Two equal positive point charges are separated by a distance $$2a$$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $$q_0$$ becomes maximum is $$\frac{a}{\sqrt{x}}$$. The value of $$x$$ is ______.

Let the two charges +q be at

$$(+a,0),(−a,0)$$

Take a point on equatorial line at distance y from center:

(0,y)

Distance from each charge to this point:

$$r=\sqrt{a^2+y^2}$$

Field due to one charge:

$$E_1=\frac{kq}{r^2}$$

Horizontal components cancel.

Vertical components add.

Vertical component from one charge:

$$E_1\cos⁡θ$$

where

$$\cos\theta=\frac{y}{r}$$

So net field

$$E=2\cdot\frac{kq}{r^2}\cdot\frac{y}{r}$$

$$=\frac{2kqy}{(a^2+y^2)^{3/2}}$$

Force on test charge

$$F=q_0E$$

To maximize force, maximize

$$f(y)=\frac{y}{(a^2+y^2)^{3/2}}$$

Differentiate:

$$\frac{df}{dy}=(a^2+y^2)^{-3/2}-\frac{3y^2}{(a^2+y^2)^{5/2}}$$

Set zero:

$$(a^2+y^2)-3y^2=0$$

$$a^2-2y^2=0$$

$$y=\frac{a}{\sqrt{2}}$$

Given distance is

$$\frac{a}{x}$$

So

$$\frac{a}{x}=\frac{a}{\sqrt{2}}$$