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Question: Work done by an external agent to move slowly a charge Q from rim of a uniformly charged horizontal ...

Work done by an external agent to move slowly a charge Q from rim of a uniformly charged horizontal disc of radius a and charge per unit area s, to center of this disc is-

A

σaQε0(2π12)\frac{\sigma aQ}{\varepsilon_{0}}\left( \frac{2}{\pi}–\frac{1}{2} \right)

B

σaQε0(121π)\frac{\sigma aQ}{\varepsilon_{0}}\left( \frac{1}{2}–\frac{1}{\pi} \right)

C

σaQε0(1π12)\frac{\sigma aQ}{\varepsilon_{0}}\left( \frac{1}{\pi}–\frac{1}{2} \right)

D

σaQε0(122π)\frac{\sigma aQ}{\varepsilon_{0}}\left( \frac{1}{2}–\frac{2}{\pi} \right)

Answer

σaQε0(121π)\frac{\sigma aQ}{\varepsilon_{0}}\left( \frac{1}{2}–\frac{1}{\pi} \right)

Explanation

Solution