Two particles A and B , each carrying a charge Q , are held fixed...

Let the origin be at the midpoint of the line joining $A and$B$. The coordinates of$A$and$B$are$-d/2$and$d/2$.<br><br>When particle$C$is displaced by a small distance$x$towards$B$, its distance from$A$becomes$d/2 + x$and from$B$becomes$d/2 - x$.<br><br>The net force on$C$is:<br><br>$F = \dfrac{1}{4\pi \varepsilon_0} \dfrac{Qq}{(d/2 + x)^2} - \dfrac{1}{4\pi \varepsilon_0} \dfrac{Qq}{(d/2 - x)^2}$<br><br>$F = \dfrac{Qq}{4\pi \varepsilon_0} \left[ \dfrac{(d/2 - x)^2 - (d/2 + x)^2}{(d^2/4 - x^2)^2} \right]<br><br>F = $\dfrac{Qq}{4\pi \varepsilon_0} \left[ \dfrac{-2dx}{(d^2/4 - x^2)^2} \right]$

Since $x \ll d$, the term $x^2$ can be neglected in the denominator:

$F \approx \dfrac{Qq}{4\pi \varepsilon_0} \left[ \dfrac{-2dx}{(d^2/4)^2} \right] = -\dfrac{8Qq}{\pi \varepsilon_0 d^3} x$

Comparing this with the standard equation of simple harmonic motion $F = -Kx$, the effective force constant is $K = \dfrac{8Qq}{\pi \varepsilon_0 d^3}$.

The time period of oscillation is given by:

$T = 2\pi \sqrt{\dfrac{m}{K}} = 2\pi \sqrt{\dfrac{\pi \varepsilon_0 m d^3}{8Qq}}$

$T = \left( \dfrac{4\pi^2 \pi \varepsilon_0 m d^3}{8Qq} \right)^{1/2} = \left( \dfrac{\pi^3 \varepsilon_0 m d^3}{2Qq} \right)^{1/2}$