A) \[\frac{4a}{3\pi {{\varepsilon }_{0}}x}\]
B) \[\frac{4q}{\sqrt{3}\,\pi {{\varepsilon }_{0}}x}\]
C) \[\frac{3q}{4\pi {{\varepsilon }_{0}}x}\]
D) \[\frac{2q}{\sqrt{3}\pi {{\varepsilon }_{0}}x}\]
Correct Answer: B
Solution :
(b) \[\frac{4q}{\sqrt{3}\,\pi {{\varepsilon }_{0}}x}\] The length of diagonal of the cube of each side x is \[\sqrt{3{{x}^{2}}}=x\sqrt{3}\] \[\therefore \]Distance between centre of cube and each vertex, \[r=\frac{x\sqrt{3}}{2}\] Now, potential, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Since, cube has 8 vertices and B charges each of value q are present there \[\therefore \,\,\,\,V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{8q}{\frac{x\sqrt{3}}{2}}=\frac{4q}{\sqrt{3}\pi {{\varepsilon }_{0}}x}\]
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