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प्रश्न
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.
उत्तर
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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