Deduce Coulomb's law from Gauss theorum.

Asked by Aarushi Srivastava | 11th May, 2022, 08:19: PM

Expert Answer:

Gauss theorem states that total electric flux over a closed surface equals enclosed charge
divided by permittivity of enclosed medium.
 
Let us consider a sphere of radius R has charge at centre of sphere.
 
Then by Gauss theorem ,  begin mathsize 14px style surface integral E times d A space equals space q over epsilon end style  .....................(1)
Where E is uniform electric field intensity over the surface of sphere , q is enclosed charge
and ε is permittivity of medium
 
Since Electric field intensity E is uniform over the surface of sphere ,
we can write the RHS of eqn.(1) as
 
begin mathsize 14px style surface integral E times d A end style = E × 4π R2 ........................(2)
where 4π R2 is surface area of sphere.
 
Using eqn.(1) and (2) , we write
 
E × 4π R2 = ( q / ε )
 
Hence Electric field E is written as
 
begin mathsize 14px style E space equals space fraction numerator q over denominator 4 pi epsilon space R squared end fraction end style
 
Above expression equals electric field at a distance R from a charge q.
 
Since electric field is force / unit charge , force experienced by a test charge q' at a distance R
from the given charge q  is written as
 
begin mathsize 14px style F space equals space q apostrophe space E space equals space fraction numerator q apostrophe space q over denominator 4 pi epsilon space R squared end fraction end style
Above expression is Columbs law that gives electrical force between two charges q and q'
that are separated by a distance R

Answered by Thiyagarajan K | 11th May, 2022, 11:13: PM