Set up the configuration

Let the two equal charges $Q$ be placed at the two ends of a line segment, each at distance $a$ from the center. A charge $q$ is placed at the center.

So the arrangement is: $Q \quad \text{---} \quad q \quad \text{---} \quad Q$ with distance $a$ between $q$ and each $Q$.

Condition for equilibrium

Since $q$ is at the center between two equal charges, the forces on $q$ due to the two outer charges are equal and opposite. Hence, net force on $q$ is automatically zero.

Therefore, for the whole system to be in equilibrium, each outer charge $Q$ must also have zero net force.

Force on one outer charge $Q$

Consider the left outer charge $Q$.

It experiences two forces:

• Repulsion due to the right outer charge $Q$

  Distance between the two outer charges is $2a$. $F_{QQ} = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{(2a)^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{4a^2}$

  This force is directed outward.

• Force due to the central charge $q$

  Distance between $Q$ and $q$ is $a$. $F_{Qq} = \frac{1}{4\pi\varepsilon_0} \frac{|Qq|}{a^2}$

  For equilibrium, this force must oppose the repulsion $F_{QQ}$. Hence $q$ must be of opposite sign to $Q$, so $q$ is negative if $Q$ is positive.

Apply equilibrium condition

For the outer charge to be in equilibrium: $F_{Qq} = F_{QQ}$

$\frac{1}{4\pi\varepsilon_0} \frac{|Qq|}{a^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{4a^2}$

Cancel common factors: $|Qq| = \frac{Q^2}{4}$

$|q| = \frac{Q}{4}$

Since $q$ must be opposite in sign to $Q$, $q = -\frac{Q}{4}$

Check options

• A: $Q/2$ ❌
• B: $-Q/2$ ❌
• C: $Q/4$ ❌
• D: $-Q/4$ ✅

Conclusion

The correct value is: $\boxed{-\frac{Q}{4}}$