The point charges -2q, -2q, and +q are put on the vertices of an equilateral triangle of side a. Find the work done by some external force in increasing the separation to 2a (in joules)

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Solution

Here work done , $W = W_{f} - W_{i}$
$W_{i} = k \frac{(- 2 q) (- 2 q)}{a} + k \frac{(- 2 q) (q)}{a} + k \frac{(q) (- 2 q)}{a} = k \frac{4 q^{2}}{a} - k \frac{4 q^{2}}{a} = 0$
$W_{f} = k \frac{(- 2 q) (- 2 q)}{2 a} + k \frac{(- 2 q) (q)}{2 a} + k \frac{(q) (- 2 q)}{2 a} = k \frac{4 q^{2}}{2 a} - k \frac{4 q^{2}}{2 a} = 0$
$∴ W = 0$

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The point charges -2q, -2q, and +q are put on the vertices of an equilateral triangle of side a. Find the work done by some external force in increasing the separation to 2a (in joules)

Here work done ,W=Wf−WiWi=k(−2q)(−2q)a+k(−2q)(q)a+k(q)(−2q)a=k4q2a−k4q2a=0Wf=k(−2q)(−2q)2a+k(−2q)(q)2a+k(q)(−2q)2a=k4q22a−k4q22a=0∴W=0