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(B) The work done required to assemble the system of charges is equal to the electrostatic potential energy $U$ of the system.The potential energy is

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$\frac{-2.6}{4\pi \varepsilon_0} \frac{q^2}{a}$

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(B) The work done required to assemble the system of charges is equal to the electrostatic potential energy $U$ of the system.The potential energy is given by the sum of the potential energies of all pairs of charges:$U = \frac{1}{4\pi\varepsilon_0} \sum_{i For the given square configuration with charges $q_1 = +q$, $q_2 = -q$, $q_3 = +q$, and $q_4 = -q$ at corners $1$, $2$, $3$, and $4$ respectively:- Four sides of length $a$: pairs $(1,2), (2,3), (3,4), (4,1)$.- Two diagonals of length $a\sqrt{2}$: pairs $(1,3), (2,4)$.Calculating the energy:$U = \frac{1}{4\pi\varepsilon_0} \left[ \frac{(+q)(-q)}{a} + \frac{(-q)(+q)}{a} + \frac{(+q)(-q)}{a} + \frac{(-q)(+q)}{a} + \frac{(+q)(+q)}{a\sqrt{2}} + \frac{(-q)(-q)}{a\sqrt{2}} \right]$$U = \frac{1}{4\pi\varepsilon_0} \left[ -\frac{q^2}{a} - \frac{q^2}{a} - \frac{q^2}{a} - \frac{q^2}{a} + \frac{q^2}{a\sqrt{2}} + \frac{q^2}{a\sqrt{2}} \right]$$U = \frac{1}{4\pi\varepsilon_0} \left[ -\frac{4q^2}{a} + \frac{2q^2}{a\sqrt{2}} \right]$$U = \frac{q^2}{4\pi\varepsilon_0 a} [ -4 + \sqrt{2} ]$Since $\sqrt{2} \approx 1.414$, the value is $-4 + 1.414 = -2.586 \approx -2.6$.Thus, $U = -\frac{2.6}{4\pi\varepsilon_0} \frac{q^2}{a}$.

The work done required to assemble the four charges at the corners of a square of side $a$, as shown in the figure, is:

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