Solve it
Asked by venkat
14th May 2018,
6:56 PM
Answered by Expert
Answer:
Potential V near earh surface, assuming earth as sphere is given by V = Q/(4πε0×r),
where r is the distance of a test point from earth centre and the test point is near the earths surface.
hence ∂V/∂r = -Q/(4πε0×r2)
At Earth surface, r=R, then we have ∂V/∂r = -Q/(4πε0×R2) = {-Q/4πR2} / ε0 = 2.5;
Q/4πR2 is surface charge density σ , hence ( σ / ε0 ) = 2.5
hence σ = 2.5×ε0 = 2.5×8.854×10-12 = 22.135 × 10-12 C/m2
Above answer is not an option given in the question. But I feel this is the method to workout the surface charge density
Answered by Expert
16th May 2018,
2:18 PM
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