In the figure shown A & B are two charged particles having charges q and – q respectively are placed on a nonconducting fixed horizontal smooth plane. B is fixed and A is attached to a non conducting massless spring of spring constant k. The other end of the spring is fixed. Mass of A is m, A and B are in equilibrium when the distance between them is r. Choose the correct options

Let x₀ = extension in the spring when A is in equilibrium. Then,
${\mathrm{kx}}_0=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2}....\left(1\right)$
Now let A be shifted by a small distance x towards B. Then the resultant force towards A is,
$\begin{array}{l}{\mathrm{F}}_{\text{res }}=\mathrm{k}\left({\mathrm{x}}_0+\mathrm{x}\right)-\frac{{\mathrm{q}}^2}{4{\mathrm{\pi ϵ}}_0(\mathrm{r}-\mathrm{x}{)}^2}=\mathrm{k}\left({\mathrm{x}}_0+\mathrm{x}\right)-\frac{{\mathrm{q}}^2}{4{\mathrm{\pi ϵ}}_0{\mathrm{r}}^2}{\left(1-\frac{\mathrm{x}}{\mathrm{r}}\right)}^{-2}\\ =\mathrm{k}\left({\mathrm{x}}_0+\mathrm{x}\right)-\frac{{\mathrm{q}}^2}{4{\mathrm{\pi ϵ}}_0{\mathrm{r}}^2}\left(1+\frac{2\mathrm{x}}{\mathrm{r}}\right);\mathrm{x}<<\mathrm{r}:\text{ Binomial expansion }\\ =\mathrm{k}\times -\frac{{\mathrm{q}}^2}{2{\mathrm{\pi ϵ}}_0{\mathrm{r}}^3}\times ;\text{ using }\left(1\right){\mathrm{F}}_{\text{res }}=\left(\mathrm{k}-\frac{{\mathrm{q}}^2}{2\mathrm{\pi }{\in }_0{\mathrm{r}}^3}\right)\times \\ \because \mathrm{F\alpha }\times \therefore \text{ SHM with }\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}-\frac{{\mathrm{q}}^2}{2{\mathrm{\pi ϵ}}_0{\mathrm{r}}^3}}}\end{array}$
For real T, $\mathrm{k}>\frac{{\mathrm{q}}^2}{2{\mathrm{\pi ϵ}}_0{\mathrm{r}}^3}\therefore {\mathrm{k}}_{min}=\frac{{\mathrm{q}}^2}{2{\mathrm{\pi ϵ}}_0{\mathrm{r}}^3}$
$\mathrm{T}=2\mathrm{\pi }\sqrt{\mathrm{k}-\frac{{\mathrm{q}}^2}{2{\mathrm{\pi ϵ}}_0{\mathrm{r}}^3}},{\mathrm{k}}_{min}=\frac{{\mathrm{q}}^2}{2{\mathrm{\pi ϵ}}_0{\mathrm{r}}^3}$