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Charges +q and -q are placed at points A and B respectively which are at a distance 2 L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is:

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$\frac{q Q}{4 π ε_{0} L}$

$\frac{q Q}{2 π ε_{0} L}$

$\frac{q Q}{6 π ε_{0} L}$

$- \frac{q Q}{6 π ε_{0} L}$

answer is D.

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Detailed Solution

W = Q(V_D – V_C)
$V_{D} = \frac{1}{4 π ε_{0}} q [\frac{1}{3 L} - \frac{1}{2 L}]$

$V_{C} = 0$

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