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(B) The initial electrostatic potential energy of the system of two charges $Q$ and $Q$ separated by distance $d$ is given by:$E = \frac{k Q^2}{d} \qu

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$-E$

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(B) The initial electrostatic potential energy of the system of two charges $Q$ and $Q$ separated by distance $d$ is given by:$E = \frac{k Q^2}{d} \quad \dots(i)$When a third charge $q_3 = -Q/2$ is placed at the midpoint,the new distance between the charges is $d/2$ for each pair.The total electrostatic potential energy $E^{\prime}$ of the system is the sum of the potential energies of all pairs:$E^{\prime} = \frac{k Q_1 Q_2}{r_{12}} + \frac{k Q_2 Q_3}{r_{23}} + \frac{k Q_1 Q_3}{r_{13}}$Substituting the values $Q_1 = Q$,$Q_2 = Q$,$Q_3 = -Q/2$,$r_{12} = d$,$r_{23} = d/2$,and $r_{13} = d/2$:$E^{\prime} = \frac{k Q^2}{d} + \frac{k Q (-Q/2)}{d/2} + \frac{k Q (-Q/2)}{d/2}$$E^{\prime} = \frac{k Q^2}{d} - \frac{k Q^2}{d} - \frac{k Q^2}{d}$$E^{\prime} = -\frac{k Q^2}{d}$Comparing this with equation $(i)$,we get $E^{\prime} = -E$.

Two equal charges of magnitude $Q$ each are placed at a distance $d$ apart. Their electrostatic energy is $E$. $A$ third charge $-Q / 2$ is brought midway between these two charges. The electrostatic energy of the system is now

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