# 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,       —–>               i.e,

Proof:- 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  is also not uniform. i.e,  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 the direction of  at a point P on the surface ds is making an angle θ w.r.to , then flux density at point ‘P’ is

—>equation 1

to get maximum flux out of the surface  and  should be in the same direction, there is a need to find out the component of  along  is

then equation 1 becomes

—>

Total flux is

∴ Total flux  = net charge enclosed Q

If there are n number of charges Q1, Q2,Q3 …..Qn then

i. for a line charge distribution                                                                       ii. for a surface charge distribution                                                             iii. for a volume charge distribution .

Closed Gaussian surface:-

The Gauss’s  law is used to find out   for symmetrical charge distributions and is used to find out  of any closed surface.

1.  is every where either normal (or) tangential to the closed surface, that is θ = π/2 or θ = 0o / π so that  is maximum or zero.
2.  is constant over the portion of the closed surface for which  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.     (1 votes, average: 5.00 out of 5) Loading... 