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(A) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all unique pairs of charges.F

Vedclass · Accepted Answer

$\frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{L}$

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(A) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all unique pairs of charges.For three charges $q_1, q_2, q_3$ at distances $r_{12}, r_{23}, r_{31}$,the potential energy is $U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right)$.In this case,$q_1 = q_2 = q_3 = q$ and $r_{12} = r_{23} = r_{31} = L$.Substituting these values,we get:$U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q^2}{L} + \frac{q^2}{L} + \frac{q^2}{L} \right)$$U = \frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{L}$

Three equal electric charges,each of magnitude $q$,are placed at the vertices of an equilateral triangle of side length $L$. The electrostatic potential energy of the system is:

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