Identical charges $(-q)$ are placed at each corner of a cube of side ' $b$ ' then the electrical potential energy of charge (+ $q$ ) which is placed at the centre of cube will be:

• A

  $\frac{-4 \sqrt{2} q^2}{\pi \varepsilon_0 b}$
• B

  $\frac{-8 \sqrt{2} q^2}{\pi \varepsilon_0 b}$
• Correct answer
  C

  $\frac{-4 q^2}{\sqrt{3} \pi \varepsilon_0 b}$
• D

  $\frac{8 \sqrt{2} q^2}{4 \pi \varepsilon_0 b}$

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 $$ \begin{aligned} & +q \text { is } v=8 \times \frac{1}{4 \pi \varepsilon_0} \frac{a(-a)}{r} \\ & =8 \times \frac{1}{4 \pi \varepsilon_0} \frac{q(-a)}{\frac{\sqrt{3}}{2} b} \\ & U=\frac{-4 q^2}{\sqrt{3 \pi \varepsilon_0 b}} \end{aligned} $$

Hence, the correct answer is option: 'C'.