Question

$\Rightarrow F' = \frac{1}{2\pi\epsilon_0} \times \frac{qQx}{[(\frac{d}{2})^2 + x^2]^{3/2}}$

For maximum force, $\frac{dF'}{dx} = 0$

$\frac{qQ}{2\pi\epsilon_0} \times [[(\frac{d}{2})^2 + x^2]^{-3/2} - x\frac{3}{2}[(\frac{d}{2})^2 + x^2]^{-5/2}2x] = 0$

Answer 1

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

Answer 2

The given equation is derived from setting the first derivative of the force $F'$ with respect to $x$ equal to zero to find the condition for maximum force. The equation is: $\frac{qQ}{2\pi\epsilon_0} \times \left[ \left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-3/2} - x\frac{3}{2}\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2}2x \right] = 0$

Since $\frac{qQ}{2\pi\epsilon_0}$ is a non-zero constant (assuming $q, Q \ne 0$), we can divide the equation by this term: $\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-3/2} - x\frac{3}{2}\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2}2x = 0$

Simplify the second term: $x\frac{3}{2}\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2}2x = 3x^2\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2}$

The equation becomes: $\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-3/2} - 3x^2\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2} = 0$

Factor out the term with the lower power, $\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2}$: $\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2} \left[ \frac{\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-3/2}}{\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2}} - 3x^2 \right] = 0$

Simplify the term inside the square bracket using the rule $\frac{a^m}{a^n} = a^{m-n}$: $\frac{\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-3/2}}{\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2}} = \left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-3/2 - (-5/2)} = \left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-3/2 + 5/2} = \left(\left(\frac{d}{2}\right)^2 + x^2\right)^{2/2} = \left(\frac{d}{2}\right)^2 + x^2$

Substitute this back into the factored equation: $\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2} \left[ \left(\frac{d}{2}\right)^2 + x^2 - 3x^2 \right] = 0$

The term $\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{-5/2} = \frac{1}{\left(\left(\frac{d}{2}\right)^2 + x^2\right)^{5/2}}$ is non-zero for real values of $x$ and $d$ (unless $d=0$ and $x=0$, where the original force is undefined). Therefore, the term in the square bracket must be zero: $\left(\frac{d}{2}\right)^2 + x^2 - 3x^2 = 0$

Combine the $x^2$ terms: $\left(\frac{d}{2}\right)^2 - 2x^2 = 0$

Solve for $x^2$: $\frac{d^2}{4} = 2x^2$ $x^2 = \frac{d^2}{4 \times 2}$ $x^2 = \frac{d^2}{8}$

Take the square root of both sides to find $x$: $x = \pm \sqrt{\frac{d^2}{8}}$ $x = \pm \frac{d}{\sqrt{8}}$ $x = \pm \frac{d}{2\sqrt{2}}$

The maximum force occurs when the test charge is located at a distance $|x| = \frac{d}{2\sqrt{2}}$ from the origin along the axis.