The arc AB with the centre C and the infinitely long wire having linear charge density $λ$ are lying in the same plane. The minimum amount of work to be done to move a point charge $q_{0}$ from point A to B through a circular path AB of radius $a$ is equal to

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$\frac{q_{0} λ}{\sqrt{2} π ε_{0}}$

$\frac{q_{0}^{2}}{2 π ε_{0}} \log (\frac{2}{3})$

$\frac{q_{0} λ}{2 π ε_{0}} \log (\frac{3}{2})$

$\frac{q_{0} λ}{2 π ε_{0}} \log (\frac{2}{3})$

answer is B.

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Detailed Solution

$dv = - \int Edr = - \int_{3 a}^{2 a} \frac{λ}{2 {πε}_{0} r} dr$

$dv = \frac{λ}{2 {πε}_{0}} \log (\frac{3}{2})$

$w = q_{0} (dv)$ $= \frac{q_{0} λ}{2 π ε_{0}} \log (\frac{3}{2})$

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