MHT CET · Physics · Electrostatics
Three charges \(Q,+q\) and \(+q\) are placed at the vertices of right-angle triangle as shown. The net electrostatic energy of the configuration is zero. The value of \(Q\) is
- A \(-2 q\)
- B \(-\frac{q}{1+\sqrt{2}}\)
- C \(+q\)
- D \(\frac{-\sqrt{2} q}{\sqrt{2}+1}\)
Answer & Solution
Correct Answer
(D) \(\frac{-\sqrt{2} q}{\sqrt{2}+1}\)
Step-by-step Solution
Detailed explanation
The potential energy of the system can be written pair-wise as follows:
\(\begin{aligned}
& U=\frac{k Q q}{l}+\frac{k q q}{l}+\frac{k Q q}{\sqrt{2} l}=0 \\
& Q=-\frac{\sqrt{2} q}{\sqrt{2}+1}
\end{aligned}\)
\(\begin{aligned}
& U=\frac{k Q q}{l}+\frac{k q q}{l}+\frac{k Q q}{\sqrt{2} l}=0 \\
& Q=-\frac{\sqrt{2} q}{\sqrt{2}+1}
\end{aligned}\)
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