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A charge Q and another charge q are placed at two points A and B which are distance l apart. Keeping q stationary at B, Q is moved to point C such that ABC forms equilateral triangle of side l. Find the net work done ?

$(A)\;\frac{1}{4 \pi \in_0} \frac{Qq}{l} \\ (B)\;\frac{1}{4 \pi \in_0} \frac{Qq}{l^2} \\ (C)\;Zero \\ (D)\;\frac{1}{4 \pi \in_0} Qql $

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Potential at A and C are equal , therefore work done in moving Q from A to C is zero.
Hence C is the correct answer.
answered Jan 8, 2014 by meena.p

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