Q:) Two charges `Q`, each are separated by a distance `d`. A third charge `q` is placed at a point on the line joining the two charges such that the system is in equilibrium. Find the magnitude of the third charge in terms of `Q`

For equilibrium for net force on each charge must be zero so Q must put middle of them and now put net force on Q zero. KQ²/d² + KQq/(d/2)² = 0 . So Q = -Q/4.Hope it clears. If u have doubts then please clearify. And if u like my answer then please approve answer

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