Gauss Law is one of the most interesting topics that engineering aspirants have to study as a part of their syllabus. There is an immense application of Gauss Law for magnetism. In addition, an important role is played by Gauss Law in electrostatics. Examiners often ask students to state Gauss Law. Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. 

Even if students are asked to state the Gauss theorem, students should know that the theorem states that the net flux through a closed surface is directly proportional to the net charge in the volume enclosed by the closed surface. This is represented by the Gauss Law formula: ϕ = Q/ϵ0, where, Q is the total charge within the given surface, and ε0 is the electric constant. Read the article for numerical problems on Gauss Law.

What is Gauss’s Law?

According to Gauss’s law, the total electric flux through a closed Gaussian surface is equal to the total charge enclosed by the surface divided by the \(ε_0\) (permittivity).
\({\phi _{{\text{closed}\;\rm{surface}}}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
Example: Let us consider a system of charge \(q_1,\;q_2,\;Q_1,\;Q_2.\)

Let the electric field due to these charges at the differential area element be, \(\overrightarrow {{E_1}} ,\,\overrightarrow {{E_2}} ,\,\overrightarrow {E'{_1}} ,\,\overrightarrow {E'{_2}} \)
\(\phi = \oint {\overrightarrow E } \cdot \overrightarrow {{\rm{d}}s}\)
\(\phi = \oint {(\overrightarrow {{E_1}} + } \overrightarrow {{E_2}} + \overrightarrow {E'{_1}} + \overrightarrow {E'{_2}} ) \cdot \overrightarrow {{\rm{d}}s}\)
\(\phi = \oint {\overrightarrow {{E_1}} \cdot \overrightarrow {{\rm{d}}s} + } \oint {\overrightarrow {{E_2}} } \cdot \overrightarrow {{\rm{d}}s} + \oint {\overrightarrow {E'{_1}} } \cdot \overrightarrow {{\rm{d}}s} + \oint {\overrightarrow {E'{_2}} } \cdot \overrightarrow {{\rm{d}}s}\)
According to the Gauss’s law, the value of this integral is equal to the net charge enclosed by the surface divided by the \(ε_0\) (permittivity).
\(\phi = \oint {\overrightarrow {{E_1}} \cdot \overrightarrow {{\rm{d}}s} + } \oint {\overrightarrow {{E_2}} } \cdot \overrightarrow {{\rm{d}}s} + \oint {\overrightarrow {E'{_1}} } \cdot \overrightarrow {{\rm{d}}s} + \oint {\overrightarrow {E'{_2}} } \cdot \overrightarrow {{\rm{d}}s} = \frac{{{q_1} + {q_2}}}{{{\varepsilon _0}}}\)

Electric Flux

Electric flux can be defined as the total amount of electric field lines (amount of electric field) passing through the given area. It is denoted by \(\phi\).
Example: Flux the electric field \(\overrightarrow {E}\) through the given area \(\overrightarrow {S}\) is defined as,
Electric flux \( = \overrightarrow E \cdot \overrightarrow S = \left| {\overrightarrow E } \right|\left| {\overrightarrow S } \right|\cos \left( \theta \right)\)
For non-uniform Electric field,
\(\phi = \int {\overrightarrow E } \cdot \overrightarrow {{\rm{d}}s} \)
It is a scalar quantity, and its unit is \(\rm{N}\;\rm{m}^2 \;\rm{C}^{-1}\) or \(\rm{V}\;\rm{m}\) (volt metre)
In many cases, Gauss’s law can be used to easily find the electric flux through a surface and avoid mathematical complexities.
Example: If a charge is inside a cube at the centre, then, mathematically calculating the flux using the integration over the surface is difficult but using the Gauss’s law, we can easily determine the flux through the surface to be, \(\frac{q}{{{\varepsilon _0}}}.\)

Electric Field Lines

Electric field lines or electric lines of force is a hypothetical concept which we use to understand the concept of Electric field.
We have the following rules, which we use while representing the field graphically.
1. Tangent drawn at any point on a field line gives the direction field at that point.
2. It emerges from a positive charge and sinks into a negative charge.

