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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 $\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~N}-\mathrm{m}^{2} \mathrm{C}^{2}\right)$
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Verified Answer
The correct answer is:
$27 \mathrm{~J}$
For a pair of charges
$$
\begin{array}{c}
U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \\
U_{\text {system }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right. \\
+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100} \\
\left.+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right] \\
=3 \times 9 \times 10^{9} \times \frac{100 \times 10^{-12} \times 100}{10} \\
=27 \mathrm{~J}
\end{array}
$$
$$
\begin{array}{c}
U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \\
U_{\text {system }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right. \\
+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100} \\
\left.+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right] \\
=3 \times 9 \times 10^{9} \times \frac{100 \times 10^{-12} \times 100}{10} \\
=27 \mathrm{~J}
\end{array}
$$
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