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Step-by-Step Solution
Step 1: Identify the Charges and Their Positions
We have a cube of side length $b$. Each of its 8 corners holds a charge $(-q)$, and there is a charge $(+q)$ placed at the center of the cube.
Step 2: Calculate the Distance from the Center to Each Corner
The body diagonal of a cube with side $b$ is $ \sqrt{3} \, b$. Since the charge $(+q)$ is at the center, the distance $d$ from the center to any corner is half of the body diagonal:
$$ d = \frac{\sqrt{3} \, b}{2}. $$
Step 3: Compute the Potential Energy Contribution from One Corner Charge
The electrostatic potential energy between two point charges $q_1$ and $q_2$ separated by distance $r$ is given by
$$ U = \frac{q_1 \, q_2}{4 \pi \varepsilon_0 \, r}. $$
Here, for one corner charge $(-q)$ and the center charge $(+q)$, the distance is $d = \frac{\sqrt{3} \, b}{2}$. Thus,
$$ U_{\text{one corner}}
= \frac{(+q)(-q)}{4 \pi \varepsilon_0 \left(\frac{\sqrt{3}\, b}{2}\right)}
= - \frac{q^2}{4 \pi \varepsilon_0} \times \frac{2}{\sqrt{3}\, b}
= - \frac{q^2}{2 \pi \varepsilon_0 \, \sqrt{3}\, b}. $$
Step 4: Sum the Contributions from All 8 Corner Charges
Since all 8 charges are identical and equidistant from the center, the total electrostatic potential energy of the center charge $(+q)$ is:
$$ U_{\text{total}} = 8 \times \left( - \frac{q^2}{2 \pi \varepsilon_0 \, \sqrt{3}\, b} \right)
= - \frac{8\, q^2}{2 \pi \varepsilon_0 \, \sqrt{3}\, b}
= - \frac{4\, q^2}{\pi \varepsilon_0 \, \sqrt{3}\, b}. $$
Step 5: Final Answer
The electrostatic potential energy of the charge $(+q)$ placed at the center of the cube is:
$$ \boxed{-\frac{4\, q^2}{\sqrt{3}\, \pi \varepsilon_0\, b}}. $$
