question_answer

A cube has point changes of magnitude q at all its vertices, electric field at the centre of the cube is:

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 :

From figure the electric field at the centre due to any charge at vertices of the cube is given by: \[E=\frac{1}{4\pi {{\varepsilon }_{0}}K}\frac{q}{{{r}^{2}}}\]   (Numerically) But the direction of electric field due to \[+q\]charge is opposite to the direction of electric field due to \[-q\]charge place diagonally opposite to each other. Hence, it is clear the net electric field at the centre of the cube due to charges at the vertices is zero.