Three point charges \( 3 n C .6 n C \) and \( 9 n C \) are placed at the comers of an equilateral triangle of side \( 0.1 \mathrm{~m} \). The potential energy of the system is

Potential energy is given as \( U=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \)
The three charges are placed at the corners of an equilateral triangle as shown in the following figure. The potential
energies of the three charges are given as

\(U_{1} =\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{3 n C \times 6 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{18 \times 10^{-18}}{0.1} \)
\( U_{2} =\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{6 n C \times 9 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{54 \times 10^{-18}}{0.1} \)
\( U_{3} =\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{9 n C \times 3 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{27 \times 10^{-18}}{0.1} \)
Therefore, potential energy of the system is given as
\(U=U_{1}+U_{2}+U_{3}=\frac{1}{4 \Pi \varepsilon_{0}}\left[\frac{18+54+27}{0.1}\right] \times 10^{-18} \)
\( =\frac{9 \times 10^{9} \times 99 \times 10^{-18}}{0.1}=8910 \times 10^{-9} \mathrm{~J}\)