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(C) Let the two fixed charges be $-q$ and the central charge be $+q$. The distance between the fixed charges is $2d$. When the central charge is displ

Vedclass · Accepted Answer

$\left(\frac{q^{2}}{2 \pi \varepsilon_{0} m d^{3}}\right)^{\frac{1}{2}}$

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(C) Let the two fixed charges be $-q$ and the central charge be $+q$. The distance between the fixed charges is $2d$. When the central charge is displaced by $x$ perpendicular to the line joining the fixed charges,the distance between each fixed charge and the displaced charge becomes $r = \sqrt{d^2 + x^2}$.The electrostatic force exerted by each fixed charge on the central charge is $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^2}{r^2} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^2}{d^2 + x^2}$.The net restoring force is $F_{	ext{net}} = -2 F \cos 	heta$,where $\cos 	heta = \frac{x}{r} = \frac{x}{\sqrt{d^2 + x^2}}$.Substituting the values,$F_{	ext{net}} = -2 \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q^2}{d^2 + x^2} ight) \left( \frac{x}{\sqrt{d^2 + x^2}} ight) = -\frac{q^2 x}{2 \pi \varepsilon_{0} (d^2 + x^2)^{3/2}}$.Since $x Using Newton's second law,$F = ma$,so $a = -\left( \frac{q^2}{2 \pi \varepsilon_{0} m d^3} ight) x$.Comparing this with the $SHM$ equation $a = -\omega^2 x$,we get $\omega = \sqrt{\frac{q^2}{2 \pi \varepsilon_{0} m d^3}}$.

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