The Gauss law is used to study the electric flux and electric field in terms of its application. The point which is necessary to keep in mind is about to the surface. The surface is closed or open it matters for physical interpretation. Using some supporting physical concepts (SPCs) like electric flux, surface types and solid angle it is tried to explain the Gauss’s Law in Electrostatics.

\dpi{120} \fn_cm \large \phi_{E}=\oint\vec{E}.\vec{dS}

This is closed surface integral. Example of it you can see in the video tutorial 🙄 .

If I write it for the open surface then its mathematical form will be

\dpi{120} \fn_cm \large \phi_{E}=\iint\vec{E}.\vec{dS}

A circular ring is an example of an open surface.

Gauss’s Law in electrostatics in Hindi you can watch here.

Gauss’s Law

For the English just click below.

Gauss’s Law

Next point is about the conversion of Gauss Law from integral to the differential form. By using the Gauss-divergence theorem. This is called Maxwell’s first equation in the Physics (Electrodynamics).

Read Also: Displacement Current and Modified Ampere’s Law 

Example: What is electric flux through a closed surface surrounding a dipole?

Ans: As you know the electric dipole is made by two equal and opposite charge. One is +q and second is -q with some fixed distance. When the electric dipole is placed in a closed surface its total charge will be zero.

So, according to the Gauss Law

\dpi{120} \fn_cm \large \phi_{E}=\oint\vec{E}.\vec{dS}=\frac{Q_{encl}}{\epsilon_{0}} = \frac{0}{\epsilon_{0}}=0

Question: What is a Gaussian surface?

Ans: Gaussian surface is a hypothetical closed surface of any shape drawn in an electric field for the purpose of solving problems regarding electric flux. The shape of the Gaussian surface depends on the symmetry of the problem. The Gaussian surface is a closed surface, for an example football, or any other ball surface.

Question: Determine electric field intensity at a point due to the uniformly charged infinite non conducting sheet by applying the Gauss law.

Answer: An artistic view of this problem is shown in the picture. A non-conducting plane sheet of yellow color has uniformly distributed charge with surface charge density σ. The intensity due to this we have to calculate at the point P and P’ shown at both the sides of the plate. The distances of P and P’ from the sheet is l. Area of each plane at P is “a”. Electric field direction is perpendicular to the plane of the cylinder and the sheet.

Gauss law application
(Gauss law application)

So, the first point is about the charge enclosed by the cylindrical Gaussian surface, that is;

q= σ a            (1)

and now the total electric flux through the Gaussian surface is

\dpi{120} \fn_cm \large \phi=\oint\oint\vec{E}.\vec{dS}            (2)   (this is closed surface integral notation)

Now the question about the total surfaces (closed and open). Check here, two are open surfaces at the ends of the cylinder and one is closed. The closed one on the sheet marked by “a” in the figure.

So writing it

\dpi{120} \fn_cm \large \phi=\oint\oint_{A}\vec{E}.\vec{dS}+\iint_{B}\vec{E}.\vec{dS}+\iint_{C}\vec{E}.\vec{dS}\\                   (3) (total electric flux)


A = Closed loop area at the sheet

B & C= Open loop area at the ends of the cylinder.

See in the figure towards P point. The surface area of the Gaussian loop is perpendicular to the direction of the electric field. So the scalar product in between E and S will be zero because of the angle i.e. 900.

So the first term contribution in the above equation (3) it will be zero.

\dpi{120} \fn_cm \large \phi=\iint_{closed}\vec{E}.\vec{dS}=0\\ \vec{E}\perp \vec{dS} \texttt{at the surface}\\ \theta=90^{0}, cos\theta=0\\        (4)

hence the equation will be a result of the electric flux at the two ends of the cylinder, that is open loop. Also, an electric field is parallel to the area of the plane. this can be written as

\dpi{120} \fn_cm \large \phi=\iint_{open}\vec{E}.\vec{dS}=\iint E dS\\ \vec{E}\parallel \vec{dS} \texttt{at both the ends}\\ \theta=0^{0}, cos\theta=1\\ EdScos\theta=EdS\\ \texttt{sum of these two terms}\\ \phi=E\iint dS \texttt{ ;E is constant at this point and }\\

The area is defined by the “a” at both ends for this plane.

\dpi{120} \fn_cm \large E\iint dS= Ea\\ \phi=Ea +Ea= 2Ea\\     ……………..(5)

Now put the value of phi and q in Gauss law from eqn (1) and (5).

\dpi{120} \fn_cm \large \phi= \frac{q}{\epsilon_{0}}\\ 2Ea= \frac{\sigma a }{\epsilon_{0}}\\ E= \frac {\sigma}{2\epsilon_{0}}

This relation shows that Electric field intensity is independent of the distance from the non-conducting sheet.

When you solve the numerical problems based on the surface area first define and verify that it is open loop or closed loop and the direction of the electric field. If it is parallel then theta angle in between the electric field and surface area will be zero. If it is perpendicular then it will be ninety degrees.

Physics Objective Questions:

1Question: The electric field due to an infinite sheet is

  1. dependent on the distance of a point of observation
  2. dependent on the location of points of observation
  3. independent of the location of a point of observation
  4. continuous at every point

2Q. The electric field below oppositely charged plates having equal charges density σ is given by

  1.    \dpi{120} \fn_cm \large \frac {\sigma}{\epsilon_{0}}\\
  2. \dpi{120} \fn_cm \large \frac {\sigma}{2\epsilon_{0}}\\
  3. \dpi{120} \fn_cm \large zero
  4. \dpi{120} \fn_cm \large \frac {2\sigma}{\epsilon_{0}}\\

3Q. The volume charge density inside a conductor is

  1. finite
  2. infinite
  3. dependent on the shape of a conductor

4Q. The position of given charges inside the Gaussian surface is changed in such a way that total charge remains constant. Then normal electric flux through the Gaussian surface

  1. Increases
  2. decreases
  3. remains unchanged
  4. changes gradually