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(D) The length of the body diagonal of a cube with side $b$ is $\sqrt{3}b$. The distance of the center of the cube from each vertex is $r = \frac{\sqr

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$\frac{-4q^2}{\sqrt{3}\pi\varepsilon_0b}$

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(D) The length of the body diagonal of a cube with side $b$ is $\sqrt{3}b$. The distance of the center of the cube from each vertex is $r = \frac{\sqrt{3}b}{2}$. The electric potential energy $U$ of a charge $q$ at the center due to $8$ charges of $-q$ at the corners is given by the sum of individual potential energies: $U = \sum_{i=1}^{8} \frac{1}{4\pi\varepsilon_0} \frac{(-q)(q)}{r}$ $U = 8 	imes \left( \frac{1}{4\pi\varepsilon_0} \cdot \frac{-q^2}{\sqrt{3}b/2} ight)$ $U = 8 	imes \left( \frac{-2q^2}{4\pi\varepsilon_0\sqrt{3}b} ight)$ $U = \frac{-16q^2}{4\pi\varepsilon_0\sqrt{3}b} = \frac{-4q^2}{\sqrt{3}\pi\varepsilon_0b}$

If identical charges $(-q)$ are placed at each corner of a cube of side $b$,then the electric potential energy of charge $(+q)$ which is placed at the center of the cube will be:

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Reason : For a system of a positive test charge and a point charge,electric potential energy $=$ electric potential.

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