For the net force on charge $q$ (placed at $x=l$) to be zero, the force due to $+4q$ (at $x=0$) must be equal and opposite to the force due to $Q$ (at $x=l/2$).

1. Force due to $+4q$: $F_1 = k \frac{(4q)(q)}{l^2}$ (directed away from origin).
2. Force due to $Q$: $F_2 = k \frac{(Q)(q)}{(l - l/2)^2} = k \frac{Q q}{(l/2)^2} = k \frac{4Qq}{l^2}$.

For equilibrium ($F_{net} = 0$): $F_1 + F_2 = 0$ $k \frac{4q^2}{l^2} + k \frac{4Qq}{l^2} = 0$

Dividing by common terms ($4kq/l^2$): $q + Q = 0 \implies Q = -q$.

(See NCERT Physics Class 12, Chapter 1, Section 1.6).