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(B) The potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i

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$-\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{3 q^2}{2 a}$

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(B) The potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.Here,we have three charges:$q_1 = q$ at $x = -a$$q_2 = -q$ at $x = 0$$q_3 = q$ at $x = a$The pairs are $(q_1, q_2)$,$(q_2, q_3)$,and $(q_1, q_3)$.The distances are:$r_{12} = a$$r_{23} = a$$r_{13} = 2a$Calculating the potential energy:$U = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{(q)(-q)}{a} + \frac{(-q)(q)}{a} + \frac{(q)(q)}{2a} \right]$$U = \frac{1}{4 \pi \varepsilon_0} \left[ -\frac{q^2}{a} - \frac{q^2}{a} + \frac{q^2}{2a} \right]$$U = \frac{1}{4 \pi \varepsilon_0} \left[ -\frac{2q^2}{a} + \frac{q^2}{2a} \right]$$U = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{-4q^2 + q^2}{2a} \right]$$U = -\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{3 q^2}{2 a}$

Three charges are placed along the $x$-axis at $x = -a$,$x = 0$,and $x = a$ as shown in the figure. The potential energy of the system is

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