Identical charges (-q) are placed at each corner of a cube of side...

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: Identical charges (-q) are placed at each corner of a cube of side ' b ' then the electrical potential energy of charge

Identical charges (-q) are placed at each corner of a cube of side ' b ' then the electrical potential energy of charge

Options

A $\frac{-4 \sqrt{2} q^2}{\pi \varepsilon_0 b}$
B $\frac{-8 \sqrt{2} q^2}{\pi \varepsilon_0 b}$
C $\frac{-4 q^2}{\sqrt{3} \pi \varepsilon_0 b}$
D $\frac{8 \sqrt{2} q^2}{4 \pi \varepsilon_0 b}$

Correct answer

C. $\frac{-4 q^2}{\sqrt{3} \pi \varepsilon_0 b}$

Step-by-step 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}

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