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elastiboysai (2327)

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Olaaa!! Perrrfect answer. 421  [532 rates]

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\mbox{The 3 charges will be 3 vertices of a tetrahedron }\\ \mbox{with the fourth vertex as the point of suspension}\\<br/>\mbox{Draw the perpendicular bisector of one side}\\<br/>\mbox{ from the point of suspension}\\<br/>\frac{F_(electr.)}{F_(gravit.)}=\frac{x/2}{\sqrt{{l^2}-\frac{x^2}{4}}}\\<br/>\mbox{approximate it as}\frac{x}{2l} (x<<l)\\<br/>\frac{2q^2}{4 \pi \epsilon_0 x^2 mg}=\frac{x}{2l}\\<br/>x=q^\frac{2}{3}(\frac{l}{\pi \epsilon_0 mg})^\frac{1}{3}\\<br/>\mbox{Now for some manipulation}\\<br/>\frac{dx}{dt}=\frac{dx}{dq} \cdot \frac{dq}{dt}\\<br/>\frac{dq}{dt}=\frac{a}{\frac{dx}{dq} \cdot \sqrt{x}}\\<br/>\frac{dx}{dq}=(\frac{l}{\pi \epsilon_0 mg})^\frac{1}{3}q^\frac{1}{3}\\<br/>\sqrt{x}=(\frac{l}{\pi \epsilon_0 mg})^\frac{1}{6}q^\frac{1}{3}\\<br/>\mbox{Substitute these and ull find}\\<br/>\frac{dq}{dt}=\frac{3a}{2}\sqrt{\frac{\pi \epsilon_0 mg}{l}}\\

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