Work performed when a point charge 2×10-8 C is transformed from infinity to a point at a distance of 1cm from the surface of the ball with a radius of 1cm and a surface charge density σ =10-8C/cm2
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a
1.1×10−4J
b
11×10−4J
c
0.11×10−4J
d
113×10−4J
answer is B.
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Detailed Solution
Potential at a distance 2cm from its centre=Q4π∈0r=4πr2σ4π∈0r′⇒r2σϵ0r′=σ2ϵ0×1100 since r=1cm and r′=2cmPDb/w the two points is equal to σ200∈0 work done =VQ=σ200ϵ0X2×10−8=11×10−4J