A) \[-\frac{4\sqrt{2}{{q}^{2}}}{\pi {{\varepsilon }_{0}}}\]
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
Solution :
Electrostatic potential energy of charge +q placed at the centre of cube is \[U=8\times \frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{\text{q}\,\text{(-q)}}{\text{half-diagonal}\,\text{distance}}\] \[=8\times \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-{{q}^{2}}}{b\frac{\sqrt{3}}{2}}\] \[=\frac{-4{{q}^{2}}}{\sqrt{3}\pi {{\varepsilon }_{0}}b}\]
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