Two fixed charges  $4Q$ (positive) and $-Q$ are located at  A and B respectively. The distance AB being 3m.

At any point to the left of charge $4Q$,  electric field due to $4Q$ (leftwards) is stronger than that due to $-Q$(rightwards). Net electric field is towards left.

In-between the charges, fields due to both the charges is towards right. Net electric field is towards right.

Null point is possible only to the right of charge $-Q$.

Let null point $P$ be at a distance of $d$ to the right of charge $-Q$

$\frac{K\left(4Q\right)}{{\left(3+d\right)}^2}=\frac{KQ}{d^2}$

$\Rightarrow d=3m$

To the right of null point, net electric field will point towards right and to its left, net electric field will point towards left.

Hence a negative charge placed at null point will be in stable equilibrium along the line joining the charges. A positive charge will be in unstable equilibrium along the line joining the charges.