3. It can be a straight line or a curved line.

4. It cannot be a closed curve.
Electric field lines cannot be closed lines because they cannot emerge and sink from the same point.
5. Two electric field lines cannot intersect.
This is because the electric field line also represents the direction of the electric field lines and at a particular point. There can be only one direction of the electric field and if it intersects then it will mean that there is a two-direction thus, electric field lines cannot intersect.

6. Number of the electric field lines that emerge or sink from a charge is proportional to the magnitude of the charge.

Application of Gauss’s Law

As the Gauss’s Law is related to the number of electric field lines, electric flux and electric field, in many cases, we can use the Gauss’s law to solve the problems related to these concepts.
Electric field due to a point charge.
Let us consider a hollow sphere as a Gaussian surface with the point charge \(q\) at the centre of the sphere.
Now from Gauss’s law we have,
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E \cdot \oint {{\rm{d}} s}\)
\( \Rightarrow \frac{q}{{{\varepsilon _0}}} = E \cdot 4\pi {r^2}\)
\( \Rightarrow E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}\)
Electric field due to uniformly charged hollow sphere or shell of radius \(R\).
• Outside or on the boundary of the shell
\(r \geqslant R\)
Let us consider a concentric hollow sphere to be a Gaussian surface with radius \(r > R.\)
Now from Gauss’s law, we have,
\( \phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
The whole charged shell is enclosed by the Gaussian surface.
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E \cdot \oint {{\rm{d}}s} \)
\( \Rightarrow \frac{q}{{{\varepsilon _0}}} = E \cdot 4\pi {r^2}\)
\( \Rightarrow E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}\)
• Inside the shell
\(r < R\)
Let us consider a concentric hollow sphere to be a Gaussian surface with radius \(r < R\).
Now from Gauss’s law, we have,
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
The charge enclosed by the shell is zero.
Therefore, the electric field inside the shell will be zero.

