Question
Repeat the previous problem if the particle C is displaced through a distance x along the line AB.
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Answer
$\text{F}_\text{AC}=\frac{\text{KQq}}{(\ell+\text{x})^2}$$\text{F}_\text{CA}=\frac{\text{KQq}}{(\ell-\text{x})^2}$
Net force $=\text{KQq}\Big[\frac{1}{(\ell-\text{x})^2}-\frac{1}{(\ell-\text{x})^2}\Big]$
$=\text{KQq}\bigg[\frac{(\ell+\text{x})^2-(\ell-\text{x})^2}{(\ell+\text{x})^2(\ell-\text{x})^2}\bigg]$
$=\text{KQq}\bigg[\frac{4\ell\text{x}}{(\ell^2-\text{x}^2)^2}\bigg]$
$\text{x}<<<\text{l}=\frac{\text{d}}{2}$ neglecting x w.r.t. $\ell$ We get
net $\text{F}=\frac{\text{KQq}4\ell\text{x}}{\ell^4}=\frac{\text{KQq}4\text{x}}{\ell^3}$ acceleration $=\frac{4\text{KQqx}}{\text{m}\ell^3}$
Time period $=2\pi\sqrt{\frac{\text{displacement}}{\text{acceleration}}}$
$=2\pi\sqrt{\frac{\text{xm}\ell^3}{4\text{KQqx}}}$
$=2\pi\sqrt{\frac{\text{m}\ell^3}{4\text{KQq}}}$
$=\sqrt{\frac{4\pi^2\text{m}\ell^34\pi\in_0}{4\text{Qq}}}$
$=\sqrt{\frac{4\pi^3\text{m}\ell^3\in_0}{\text{Qq}}}$
$=\sqrt{4\pi^3\text{md}^3\in_08\text{Qq}}$
$=\bigg[\frac{\pi^3\text{md}^3\in_0}{2\text{Qq}}\bigg]^{\frac{1}{2}}$
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