ϕ = 2 E A**

So,

**2 E A = σ A ⁄ ε**_{0}

Hence, Electric Field Due to Infinite Plain Sheet is given as

E = σ ⁄ 2 ε_{0}

The above formula shows that the electric field generated by an infinite plane sheet is independent of the cross-sectional area A.

## Electric Field due to Thin Spherical Shell

Consider a thin spherical shell with a radius “** R**”, charge

**, and a surface charge density of**

**q****(such that σ = q / 4 π r**

**σ**^{2}). The shell possesses spherical symmetry. The electric field owing to the spherical shell can be calculated in two ways:

- Outside the spherical shell
- Inside the spherical shell

**Electric Field Outside the Spherical Shell**

**Electric Field Outside the Spherical Shell**

Take a point P outside the spherical shell at a distance r from the center of the spherical shell to get the electric field. We use a Gaussian spherical surface with a radius of r and center ‘O’. Because all points are equally spaced “r” from the sphere’s center, the Gaussian surface will pass through P and experience a constant electric field E all around. So, Therefore, the total electric flux:

**ϕ = q ⁄ ε**_{0}** = E × 4 π r**^{2}

Charge contained inside the surface, q = σ × 4 π R^{2}

**E × 4 π r**^{2}** = σ × 4 π R**^{2}** ⁄ ε**_{0}

**E = σ R**^{2}** ⁄ ε**_{0}** r**^{2}

Electric field can also be written in the form of charge as (σ = q / 4 π r^{2})

Hence, Electric Field Outside Shpherical shell is given as

E = k q ⁄r^{2}

It’s important to keep in mind that if the surface charge density σ is negative, the electric field will be radially inward.

**Electric Field Inside the Spherical Shell**

**Electric Field Inside the Spherical Shell**

Let’s look at point P inside the spherical shell to see how the electric field there is. We may use symmetry to create a spherical Gaussian surface that passes through P, is centered at O, and has a radius of r. Now, based on Gauss’s Law,

ϕ = q ⁄ ε_{0}= E × 4πr^{2}

Since surface charge density is spread outside the surface, there is no charge contained inside the shell. Therefore, the electric field inside shell** **from the above formula is also zero, i.e.,

E = 0 (since q = 0)

**Also, Check**

## Examples on Application of Gauss Law

**Example 1: A hemispherical bowl of radius r is placed in a region of space with a uniform electric field E. Find out the electric flux through the bowl.**

**Solution:**

The surface area of the given bowl, dA = 2 π r

^{2}The field lines are parallel the axis of the plane of the bowl,i.e., θ = 0°

The electric flux, ϕ = E (dA) cosθ

= E (2 π r

^{2}) cos0°= E (2 π r

^{2})Hence, the electric flux through the bowl is

E (2 π r^{2}.)

**Example 2: How does the electric flow via the Gaussian surface vary if the radius of the Gaussian surface containing a charge is halved?**

**Solution:**

Even when the radius is half, the total charge contained by the Gaussian surface stays the same. As a result, according to Gauss’ theory, total electric flux remains constant. The electric flux will not vary as it passes through the Gaussian surface.

**Example 3: A cylindrical surface of radius r and length l, encloses a thin straight infinitely long conduction wire with charge density whose axis coincides with the surface. Find the formula for the electric flux through the cylinder’s surface.**

**Solution:**

A thin straight infinitely long wire has a uniform linear charge distribution. Consider the charge enclosed by the cylindrical surface be q.

Linear charge density, λ = q ⁄ l

Therefore, charge enclosed by the surface, q = λ l

The total electric flux through the surface of cylinder, ϕ = q ⁄ ε

_{0}= λ l ⁄ ε

_{0}Hence, the formula for electric flux through the cylinder’s surface is

λ l ⁄ ε._{0}

**Example 4: A charge of 2×10**^{-8}** C is distributed uniformly on the surface of a sphere of radius 2 cm. It is covered by a concentric, hollow conducting sphere of radius 5 cm. Find the electric field at a point 3 cm away from the center.**

**Solution:**

Let us consider the below figure.

Assume we need to locate the field at point P. P should be used to draw a concentric spherical surface. All of the points on this surface are comparable, and the field at each of them will be equal in magnitude and radial in direction due to symmetry.

The flux through this surface = ∮ E dS

= E ∮ dS

= E (4 π r

^{2})where r = 3 cm = 3 × 10

^{-2}mThis flux is equal to the charge q contained within the surface divided by ε

_{0}according to Gauss’ law.Thus,

E (4 π r

^{2}) = q/ε_{0}E = q ⁄ 4 π ε

_{0}r^{2}= ( 9 × 10

^{9}) × [(2 × 10^{-8})/(9 × 10^{-4})]= 2 × 10

^{5}N ⁄ CThe electric field at a point 3 cm away from the centre is

2 × 10^{5}.N ⁄ C

## FAQs on Application of Gauss Law

### 1. What is Gauss Law?

Gauss Law is one of the fundamental law of the electrostatics that states, “Total flux associated with a closed surface is 1/ε

_{0}times the charge enclosed by the closed surface.”

**2. What is the Application of Gauss law?**

**2. What is the Application of Gauss law?**

According to Gauss Law, the total electric flux coming out of a closed surface is equal to the charge enclosed inside the surface divided by the permittivity. Gauss Law is used to find the electric field of various objects.

**3. What are Real-Life Applications of Gauss law?**

**3. What are Real-Life Applications of Gauss law?**

Real-life applications of Gauss Law involve solving complex electrostatic problems with unique symmetries such as cylindrical, spherical, or planar symmetry

### 4. What is a Gaussian Surface?

Gaussian surface is the surface of the on which the Gauss Law is applied. It is the 3-D surface that is used for finding electric field and explain various things.

**5. Is Gauss Law Applicable to Non-Uniform Electric Field?**

**5. Is Gauss Law Applicable to Non-Uniform Electric Field?**

Yes, we can easily apply the Gauss law to non-uniform electric fields.