MHT CET · Physics · Electrostatics
Three particles, each having a charge of \(10 \mu \mathrm{C}\) are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is \(\left(\right.\) Given \(\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~N}-\mathrm{m}^{2} \mathrm{C}^{2}\right)\)
- A Zero
- B \(\infty\)
- C \(27 \mathrm{~J}\)
- D \(100 \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(27 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
For a pair of charges
\(U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \)
\( U_{\text {system }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right. \) \( +\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100} \) \( \left.+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right] \)
\( =3 \times 9 \times 10^{9} \times \frac{100 \times 10^{-12} \times 100}{10} \)
\( =27 \mathrm{~J}\)
\(U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \)
\( U_{\text {system }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right. \) \( +\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100} \) \( \left.+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right] \)
\( =3 \times 9 \times 10^{9} \times \frac{100 \times 10^{-12} \times 100}{10} \)
\( =27 \mathrm{~J}\)
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