Gauss’s law of electrostatics to find electric field

In other articles, we learned that an electric charge can produce an electric field around it. Then we studied its effects and properties. Also, we became to know that charges feel Coulomb’s force due to the existence of the electric field. Now, we need a formula or equation to find the electric field produced by the charged conductors around them. Gauss’s law of electrostatics is that kind of law which can be used to find the electric field due to charged conductors like sphere, wire, and plate. One can also use the Coulomb’s law in this purpose. But use of Gauss’s law makes the calculation easy. In this article, we are going to discuss the Gauss’s law, its formula or equation and applications.

Contents in 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 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.

Formula of Gauss’s law

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 the Gauss’s law. It is the integral form of the Gauss’s law. 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 the 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, the 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 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 bound volume charge. There will be no bound surface charge in a Gaussian surface inside a dielectric.

Then from the 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 the Gauss’s law inside a dielectric medium.

Derivation of Gauss’s law from Coulomb’s law

Let a test charge q1 is 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 total charge enclosed by the surface. This is nothing but the Gauss’s law in electrostatics.

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

Applications of Gauss’s law

The main purpose of the Gauss’s law in electrostatics is to find the electric field for different type 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

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 Sphere, cylinder, straight wire, plane sheet, etc. To use this law all conductors should have some charge inside it. One can use Gauss’s law to find electric field due to a point charge, but this law cannot be used to find electric field for an electric dipole and other irregular shaped conductors. To find the electric field for an electric dipole we need to use the Coulomb’s law.

This is all from this article. If you have any question on this topic you can ask me in the comment section.

Thank you!

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