A) \[-\frac{Q}{2}\]
B) \[+\frac{Q}{2}\]
C) \[-\frac{Q}{4}\]
D) \[+\frac{Q}{4}\]
Correct Answer: C
Solution :
Let \[{{V}_{1}}\]be the potential at the centre of the cube due to one charge. \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{x}\] and \[x=\frac{a\sqrt{3}}{2}\] Potential due to all eight comers of the cube \[\Rightarrow \] \[V=8{{V}_{1}}\] \[=8\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{\sqrt{\frac{3}{2}}\times a} \right]\] \[=\frac{16Q}{4\pi {{\varepsilon }_{0}}\sqrt{3}\times a}\] \[=\frac{4Q}{a\sqrt{3}\pi {{\varepsilon }_{0}}}\]
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