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 positive charges $$+q$$ be placed at $$x = -a$$ and $$x = +a$$.

Let the test charge $$q_0$$ be placed at a distance $$y$$ from the center along the perpendicular bisector (the y-axis).

The distance from either charge to $$q_0$$ is $$r = \sqrt{a^2 + y^2}$$.

The magnitude of the force exerted by one charge is $$F' = \frac{k q q_0}{r^2} = \frac{k q q_0}{a^2 + y^2}$$

By symmetry, the horizontal components of the two forces cancel each other. The vertical components add together.

$$F_{\text{net}} = 2F' \cos\theta$$

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

$$F_{\text{net}} = 2 \left( \frac{k q q_0}{a^2 + y^2} \right) \left( \frac{y}{\sqrt{a^2 + y^2}} \right)$$

$$F_{\text{net}} = \frac{2 k q q_0 y}{(a^2 + y^2)^{3/2}}$$

To find the distance $$y$$ where the force is maximum, we differentiate $$F_{\text{net}}$$ with respect to $$y$$ and set it to zero ($$\frac{dF_{\text{net}}}{dy} = 0$$):

Since $$2 k q q_0$$ is a constant, we only need to maximize the term $$y(a^2 + y^2)^{-3/2}$$:

$$\frac{d}{dy} \left[ y(a^2 + y^2)^{-3/2} \right] = 0$$

$$1 \cdot (a^2 + y^2)^{-3/2} + y \left( -\frac{3}{2} \right)(a^2 + y^2)^{-5/2}(2y) = 0$$

$$(a^2 + y^2)^{-3/2} - 3y^2(a^2 + y^2)^{-5/2} = 0$$

$$(a^2 + y^2)^1 - 3y^2 = 0$$

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

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