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(C) The distance of each corner of the cube from its center is $r = \frac{\sqrt{3}b}{2}$.There are $8$ charges of magnitude $(-q)$ at the corners.The

Vedclass · Accepted Answer

$\frac{-4q^2}{\sqrt{3}\pi \epsilon_0 b}$

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(C) The distance of each corner of the cube from its center is $r = \frac{\sqrt{3}b}{2}$.There are $8$ charges of magnitude $(-q)$ at the corners.The potential energy $U$ of a charge $q_0$ at the center is given by $U = \sum \frac{1}{4\pi \epsilon_0} \frac{q_i q_0}{r}$.Here,$q_i = -q$,$q_0 = q$,and $r = \frac{\sqrt{3}b}{2}$.$U = 8 	imes \left[ \frac{1}{4\pi \epsilon_0} \frac{(-q)(q)}{\frac{\sqrt{3}b}{2}} ight]$.$U = 8 	imes \left[ \frac{-q^2}{4\pi \epsilon_0} 	imes \frac{2}{\sqrt{3}b} ight]$.$U = \frac{-16q^2}{4\sqrt{3}\pi \epsilon_0 b} = \frac{-4q^2}{\sqrt{3}\pi \epsilon_0 b}$.

If identical charges $(-q)$ are placed at each corner of a cube of side '$b$',what will be the $E.P.E$ (Electrostatic Potential Energy) of a charge $(+q)$ placed at the center of the cube?

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