A thin spherical shell of radius R has a fixed ch

Question

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 with
time 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 radius
rp

Arun · Accepted Answer

Gauss's Law : ∮Er​.dS=ϵo​qenclosed​​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 rROutside the shell : r≥RCharge enclosed by the Gaussian surface qenclosed​=QFrom Gauss's law Er​∮dS=ϵo​qenclosed​​We get Er​×4πr2=ϵo​Q​Electric field at a point outside the shell Er​=4πϵo​r2Q​ for r≥RThus electric field decreases as r21​

Electrostatics> A thin spherical shell of radius R has a ...

ats , 5 Years ago

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