Gauss theorm: its definition , verificstion and application

Asked by rubyjha6650 | 16th Feb, 2020, 12:59: PM

Expert Answer:

Gauss theorem states that total electric flux over a closed surface equals the charge enclosed by the closed surface
divided by permitivity of the enclosed medium. Mathematically it is stated as
 
begin mathsize 14px style surface integral E times d A space equals space q over epsilon end style
where E is electric field and E•dA is the electric flux passing through area dA , q is the enclosed charge and
ε is the permitivity of enclosed medium
 
verification :-
Let us assume a charge +q is at centre of an sphere of radius R. 
 
Electric field at P  =  q / [ 4πεo R2 ]
 
Due to spherical symmetry, electric field will be same at any point on the surface of sphere.
 
Hence electrcic flux over the closed surface of sphere is obtained as
 
begin mathsize 14px style surface integral E times d A space equals space surface integral fraction numerator q over denominator 4 pi epsilon subscript o R squared end fraction space d A space equals fraction numerator q over denominator 4 pi epsilon subscript o R squared end fraction surface integral space d A space equals space fraction numerator q over denominator 4 pi epsilon subscript o R squared end fraction cross times 4 pi R squared space equals space q over epsilon subscript o end style
Hence gauss theorem is verified
---------------------------------------------------------------
 
Application
 
Gauss theorem is used to get electric field when electric field is uniform over the closed surface
so that surface integration is performed easily.
 
Let us consider an infinite rod having linear charge density ( charge per unit length ) λ.
Due to its infinite length, electric field near the rod will be uniform and direction of the field is normal to the rod.
 
let us consider an imaginary cylinder of radius r and length l is surrounded part of the rod length l.
Let us assume the cylinder is symmetrically placed with respect to rod as shown in figure.
 
By Gauss law, total flux over the closed surface of cylinder = E × ( 2πr l ) = λl / εo
 
Hence electric field near rod of infinite length ,  E = λ / [ (2πr) εo ]
 

Answered by Thiyagarajan K | 16th Feb, 2020, 09:46: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.