Identical charges $(-q)$ are placed at each corner of a cube of side ' $b$ ' then the electrical potential…
- $\frac{-4 \sqrt{2} q^2}{\pi \varepsilon_0 b}$
- $\frac{-8 \sqrt{2} q^2}{\pi \varepsilon_0 b}$
- $\frac{-4 q^2}{\sqrt{3} \pi \varepsilon_0 b}$
- $\frac{8 \sqrt{2} q^2}{4 \pi \varepsilon_0 b}$
Solution
Distance between the charge $-q$ and $+q$ is
$r=\sqrt{\frac{(\sqrt{2} b)^2+(b)^2}{2}}=\frac{\sqrt{3}}{2} b$
Electric potential energy of charge
$\begin{aligned}
& +q \text { is } v=8 \times \frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{r} \\
& =8 \times \frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{\frac{\sqrt{3}}{2} b} \\
& U=\frac{-4 q^2}{\sqrt{3 \pi \varepsilon_0 b}}
\end{aligned}$Asked in: NEET 2002