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Three charges $-q,+Q$ and $-q$ are placed at equal distances along a straight line. If the total $P E$ of the system is zero, then the ratio $Q / q$ becomes
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Verified Answer
The correct answer is:
$\frac{1}{4}$
The given situation is shown below
Total potential energy of the system $=0$
$$
\Rightarrow \frac{1}{4 \pi \varepsilon_{0}}\left[\frac{(-q) Q}{a}+\frac{(+Q)(-q)}{a}+\frac{(-q)(-q)}{2 a}\right]=0
$$
$$
\begin{aligned}
&\Rightarrow \quad-\frac{q Q}{a}-\frac{q Q}{a}+\frac{q^{2}}{2 a}=0 \Rightarrow \frac{2 q Q}{a}=\frac{q^{2}}{2 a} \\
&\Rightarrow \quad 2 Q=\frac{q}{2} \Rightarrow \frac{Q}{q}=\frac{1}{4}
\end{aligned}
$$
Total potential energy of the system $=0$
$$
\Rightarrow \frac{1}{4 \pi \varepsilon_{0}}\left[\frac{(-q) Q}{a}+\frac{(+Q)(-q)}{a}+\frac{(-q)(-q)}{2 a}\right]=0
$$
$$
\begin{aligned}
&\Rightarrow \quad-\frac{q Q}{a}-\frac{q Q}{a}+\frac{q^{2}}{2 a}=0 \Rightarrow \frac{2 q Q}{a}=\frac{q^{2}}{2 a} \\
&\Rightarrow \quad 2 Q=\frac{q}{2} \Rightarrow \frac{Q}{q}=\frac{1}{4}
\end{aligned}
$$
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