# Gauss’s law of electrostatics – formula & derivation

Electrostatics Lecture – 6:

In the third article on electrostatics, we became to know that an electric charge can produce an electric field around it. Then we studied its properties and other things related to it. Since there are various types of charge distribution in different conductors, the formula for the electric field will be different for those. Gauss’s law of electrostatics is that kind of law that can be used to find the electric field due to symmetrically charged conductors like spheres, wires, and plates. One can also use Coulomb’s law for this purpose. But the use of Gauss’s law formula makes the calculation easy. In this article, I’m going to discuss the Gauss law formula, its derivation and applications.

Contents of this article:

1. Statement of Gauss’s law
2. Formula of Gauss’s law
3. Differential form of Gauss’s law
4. Gauss’s law in Dielectric medium
5. Proof of Gauss’s law formula from Coulomb’s law
6. Applications of Gauss’s law
7. Limitation of Gauss’s law

## Statement of Gauss’s law

Gauss’s law in electrostatics states that the electric flux passing through a closed surface is equal to the $\small \frac{1}{\epsilon _{0}}$ times total charge enclosed by the surface. Gauss’s law gives the expression for electric field for charged conductors. This law has a wide use to find the electric field at a point.

## Gauss law formula

Let a closed surface is containing q amount of charge inside it. Now, according to Gauss’s law of electrostatics, total electric flux passing through the closed surface is, $\small \phi =\frac{q}{\epsilon _{0}}$ ……………(1)

Now, the electric flux through a surface S in the electric field E is, $\small \phi =\oint \vec{E}.d\vec{S}$…………..(2)

Then from equation-(1) and equation-(2) we get, $\small \oint \vec{E}.d\vec{S}=\frac{q}{\epsilon _{0}}$………….(3)

This is the equation or formula for Gauss’s law. It is the integral form of Gauss’s law equation. Using this formula one can find the electric field for symmetrically charged conductors.

## Differential form of Gauss’s law

We have the integral form of Gauss’s law as $\small \oint \vec{E}.d\vec{S}=\frac{q}{\epsilon _{0}}$

Now, if $\small \rho$ be the volume charge density then charge, $\small q=\int \rho dV$

Again, from Gauss’s divergence theorem, $\small \int \vec{E}.d\vec{S}=\int \triangledown .\vec{E} dV$

Then equation-(3) can be written as, $\small \int \triangledown .\vec{E} dV = \int \frac{\rho }{\epsilon _{0}}dV$

or, $\small \int [\triangledown .\vec{E} - \frac{\rho }{\epsilon _{0}} ]dV = 0$

or, $\small \triangledown .\vec{E} - \frac{\rho }{\epsilon _{0}} = 0$

Then, $\small \triangledown .\vec{E} = \frac{\rho }{\epsilon _{0}}$ ………………..(4)

This is the differential form of Gauss’s law of electrostatics.

## Equation of Gauss law inside a Conductor

The electric field inside a conductor is zero. Because the net electric charge inside the conductor becomes zero. Hence, no electric flux is enclosed inside the conductor. Therefore, Gauss’s law inside a conductor can be written as, $\small \phi =\oint \vec{E}.d\vec{S}=0$

## Formula of Gauss’s Law in Dielectric medium

There can be two types of charges inside a dielectric medium – free charges and bound charges. If qf and qb be the total free charge and bound charge respectively, then the total charge inside a dielectric medium is, q = qf + qb.

Again, if P be the polarization vector, then bound charge, qb= – $\small \int \vec{P}.d\vec{S}$. This is the bound volume charge. There will be no bound surface charge in a Gaussian surface inside a dielectric.

