AP EAMCET · PHYSICS · Electrostatics
Three equal electric charges of each charge ' \(q\) ' are placed at the vertices of an equilateral triangle of side of length ' \(L\) ', then potential energy of the system is
- A \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{L}\)
- B \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{3 L}\)
- C \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 q^2}{3 L}\)
- D \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{L}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{L}\)
Step-by-step Solution
Detailed explanation
The potential energy of the charge system is \(\mathrm{U}=3\left(\frac{\mathrm{kq}}{\mathrm{~L}}\right)=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{3 \mathrm{q}^2}{\mathrm{~L}}\)
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