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(C) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{1}{4 \pi \epsilon_0} \frac{q_i q_j}{r_{ij}}$.For three identical

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$4.1$

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(C) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{1}{4 \pi \epsilon_0} \frac{q_i q_j}{r_{ij}}$.For three identical charges $q$ placed at the vertices of an equilateral triangle of side $r$,the total potential energy is $U = 3 	imes \left( \frac{1}{4 \pi \epsilon_0} \frac{q^2}{r} ight)$.Given: $q = 3 \mu C = 3 	imes 10^{-6} \ C$,$r = 6 \ cm = 0.06 \ m$,and $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 \ N \cdot m^2/C^2$.Substituting the values:$U = 3 	imes \left( 9 	imes 10^9 	imes \frac{(3 	imes 10^{-6})^2}{0.06} ight)$$U = 3 	imes \left( 9 	imes 10^9 	imes \frac{9 	imes 10^{-12}}{0.06} ight)$$U = 3 	imes \left( \frac{81 	imes 10^{-3}}{0.06} ight)$$U = 3 	imes 1.35 = 4.05 \ J$.Rounding to the nearest value,we get $4.1 \ J$.

Three charges,each of magnitude $3 \mu C$,are placed on the vertices of an equilateral triangle of side $6 \ cm$. The net potential energy of the system will be nearly $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ SI \ unit\right]$. (in $J$)

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