APPLICATION OF GAUSS LAW

 

FIELD DUE TO UNIFORMLY CHARGED THIN SPHERICAL SHELL :

(1)Field out side the shell

Flux through Gaussian surface = E×4πr² 

Charge enclosed =σ×4πR²       

By Gaussian’slaw                                                                           

E×4πr² = σ /є₀ ×4Rπ² 

E = σ R²/є₀ r²= q/ 4 є₀π r²

           Where  q = 4πR² σ  is the total charge on the spherical                                Vectorially                                                  

                 E = q/ 4є₀ πr² r^

(2) Field inside the shell

However in this case , the Gaussian surface enclosed no charge

                           Diagram

 According to  Gauss’s law 

                                E×4_r² = 0

                                E = 0                        (r<R)