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Two free positive charges 4q and q are a distance l apart. What charge Q is needed to achieve equilibrium for the entire system and where should it be placed form charge q? Why we do not equate equilibrium at Q rather than at q or 4q for magnitude of Q
4qandqare given to be free positive charges placed at a distancelapart. So, chargeQneeded to achieve equilibrium for the entire system, should be of negative sign. LetQis at a distancexfromqcharge, and therefore(l−x)from4qcharge.From Coulomb's Law, force between two charges isr2kq1q2So, for equilibrium, force betweenqandQshould be equal to force between4qandQi.e, x2kqQ=(l−x)2k4qQ ...(i)(l−x)2=4x2Taking square root on both sides:l−x=2xl=3xx=3li.e, the chargeQshould be placed at a distance3lfrom chargeq.Now, applying the condition of equilibrium on+qcharge,⇒(3l)2kQq=l2k(4q)q⇒l2Q=4q(3l)2⇒Q=9×l24ql2⇒Q=94qSinceQshould be negative as explained above, soQ=9−4q
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