Hermione Weasley
Last Activity: 9 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)2 = 4k(Q1)(q)/r2
Since the forces F1 and F2 act in the same direction (towards the the charge Q1),
F1 + F2 =
=
=
= F1 + F2 = k(Q1)(Q2)/r
2 + 4k(Q1)(q)/r
2 = k(Q1)/r
2 * (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
