A spherical charged conductor has σ as the surface density of charge. The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?
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a
E/4
b
E/2
c
E
d
2E
answer is C.
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Detailed Solution
E=14πε0⋅QR2=14πε0×4πR2σR2=σε0This is independent of radius and depends on σ. Hence, the electric field on the surface of the new sphere will be E.