Three charges $-\mathrm{q}, \mathrm{Q}$ and $-\mathrm{q}$ are placed at equal distances on a straight line.…
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- 1 : 1
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Solution
Potential energy of the system is zero.
$\begin{aligned}
& \therefore \frac{1}{4 \pi \varepsilon_0}\left(\frac{-\mathrm{q} \cdot \mathrm{Q}}{\mathrm{x}}+\frac{-\mathrm{q} \cdot \mathrm{Q}}{\mathrm{x}}+\frac{(-\mathrm{q})(-\mathrm{q})}{2 \mathrm{x}}\right)=0 \\
& \therefore-2 \mathrm{qQ}+\frac{\mathrm{q}^2}{2}=0 \\
& \therefore 2 \mathrm{Q}=\frac{\mathrm{q}}{2} \text { or } \frac{\mathrm{Q}}{\mathrm{q}}=\frac{1}{4}
\end{aligned}$Asked in: MHT CET 2021 (24 Sep Shift 1)