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Question

(A) Let the charge $q_0$ be displaced by a small distance $x$ from the midpoint towards one of the charges $Q$.The net force on $q_0$ is $F = F_1 - F_

Vedclass · Accepted Answer

$\sqrt{\frac{4 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}$

Vedclass · Answer

(A) Let the charge $q_0$ be displaced by a small distance $x$ from the midpoint towards one of the charges $Q$.The net force on $q_0$ is $F = F_1 - F_2 = \frac{1}{4\pi\varepsilon_0} \frac{Qq_0}{(a-x)^2} - \frac{1}{4\pi\varepsilon_0} \frac{Qq_0}{(a+x)^2}$.$F = \frac{Qq_0}{4\pi\varepsilon_0} \left[ \frac{(a+x)^2 - (a-x)^2}{(a^2-x^2)^2} \right] = \frac{Qq_0}{4\pi\varepsilon_0} \left[ \frac{4ax}{(a^2-x^2)^2} \right]$.For small $x$,$x^2 \approx 0$,so $F \approx \frac{Qq_0}{4\pi\varepsilon_0} \frac{4ax}{a^4} = \frac{Qq_0 x}{\pi\varepsilon_0 a^3}$.Since $F = -ma$ (restoring force),$a = -\left( \frac{Qq_0}{\pi\varepsilon_0 m a^3} \right) x$.Comparing with $a = -\omega^2 x$,we get $\omega^2 = \frac{Qq_0}{\pi\varepsilon_0 m a^3}$,so $\omega = \sqrt{\frac{Qq_0}{\pi\varepsilon_0 m a^3}}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\pi\varepsilon_0 m a^3}{Qq_0}} = \sqrt{\frac{4\pi^3\varepsilon_0 m a^3}{Qq_0}}$.

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