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(C) The potential energy of a system of point charges is given by $U = \sum \frac{1}{4\pi \varepsilon_0} \frac{q_i q_j}{r_{ij}}$.Let the charges be $q

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$ -\frac{7q^2}{8\pi \varepsilon_0 a} $

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(C) The potential energy of a system of point charges is given by $U = \sum \frac{1}{4\pi \varepsilon_0} \frac{q_i q_j}{r_{ij}}$.Let the charges be $q_1 = q$ at $x = -a$,$q_2 = q$ at $x = a$,and $q_3 = -2q$ at $x = 0$.The distances between the charges are $r_{12} = 2a$,$r_{13} = a$,and $r_{23} = a$.The total potential energy is $U = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right)$.Substituting the values: $U = \frac{1}{4\pi \varepsilon_0} \left( \frac{q \cdot q}{2a} + \frac{q \cdot (-2q)}{a} + \frac{q \cdot (-2q)}{a} \right)$.$U = \frac{1}{4\pi \varepsilon_0} \left( \frac{q^2}{2a} - \frac{2q^2}{a} - \frac{2q^2}{a} \right) = \frac{1}{4\pi \varepsilon_0} \left( \frac{q^2 - 4q^2 - 4q^2}{2a} \right)$.$U = \frac{1}{4\pi \varepsilon_0} \left( -\frac{7q^2}{2a} \right) = -\frac{7q^2}{8\pi \varepsilon_0 a}$.

Two equal charges $q$ are placed at a distance of $2a$ and a third charge $-2q$ is placed at the midpoint. The potential energy of the system is:

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