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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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$1 .1 \times 10^{- 4} J$

$11 \times 10^{- 4} J$

$0.11 \times 10^{- 4} J$

$113 \times 10^{- 4} J$

answer is B.

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Detailed Solution

Potential at a distance 2cm from its centre

$= \frac{Q}{4 π \in_{0} r} = \frac{4 {πr}^{2} σ}{4 π \in_{0} r^{'}} \Rightarrow \frac{r^{2} σ}{ϵ_{0} r^{'}} = \frac{σ}{2 ϵ_{0}} \times \frac{1}{100}$

$since r = 1 cm and r^{'} = 2 cm$

$PDb / w the two points is equal to \frac{σ}{200 \in_{0}}$

$work done = VQ = \frac{σ}{200 ϵ_{0}} X_{2 \times 10^{- 8}} = 11 \times 10^{- 4} J$

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