KCET · Physics · Electrostatics
The charges \(+q\) and \(+q\) are placed at the vertices of an equilateral triangle of side If the net electrostatic potential energy of the system is zero, then is equal to
- A \(-\frac{q}{2}\)
- B \(-q\)
- C \(\frac{+q}{2}\)
- D Zero
Answer & Solution
Correct Answer
(A) \(-\frac{q}{2}\)
Step-by-step Solution
Detailed explanation
Potential energy of the system
\(\mathrm{U}=\frac{\mathrm{KQq}}{1}+\frac{\mathrm{Kq}^{2}}{1}+\frac{\mathrm{KqQ}}{1}=0 \)
\( \Rightarrow \frac{\mathrm{Kq}}{1}(\mathrm{Q}+\mathrm{q}+\mathrm{Q}) =0 \)
\( \Rightarrow \mathrm{Q} =-\frac{\mathrm{q}}{2}\)
\(\mathrm{U}=\frac{\mathrm{KQq}}{1}+\frac{\mathrm{Kq}^{2}}{1}+\frac{\mathrm{KqQ}}{1}=0 \)
\( \Rightarrow \frac{\mathrm{Kq}}{1}(\mathrm{Q}+\mathrm{q}+\mathrm{Q}) =0 \)
\( \Rightarrow \mathrm{Q} =-\frac{\mathrm{q}}{2}\)
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