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# If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is what?

Hermione Weasley
12 Points
5 years ago
For the convenience of answering this question, consider the two equal charges to be Q1 and Q2 such that Q1 = Q2 = Q,
Q1 and Q2 are placed at a distance of ‘r’ from each other.
Let the charge q be placed at the midpoint of the line joining the equal charges, Q1 and Q2,
Therefore, q is placed at a distance of r/2 from both, Q1 and Q2,
Consider one of the charges Q1.
The force exerted on Q1 due to Q2 = F1 = k(Q1)(Q2)/r2
The force exerted on Q1 due to q = F2 = k(Q1)(q)/(r/2)  = 4k(Q1)(q)/r2
Since the forces F1 and F2 act in the same direction (towards the the charge Q1),
F1 + F2 = $\sqrt{F1^{2} + F2^{2} + 2(F1)(F2)cos\theta }$ = $\sqrt{F1^{2} + F2^{2} + 2(F1)(F2)cos0 }$
= $\sqrt{(F1 + F2)^{2} }$ = F1 + F2 = k(Q1)(Q2)/r+  4k(Q1)(q)/r2 = k(Q1)/r2 * (Q2 + 4q)
Since Q1 = Q2 = Q,
F1 + F2 = k(Q)/r2 * (Q + 4q)
But since the system is at equilibrium, F1 + F2 must be equal to zero.
Thus,
k(Q)/r2 * (Q + 4q) = 0
=> Q + 4q = 0
=> q = -Q/4