# Gauss’s law and its applications

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

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 $\overrightarrow{D}$ 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 $\overrightarrow{ds_{n}}=&space;ds.\overrightarrow{a_{n}}$the direction of $\overrightarrow{D}$ at a point P on the surface ds is making an angle θ w.r.to $\overrightarrow{ds_{n}}$, then flux density at point ‘P’ is $D&space;=&space;\frac{d\psi&space;}{ds}$

$d\psi&space;=&space;D.ds$—>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 $D_{s}cos\theta&space;=&space;D_{s&space;normal}$

then equation 1 becomes $d\psi&space;=&space;D_{snormal}&space;ds$

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

Total flux is $\psi&space;=&space;\oint_{s}d\psi&space;=\oint_{s}\overrightarrow{D}.\overrightarrow{ds}$

$\therefore&space;\psi&space;=\oint_{s}\overrightarrow{D}.\overrightarrow{ds}$

∴ Total flux $\psi$ = net charge enclosed Q

$\therefore&space;Q&space;_{enclosed}=\oint_{s}\overrightarrow{D}.\overrightarrow{ds}$

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

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

Closed Gaussian surface:-

The Gauss’s  law is used to find out  $\overrightarrow{E}&space;or&space;\overrightarrow{D}$ for symmetrical charge distributions and is used to find out $\psi&space;or&space;Q$ 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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