Then from Gauss’s law in equation-(3) we get, $\small \oint \vec{E}.d\vec{S} =\frac{q}{\epsilon _{0}}$

or, $\small \oint \vec{E}.d\vec{S} =\frac{q_{f} - \int \vec{P}.d\vec{S}}{\epsilon _{0}}$

or, $\small \int [\epsilon _{0}\vec{E} + \vec{P}] .d\vec{S}$ = qf ………..(5)

Now, if $\small \rho _{f}$ be the density of the free charge then, $\small q_{f} =\int \rho _{f} dV$ ………..(6)

And from Gauss’s divergence theorem, $\small \int [\epsilon _{0}\vec{E} + \vec{P}] .d\vec{S}$ = $\small \int \triangledown .[\epsilon _{0}\vec{E}+\vec{P}]dV$ …………….(7)

Now, putting equation-(6) and (7) in the equation-(5) we get, $\small \int \triangledown .[\epsilon _{0}\vec{E}+\vec{P}]dV$=$\small \int \rho _{f} dV$

or, $\small \int [\triangledown .(\epsilon _{0}\vec{E}+\vec{P}) -\rho _{f}] dV$=0

or, $\small \triangledown .(\epsilon _{0}\vec{E}+\vec{P}) -\rho _{f}$=0

Hence, $\small \triangledown .(\epsilon _{0}\vec{E}+\vec{P}) =\rho _{f}$ ……………..(8)

Now, we introduce a new physical quantity, Displacement vector, $\small \vec{D} = \triangledown .(\epsilon _{0}\vec{E}+\vec{P})$

Then we can write, $\small \triangledown .\vec{D} =\rho _{f}$ …………………(9)

This is the formula or equation for Gauss’s law inside a dielectric medium.

## Gauss law derivation from Coulomb’s law

Let a test charge q1 be placed at r distance from a source charge q. Then from Coulomb’s law of electrostatics we get,

The electrostatic force on the charge q1 due to charge q is, $\small F=\frac{qq_{1}}{4\pi \epsilon _{0}r^{2}}$

Thus, the electric field at the position of q1 due to the charge q is, $\small E=\frac{q}{4\pi \epsilon _{0}r^{2}}$

Now, we imagine a closed spherical surface of radius r around the source charge q. Then the charge enclosed by the surface is q.

Now, the electric flux through the entire spherical surface is, $\small \phi =\oint \vec{E}.d\vec{S}$

or, $\small \phi =ES$

or, $\small \phi = \frac{q}{4\pi \epsilon _{0}r^{2}}.4\pi r^{2}$ = $\frac{q}{ \epsilon _{0}}$

Thus, electric flux through the closed surface is equal to the $\small \frac{1}{\epsilon _{0}}$ times the total charge enclosed by the surface. This is nothing but Gauss’s law in electrostatics.

In this way, one can derive Gauss’s law from Coulomb’s law.

## Applications of Gauss’s law

The main purpose of Gauss’s law in electrostatics is to find the electric field for different types of conductors. This law can be used to find the electric field for a

• point charge
• Uniformly charged infinitely long wire
• charged spherical conductor
• cylindrical conductor
• infinitely charged plane
• Inside a capacitor

Also, one can find the electric flux through a closed surface by using this law. In this case, the total charge inside the surface should be known.

## Limitation of Gauss Law

The limitations of Gauss law are as followings –

• One cannot find the electric field for any arbitrary conductor. Gauss’s law is useful for calculating electric fields only for some symmetrically charged conductors like spheres, cylinders, straight wires, plane sheets, etc.
• To use this law all conductors should have some charge inside them.
• One can use Gauss’s law to find the electric field due to a point charge, but this law cannot be used to find the electric field for an electric dipole and other irregularly shaped conductors. To find the electric field for an electric dipole we need to use Coulomb’s law.

This is all from this article on the Derivation of Gauss’s law formula in electrostatics. If you have any questions on this topic you can ask me in the comment section. Click on the Next Article button to read an article on Electrostatic Potential.

Thank you!

Related Posts:

1. Coulomb’s law of electrostatics
2. Electrostatic Potential
3. Electrostatic Potential energy
4. Electric field and electric field intensity

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