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Question
Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side ‘a’ as shown below.
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Solution
Work done to keep the system bound is
`W=(Kq_1q_2)/a+(Kq_2q_3)/a+(Kq_1q_3)/a`
`W=(Kq(-4q))/a+(K(-4q)2q)/a+(Kq(2q))/a`
`W=(-K4q^2)/a-(K8q^2)/a+(K2q^2)/a`
`W=(Kq^2)/a(-4-8+2)=(-10Kq^2)/a`
Therefore, the work done to dissociate the system is
`W_d=-W=(10Kq^2)/a`
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