Two fixed charges 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3 m.

The point $\mathrm{P}$ where field due to both charges is equal and opposite is on AB outside because charges are opposite in nature and it will be near to the charge having lesser magnitude.

The charge at P will execute oscillation if it is in stable equilibrium. 

$\Rightarrow \frac{\mathrm{dU}}{\mathrm{dx}}=0 \mathrm{and} \frac{{\mathrm{d}}^2\mathrm{U}}{{\mathrm{dx}}^2}>0 \left[\mathrm{where} \mathrm{x} =\mathrm{BP}\right]$

let the charge at P is q.

$$\begin{gathered} \because {\mathrm{F}}_{\mathrm{q}}=-\frac{\mathrm{dU}}{\mathrm{dx}} \\ \therefore \frac{\mathrm{dU}}{\mathrm{dx}}=-{\mathrm{F}}_{\mathrm{q}}=\mathrm{kQq}\left(\frac{1}{{\mathrm{x}}^2}-\frac{4}{{\left(\mathrm{r}+\mathrm{x}\right)}^2}\right) \\ \Rightarrow \frac{\mathrm{dU}}{\mathrm{dx}}=0 \Rightarrow \mathbf{x}\mathbf{=}\mathbf{r} \end{gathered}$$

$\mathrm{again}, {\left[\frac{{\mathrm{d}}^2\mathrm{U}}{{\mathrm{dx}}^2}\right]}_{\mathrm{x}=\mathrm{r}}=-\frac{\mathrm{kQq}}{{\mathrm{r}}^3}$

For stable equilibrium, $\frac{{\mathrm{d}}^2\mathrm{U}}{{\mathrm{dx}}^2}>0$

 $$\begin{gathered} \Rightarrow -\frac{\mathrm{kQq}}{{\mathrm{r}}^3}>0 \\ \Rightarrow \mathrm{q}<0 \end{gathered}$$