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?

Question

Hermione Weasley · Answer

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)/r 2 The force exerted on Q1 due to q = F2 = k(Q1)(q)/(r /2) 2 = 4k(Q1)(q)/r 2 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)/r 2 * (Q + 4q) But since the system is at equilibrium, F1 + F2 must be equal to zero. Thus, k(Q)/r 2 * (Q + 4q) = 0 => Q + 4q = 0 => q = -Q/4

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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?

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?

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