Electric field due to solid charged sphere.
• Outside or the boundary of the sphere
\(r ≥ R\)
Let us consider a concentric hollow sphere to be a Gaussian surface with radius \(r > R\).
Now from Gauss’s law, we have,
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
The whole charged sphere is enclosed by the Gaussian surface.
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E \cdot \oint {{\rm{d}} s} \)
\( \Rightarrow \frac{q}{{{\varepsilon _0}}} = E \cdot 4\pi {r^2}\)
\( \Rightarrow E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}\)
• Inside the sphere
\(r < R\)
Let us consider a hollow sphere to be a Gaussian surface with radius \(r < R\) and concentric with the given sphere.
Now from Gauss’s law, we have,
\( \Rightarrow \phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
Now, since the sphere is uniformly charged, the volume charge density is given to be,
\(\rho = \frac{q}{{4\pi {R^2}}}\)
Therefore, the charge enclosed by the Gaussian surface will be,
\({q_{{\text{enclosed}}}} = \frac{{3q}}{{4\pi {R^3}}} \times \frac{4}{3}\pi {r^3}\)
\(\Rightarrow \phi = \frac{{\frac{q}{{{R^3}}}{r^3}}}{{{\varepsilon _0}}} = E.\oint {{\rm{d}}s}\)
\( \Rightarrow \frac{{\frac{q}{{{R^3}}}{r^3}}}{{{\varepsilon _0}}} = E.4\pi {r^2}\)
\( \Rightarrow E = \frac{{qr}}{{4\pi {\varepsilon _0}{R^3}}}\)
Electric field due to uniformly charged infinite long hollow cylinder
• Outside the cylinder
\(r ≥ R\)
Let us consider a hollow cylinder of length \(l\) and radius \(r > R\), to be the Gaussian surface which is concentric with the given cylinder.
From Gauss’s law, we have,
\( \phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E\oint { \cdot {\rm{d}}s} \)
As the cylinder is infinitely long, electric field will be along radial direction only, and hence flux through upper and lower caps of the Gaussian surface will be zero.
\( \Rightarrow \frac{{\lambda l}}{{{\varepsilon _0}}} = E.2\pi rl\)
\( \Rightarrow E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}\)
Also, if instead of the hollow cylinder we have a charged thread the expression for Electric field remains same.
Electric field due to a uniform charged thread is,
\(E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}\)
• Inside the cylinder
\(r < R\)
If the Gaussian surface is inside of the hollow charged cylinder the net charge enclosed by it is zero. Therefore, electric field will be zero as there are no other charge in the system.
Electric field due to uniformly charged infinite solid cylinder
• Outside the cylinder
\(r ≥ R\)
Let us consider a solid cylinder to be the Gaussian surface which is concentric with the given cylinder.
From Gauss’s law, we have,
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E \cdot \oint {{\rm{d}}s} \)
As the cylinder is infinitely long, an electric field will be along radial direction only, and hence flux through upper and lower caps of the Gaussian surface will be zero.
\( \Rightarrow \frac{{\rho \pi {R^2}l}}{{{\varepsilon _0}}} = E.2\pi rl\)
\( \Rightarrow E = \frac{{\rho {R^2}}}{{2{\varepsilon _0}r}}\)
• Inside the cylinder
\(r < R\)
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E \cdot \oint {{\rm{d}}s} \)
As the cylinder is infinitely long, an electric field will be along radial direction only, and hence flux through upper and lower caps of the Gaussian surface will be zero.
\( \Rightarrow \frac{{\rho \pi {r^2}l}}{{{\varepsilon _0}}} = E.2\pi rl\)
\( \Rightarrow E = \frac{{\rho r}}{{2{\varepsilon _0}}}\)
Electric field due to uniformly charged plate
• Outside of the plate
\(x ≥ d\)
Let us consider a cylinder of height \(2x\) as a Gaussian surface which passes horizontally through the charged plate and whose lateral mid-point is the same as the mid-point of the plate. As shown in the figure.
From Gauss’s law, we have,
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E \cdot \oint {{\rm{d}}s} \)
\( \Rightarrow \frac{{\rho Sd}}{{{\varepsilon _0}}} = 2ES\)
\( \Rightarrow E = \frac{{\rho d}}{{2{\varepsilon _0}}}\)
• Inside of the plate
\(x < d\)
Let us consider a cylinder of height \(2x\) as a Gaussian surface which passes horizontally through the charged plate and whose lateral mid-point is the same as the mid-point of the plate. As shown in the figure.
From Gauss’s Law,
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
\( \Rightarrow \phi = \frac{q}{{{\varepsilon _0}}} = E \cdot \oint {{\rm{d}}s} \)
\( \Rightarrow \frac{{\rho S2x}}{{{\varepsilon _0}}} = 2ES\)
\( \Rightarrow E = \frac{{\rho x}}{{{\varepsilon _0}}}\)

Sample Problems

Q.1: In the given figure, we have a uniformly charged cylinder of radius ‘\(a\)’ with a cylindrical cavity of radius ‘\(b\)’ located at the distance ‘\(c\)’ from the centre as shown in the figure. Find the electric field inside the cavity.

Ans: We know that the electric field inside a cylindrical charged body is given by,
\( \Rightarrow E = \frac{{\rho r}}{{2{\varepsilon _0}}}\)
Let us consider a point ‘\(P\)’ inside of the cavity which is at a distance \(x\) from the centre of the solid cylinder and da distance of \(y\) from the centre of the cavity.
Electric field at the point ‘\(P\)’ will equal to the difference of, the electric field at ‘\(P\)’ due to complete solid cylinder and the electric field due to solid cylinder of radius equal to that of the cavity and same location.
\( \Rightarrow \overrightarrow E = \frac{{\rho \overrightarrow x }}{{2{\varepsilon _0}}} – \frac{{\rho \overrightarrow y }}{{2{\varepsilon _0}}}\)
\( \Rightarrow \overrightarrow E = \frac{\rho }{{2{\varepsilon _0}}}\left( {\overrightarrow x – \overrightarrow y } \right)\)

