# A thin spherical shell of radius R has a fixed charge +q distributed uniformly over its surface.Now the sphere's angular velocity increases linearly withtime w = w0 + kt.The line integral of the electric field (∫E.dl ) around the circular path P located at the centre of the sphere is (Assume that the normal to the plane containing the path is along the +z axis and that the radiusrp

Arun
25757 Points
2 years ago
Gauss's Law :   Er.dS=ϵoqenclosed
As the charge lies on the surface of shell, thus no charge is enclosed by the Gaussian surface lying inside shell. Hence from Gauss's law, net electric field inside the shell is zero.
Er=0   for  rR
Outside the shell :  rR
Charge enclosed by the Gaussian surface  qenclosed=Q
From Gauss's law  ErdS=ϵoqenclosed
We get  Er×4πr2=ϵoQ
Electric field at a point outside the shell  Er=4πϵor2Q  for rR
Thus electric field decreases as r21