Can Coulomb's law be used to derive Gauss's law? How?

1 Answer
Mar 25, 2017

Answer:

I shall outline a better method. Not just by assuming the surface is spherical with area #4pir^2#.

So, to start with I shall prove it for the field due to a single point charge. (At O)
Since superposition principle applies to electric fields, this can be generalised to any number of charges quite easily.

There it is, given below.

Explanation:

Consider a charged particle #q# placed at O and is being surrounded and enclosed by a closed surface #S#. (Of arbitrary size and shape).

Take an arbitrary surface element #dS# on the surface at a distance #r# from O.

Let the normal on #dS# make an angle #theta# with the electric field of #q# passing through #dS#.

Thus, #vec E*dvecS = EdS Cos theta# which is equal to the flux through the small area.

Now, by Coulomb's law,

#E = q/(4piepsilon_0r^2)#

Which gives, #dphi = q/(4piepsilon_0)*dScos theta/(r^2)#

Now, #dS Cos theta/(r^2)# is the solid angle which can be denoted as, #domega#

Thus, the flux through #dS# is,

#dphi = q/(4piepsilon_0)*domega#

Integrating over the entire surface,

#int int_S vecE*dvecS = q/(4piepsilon_0) int domega#

But, the net solid angle subtended by a closed surface at an internal point is always #4pi#.

#int int_S vecE*dvecS = q/(4piepsilon_0)*4pi#

Thus,
#int int_S vecE*dvecS = q/epsilon_0#

Which is Gauss law in integral form.

One may derive the differential form from Coulomb's law as well by taking the Divergence of the electric field and then proceeding with vector calculus methods.

The differential form looks something like this, #nabla*vec E = rho/epsilon_0# where #rho# is the charge density.