Two point charges $$Q$$ each are placed at a distance $$d$$ apart. A third point charge $$q$$ is placed at a distance $$x$$ from mid-point on the perpendicular bisector. The value of $$x$$ at which charge $$q$$ will experience the maximum Coulomb's force is:

Let the two identical charges $$Q$$ be positioned along the horizontal axis, separated by a distance $$d$$. The midpoint acts as the origin, placing the charges at $$(-d/2, 0)$$ and $$(d/2, 0)$$.

The third charge $$q$$ is placed on the perpendicular bisector at a distance $$x$$ from the origin, at coordinates $$(0, x)$$.

The distance $$r$$ from either charge $$Q$$ to the charge $$q$$ is:    $$r = \sqrt{x^2 + \left(\frac{d}{2}\right)^2}$$

The magnitude of the Coulomb force exerted by one charge $$Q$$ on $$q$$ is:   $$F_1 = \frac{kQq}{r^2} = \frac{kQq}{x^2 + \left(\frac{d}{2}\right)^2}$$

By symmetry, the horizontal components of the forces cancel each other out, while the vertical components along the bisector add up:  $$F_{\text{net}} = 2F_1\cos\theta$$

Where $$\cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2 + \left(\frac{d}{2}\right)^2}}$$. Substituting this back gives:   $$F_{\text{net}} = \frac{2kQqx}{\left[x^2 + \left(\frac{d}{2}\right)^2\right]^{3/2}}$$

To find the value of $$x$$ that yields the maximum force, we set the first derivative with respect to $$x$$ equal to zero:

$$\frac{d}{dx}\left[\frac{x}{\left(x^2 + \left(\frac{d}{2}\right)^2\right)^{3/2}}\right] = 0$$ $$\implies $$ $$\frac{\left(x^2 + \left(\frac{d}{2}\right)^2\right)^{3/2}(1) - x \cdot \frac{3}{2}\left(x^2 + \left(\frac{d}{2}\right)^2\right)^{1/2}(2x)}{\left(x^2 + \left(\frac{d}{2}\right)^2\right)^3} = 0$$

$$\left(x^2 + \left(\frac{d}{2}\right)^2\right) - 3x^2 = 0$$ $$\implies$$ $$\left(\frac{d}{2}\right)^2 = 2x^2$$ $$\implies$$ $$x^2 = \frac{d^2}{8}$$

$$\implies x = \frac{d}{2\sqrt{2}}$$