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Q. Three charges $-q,- Q$ and $+q$ are placed in a straight line as shown. If the total potential energy of the system is zero, then the ratio $\frac{q}{Q}$ is
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Electrostatic Potential and Capacitance

Solution:

Potential energy of the system
$\frac{-k q Q}{x}-\frac{k Q q}{x}+\frac{k q^{2}}{2 x}=0$
$\Rightarrow \frac{-4 k q Q+k q^{2}}{2 x}=0$
$ \Rightarrow k q^{2}=4 k Q q$
$\Rightarrow \frac{q}{Q}=4$