 Solution

Put $+q$ at $x=0$ and $+4q$ at $x=l$, and write $k_e=1/(4\pi {ϵ}_0)$. The electric fields of the two positive point charges can cancel only between them. For $Q\ne 0$, zero Lorentz force on $Q$ therefore requiresThe force of $+4q$ on $+q$ points left, so $Q$ must be negative to pull $+q$ right. Equating these forces givesFor $+4q$, the attraction to $Q$ at separation $2l/3$ has magnitude $k_e(4q)(4q/9)/(2l/3{)}^2=4k_e{q}^2/l^2$, exactly balancing its repulsion from $+q$. Thus all three forces vanish, not only the force on the inserted charge.