Question

State gauss law in electrostatics.

Answer 1

Hint: Gauss law was published in the year 1867 as a part of the work done by the German Mathematician Carl Gauss. Surprisingly it was published after the death of the German Mathematician.

Complete step by step solution:
Gauss Law is given as the total electric flux passing over a closed surface in vacuum is $1/{\varepsilon _o}$ times the total charge inside the surface.
${\phi _{net}} = \oint {\overrightarrow E .\overrightarrow {ds} } = \dfrac{{{q_{in}}}}{{{\varepsilon _o}}}$ ;
Here:
E = Electric field;
q = charge;
${\varepsilon _o}$= Permittivity of free space;
ds = Small area;
${\phi _{net}}$= net flux;
To know the strength of a field passing through a surface we need to know its flux. In other words flux tells us about the strength of the field and electric flux density can be considered as Electric Field. As per the above formula the net flux is equal to the net electric field inside a closed surface “${\phi _{net}} = \oint {\overrightarrow E .\overrightarrow {ds} } $”. The electric flux density tells us the number of electric field lines passing through a given area. If there are a greater number of electric field lines passing through a closed surface then the electric field is denser and if there are a smaller number of electric field lines passing through a particular area then the electric field is less dense.

Note: By the help of gauss law we can calculate the electric flux through any surface and hence we can get an idea of electric field passing through that particular surface. There are few conditions that are required to be fully filled while applying gaussian surface to find the flux.