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If a charge q is rotated around charge Q in a circle of radius y by $90^{\circ}$, the work done will be

$(A)\;q \times 2 \pi r \\ (B)\;0 \\ (C)\; q \times \large\frac{\pi}{2} r \\ (D)\;q \times \large\frac{2 \pi Q}{r} $
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Since q moves on equipotential surface therefore work done is zero.
=> $ q \times \large\frac{2 \pi Q}{r} $
Hence D is the correct answer.
answered Jan 6, 2014 by meena.p

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