From figure we have,
\(\left( {\overrightarrow x – \overrightarrow y } \right) = \overrightarrow c \)
Therefore, the electric field inside the cavity will be,
\(\overrightarrow E = \frac{\rho }{{2{\varepsilon _0}}}\overrightarrow c .\)

Q.2: Electric field in space is given by, \(\overrightarrow E = {E_0}{x^2}\widehat i\). Find the charge enclosed by the cube of side ‘\(l\)’ kept at \(x = l\) to \(x = 2l\) as shown in the figure.

Ans: We know that from Gauss’s law, the flux through a closed surface is given by,
\(\phi = \oint {\overrightarrow E \cdot \overrightarrow {{\rm{d}}s} } = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
\(\Rightarrow {q_{{\text{enclosed}}}} = {\varepsilon _0}\phi\)
Electric field at \(x = l\) is \(\overrightarrow {{E_l}} = {E_0}{l^2}\widehat i\)
Electric field at \(x = 2l\) is \(\overrightarrow {{E_{2l}}} = 4{E_0}{l^2}\widehat i\)
Area of the face through which the electric field will cross is \(l^2\)
Flux through the face located at \(x = l\) is,
\({\phi _l} = – {E_0}{l^4}\)
Flux through the face located at \(x = 2l\) is,
\({\phi _{2l}} = 4 {E_0}{l^4}\)
The net flux is the sum of the two fluxes,
\({\phi_{{\text{net}}}} = {\phi _l} + {\phi_{2l}} = 3{E_0}{l^4}\)
Therefore, the charge enclosed by the surface is,
\({q_{{\text{enclosed}}}} = {\varepsilon _0}\phi \)
\( \Rightarrow {q_{{\text{enclosed}}}} = 3{\varepsilon _0}{E_0}{l^4}.\)

Summary

In the above article, we learned that Gauss’s law for electrostatics is one of the four equations that govern electromagnetics, and it states that the total electric flux through a closed Gaussian surface is equal to the total charge enclosed by the surface divided by the permittivity \(({\varepsilon _0}).\)
\({\phi_{{\text{closed}\;\text{surface}}}} = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}}\)
Electric flux is the amount of electric field passing through the given surface.
Electric field lines are a hypothetical concept that is introduced to analyse the electric field.
We can use Gauss’s law to determine the value of an electric field in various cases.

Frequently Asked Questions on Gauss Law

Q.1: What is Gauss’s Law?
Ans: Gauss’s Law for electrostatics states that the electric flux through any closed surface is equal to the charge enclosed by the closed surface divided by the permittivity of the space.
We also have Gauss’s law for magnetics which states that magnetic flux through any closed surface will be zero.

Q.2: From where do the electric field lines emerge, and where do they sink?
Ans: Electric field lines emerge from a positive charge and sink into negative charges.

Q.3: What is electric flux?
Ans: Electric flux through a surface is equal to the amount of electric field passing through it.
Electric flux \(\phi = \overrightarrow E \cdot \overrightarrow s .\)

Q.4: Why do we have zero electric fields inside a charged shell?
Ans: If we consider a hollow sphere inside of the shell as a Gaussian surface, then the net charge enclosed by the surface is zero and since there is point symmetry the magnitude of the electric field must be the same at all the points, therefore, the only way this is possible is to have the magnitude of the electric field to be zero.

Q.5: Is Gauss’s law valid for any surface?
Ans: Yes, Gauss’s law is valid for any surface, but we cannot verify it for each and every surface due to mathematical constraints.

PRACTICE QUESTIONS ON ELECTROSTATIC FORCE & COULOMB’S LAW HERE

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