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(C) Let the two fixed charges $Q$ be located at positions $-d$ and $+d$.When the charge $q$ is displaced by a distance $x$ from the origin,the net for

Vedclass · Accepted Answer

$2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Q q}}$

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(C) Let the two fixed charges $Q$ be located at positions $-d$ and $+d$.When the charge $q$ is displaced by a distance $x$ from the origin,the net force acting on it is:$F = F_{left} - F_{right} = \frac{1}{4\pi\varepsilon_{0}} \frac{Qq}{(d-x)^{2}} - \frac{1}{4\pi\varepsilon_{0}} \frac{Qq}{(d+x)^{2}}$$F = \frac{Qq}{4\pi\varepsilon_{0}} \left[ \frac{(d+x)^{2} - (d-x)^{2}}{(d^{2}-x^{2})^{2}} \right] = \frac{Qq}{4\pi\varepsilon_{0}} \left[ \frac{4dx}{(d^{2}-x^{2})^{2}} \right]$Since $x \ll d$,we can approximate $(d^{2}-x^{2})^{2} \approx d^{4}$:$F \approx \frac{Qq}{4\pi\varepsilon_{0}} \cdot \frac{4dx}{d^{4}} = \frac{Qqx}{\pi\varepsilon_{0}d^{3}}$Since the force is directed towards the mean position,$F = -kx$,where $k = \frac{Qq}{\pi\varepsilon_{0}d^{3}}$.The time period $T$ is given by:$T = 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M \pi \varepsilon_{0} d^{3}}{Qq}} = 2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Qq}}$

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