Question

A charge \( q \) is placed at the centre of the line joining two equal charges \( Q \). The system of the three charges will be in equilibrium if \( q \) is equal to:

• A. \( -\frac{Q}{2} \)
• B. \( -\frac{Q}{4} \)
• C. \( \frac{Q}{2} \)
• D. \( \frac{Q}{4} \)

Hint:

To achieve equilibrium in a system of charges, the net force on each charge must be zero. Use Coulomb's law to calculate the forces.

Answer 1

The Correct Option is A

Step 1: Set up the positions.
Place the two equal charges \( Q \) at \( x = -a \) and \( x = +a \), with the unknown charge \( q \) sitting at the centre, \( x = 0 \).

Step 2: Check the centre charge first.
By symmetry, the charge \( Q \) on the left and the charge \( Q \) on the right are the same distance \( a \) from the centre, so they pull or push \( q \) equally in opposite directions. These two forces cancel automatically, for any value of \( q \) and any position as long as it stays exactly in the middle. So the centre charge does not fix the value of \( q \); the outer charges do.

Step 3: Balance the forces on one outer charge.
Look at the charge \( Q \) sitting at \( x = +a \). It feels two forces: repulsion from the other charge \( Q \) at \( x = -a \), a distance \( 2a \) away, and a force from the centre charge \( q \), a distance \( a \) away.
Repulsion from the far charge:
\[ F_{QQ} = \frac{kQ^2}{(2a)^2} = \frac{kQ^2}{4a^2} \]
Force from the centre charge:
\[ F_{Qq} = \frac{kQ|q|}{a^2} \]

Step 4: Equate the two forces and solve.
For equilibrium, these must be equal in size:
\[ \frac{kQ^2}{4a^2} = \frac{kQ|q|}{a^2} \]
\[ |q| = \frac{Q}{4} \]
Since the force from \( q \) needs to pull the outer charge inward, opposing the outward push from the other outer charge, \( q \) must be negative.

Step 5: Final Answer.
\[ \boxed{q = -\frac{Q}{4}} \]

Answer 2

There is a general pattern behind these three-charge equilibrium problems that makes the answer quick to spot once it's recognised, and checking each option against it settles the question.

1. \( -\frac{Q}{2} \): For a charge placed exactly halfway between two equal charges \( Q \) a distance \( 2a \) apart, the general balancing condition works out to a quarter of \( Q \) in size, not a half. A charge this large would pull the outer charges in far more than needed to cancel their mutual repulsion.
2. \( -\frac{Q}{4} \): The general result for two equal charges \( Q \) with a third charge exactly at the midpoint is that the middle charge must equal \( -\frac{Q}{4} \) for the whole line to sit in equilibrium. This comes from the fact that the midpoint charge is twice as close to each outer charge as the outer charges are to each other, and force falls off with the square of distance, so a charge one quarter the size at half the distance produces a matching force.
3. \( \frac{Q}{2} \): Wrong on both size and sign for the same reasons as \( -\frac{Q}{2} \), just with the sign flipped, which would add to the repulsion instead of cancelling it.
4. \( \frac{Q}{4} \): The magnitude here matches the correct answer, but the sign is wrong. A positive charge at the centre would repel both outer charges rather than pulling them inward, so it cannot hold the system in equilibrium.

The distance-squared relationship in Coulomb's law is what fixes the middle charge at exactly a quarter of the outer charge's magnitude, and the requirement that it pulls rather than pushes fixes the sign as negative.

The correct answer is \( -\frac{Q}{4} \).