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(C) The potential energy $U$ of a system of point charges is given by the sum of potential energies of all unique pairs of charges: $U = \sum \frac{1}

Vedclass · Accepted Answer

$\frac{1}{4\pi \varepsilon_0} \frac{3q^2}{l}$

Vedclass · Answer

(C) The potential energy $U$ of a system of point charges is given by the sum of potential energies of all unique pairs of charges: $U = \sum \frac{1}{4\pi \varepsilon_0} \frac{q_i q_j}{r_{ij}}$.For an equilateral triangle with $3$ charges $q$ at each vertex and side length $l$,there are $3$ pairs of charges,each separated by distance $l$.The potential energy for each pair is $U_{pair} = \frac{1}{4\pi \varepsilon_0} \frac{q^2}{l}$.Since there are $3$ such pairs,the total net potential energy is $U_{net} = 3 	imes \frac{1}{4\pi \varepsilon_0} \frac{q^2}{l} = \frac{1}{4\pi \varepsilon_0} \frac{3q^2}{l}$.

If $3$ charges of magnitude $q$ each are placed at the vertices of an equilateral triangle of side $l \ cm$,what is the net potential energy of the system?

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