Gauss’s law and its applications-(2)

Gauss’s law:-

Statement:- Gauss’s law states that the total flux coming out of a closed surface is equal to the net charge enclosed by that surface.

i.e, \oint_{s}\overrightarrow{D}.\overrightarrow{ds}= Q_{enclosed}      —–>               i.e, \psi _{e} = Q_{enclosed}


Consider a closed surface of any shape with charge distribution as shown in the given figure. Now this surface is enclosed by a charge of Q coulombs , then a flux of Q coulombs will pass through the enclosing surface.

Now divide the entire area S into small pieces of differential areas ds or Δs.

Now at point P , the flux density is Ds and is direction is as shown in the figure. The surface S is chosen is an irregular surface and \overrightarrow{D} is also not uniform. i.e, \overrightarrow{D} direction as well as it’s magnitude is going to change from point to point.

now for the surface area ds the normal vector to the surface is \overrightarrow{ds_{n}}= ds.\overrightarrow{a_{n}}the direction of \overrightarrow{D} at a point P on the surface ds is making an angle θ \overrightarrow{ds_{n}}, then flux density at point ‘P’ is D = \frac{d\psi }{ds}

d\psi = D.ds—>equation 1

to get maximum flux out of the surface \overrightarrow{D} and \overrightarrow{ds_{n}} should be in the same direction, there is a need to find out the component of \overrightarrow{D} along \overrightarrow{ds_{n}} is D_{s}cos\theta = D_{s normal}

then equation 1 becomes d\psi = D_{snormal} ds

d\psi =D _{s}cos\theta ds —> d\psi = \overrightarrow{D}.\overrightarrow{ds}

Total flux is \psi = \oint_{s}d\psi =\oint_{s}\overrightarrow{D}.\overrightarrow{ds}

\therefore \psi =\oint_{s}\overrightarrow{D}.\overrightarrow{ds}

∴ Total flux \psi = net charge enclosed Q

\therefore Q _{enclosed}=\oint_{s}\overrightarrow{D}.\overrightarrow{ds}

If there are n number of charges Q1, Q2,Q3 …..Qn then Q= \sum Q_{n}

i. for a line charge distribution Q=\int_{l}\rho _{l}dl                                                                      ii. for a surface charge distribution Q=\int_{s}\rho _{s}ds                                                            iii. for a volume charge distribution Q=\int_{v}\rho _{v}dv.

Closed Gaussian surface:-

The Gauss’s  law is used to find out  \overrightarrow{E} or \overrightarrow{D} for symmetrical charge distributions and is used to find out \psi or Q of any closed surface.

  1. \overrightarrow{D} is every where either normal (or) tangential to the closed surface, that is θ = π/2 or θ = 0o / π so that \overrightarrow{D}.\overrightarrow{ds} is maximum or zero.
  2. \overrightarrow{D} is constant over the portion of the closed surface for which \overrightarrow{D}.\overrightarrow{ds} is not zero.

To apply Gauss’s law to a charge distribution, surface is required to be a Gaussian surface that encloses the  charge distribution or charged body.

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.