Sunday, December 30, 2012

Gauss’s Theorem and its Proof


The surface integral of electrostatic field produce by any source over any closed surface S enclosing a volume V in vacuum i.e. total electric flux over the closed surface S in vacuum is 1/epsilon times the total charge Q contained inside S.
The surface chosen to calculate the surface integral is called Gaussian surface.


The flux through area element ΔS is
The unit vector rˆ is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and rˆ have the same direction. Therefore,

 Since the magnitude of a unit vector is 1. The total flux through the sphere is obtained by adding up flux through all the different area elements:

 Since each area element of the sphere is at the same distance r from the charge,


(i)                 Gauss’s law is true for any closed surface, no matter what its shape or size.

(ii)               The term q on the right side of Gauss’s law includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.

(iii)             In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law.

(iv)              Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface.

(v)                Finally, Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law. 


  1. Great notes! Thanks!

  2. Good evening Sir Me atul Rawat hun Ghaziabad wala
    aapke hi notes chiye the mujhe jo yaha pe mil gaye. Thanks a Lot.

  3. Thank you so much ��

  4. not to good....

  5. Prove is using surface as sphere(Special Case). General Proof-

  6. This comment has been removed by the author.

  7. why flux through a closed surface containing charge q is equal to q by E0 for any surface??