Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0).…

The net force on q at origin is

$\vec{F}=\vec{F}_1+\vec{F}_2=\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{r^2} \hat{i}+\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{r^2}(-\hat{i})=\hat{0}$

The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is

U = $\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{x}+\frac{1}{4 \pi \varepsilon_0} . \frac{q^2}{\left(a^2+x\right)}$

$=\frac{1}{4 \pi \varepsilon_0} . q^2\left[\frac{1}{a-x}+\frac{1}{a+x}\right]$

$\frac{d U}{d x} =\frac{q^2}{4 \pi \varepsilon_0}\left[\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\right]$

For U to be minimum

$\frac{d U}{d x}=0, \quad \frac{d^2 U}{d x^2}>0$,

$\Rightarrow(a-x)^2=(a+x)^2$

$\Rightarrow a+x= \pm(a-x)$

⇒ x = 0, because other solution is irrelevent.

Thus the charged particle at the origin will have minimum force and minimum P.E.

∴ (D).