A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{6q}{3{{a}^{2}}}\]
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{8q}{{{a}^{2}}}\]
C) zero
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-8q}{{{a}^{2}}}\]
Correct Answer: C
Solution :
opposite to it. So, the net electric field at centre of the cube is zero.You need to login to perform this action.
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