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(B) Let the two charges $Q$ be at $(-a, 0)$ and $(a, 0)$. The charge $-q$ is at the origin $(0, 0)$.When $-q$ is displaced by a small distance $x$ alo

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{{2m{a^3}\pi\, { \in _0}}}{{Qq}}} $

Vedclass · Answer

(B) Let the two charges $Q$ be at $(-a, 0)$ and $(a, 0)$. The charge $-q$ is at the origin $(0, 0)$.When $-q$ is displaced by a small distance $x$ along the $y$-axis,the distance $r$ from each charge $Q$ is $r = \sqrt{a^2 + x^2}$.The force exerted by each charge $Q$ on $-q$ is $F = \frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2} = \frac{1}{4\pi \epsilon_0} \frac{Qq}{a^2 + x^2}$.The component of force along the $y$-axis is $F_y = -2F \sin 	heta$,where $\sin 	heta = \frac{x}{r} = \frac{x}{\sqrt{a^2 + x^2}}$.So,$F_{net} = -2 \left( \frac{1}{4\pi \epsilon_0} \frac{Qq}{a^2 + x^2} ight) \frac{x}{\sqrt{a^2 + x^2}} = -\frac{2Qqx}{4\pi \epsilon_0 (a^2 + x^2)^{3/2}}$.For small $x$,$a^2 + x^2 \approx a^2$,so $F_{net} \approx -\frac{2Qqx}{4\pi \epsilon_0 a^3} = -\left( \frac{Qq}{2\pi \epsilon_0 a^3} ight) x$.Comparing with $F = -m\omega^2 x$,we get $\omega^2 = \frac{Qq}{2\pi \epsilon_0 m a^3}$,so $\omega = \sqrt{\frac{Qq}{2\pi \epsilon_0 m a^3}}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2\pi \epsilon_0 m a^3}{Qq}} = 2\pi \sqrt{\frac{2m a^3 \pi \epsilon_0}{Qq}}$.

Two charges each of magnitude $Q$ are fixed at a distance of $2a$ apart. $A$ third charge ($-q$ of mass $m$) is placed at the midpoint of the two charges. If the $-q$ charge is slightly displaced perpendicular to the line joining the charges,find its time period.

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