question_answer

A charge Q\[\mu \]C is placed at the centre of cube, the flux coming out from any surface will be :

A) \[\frac{Q}{24{{\varepsilon }_{0}}}\]                                        

B) \[\frac{Q}{8{{\varepsilon }_{0}}}\]

C) \[\frac{Q}{6{{\varepsilon }_{0}}}\times {{10}^{-6}}\]                       

D) \[\frac{Q}{6{{\varepsilon }_{0}}}\times {{10}^{-3}}\]

Correct Answer: C

Solution :

From Gauss theorem, the electric flux through any closed surface is equal to \[\frac{1}{{{\varepsilon }_{0}}}\] times the net charge Q enclosed by the surface, that is, \[\text{o}{{\text{ }\!\!|\!\!\text{ }}_{E}}=\int_{{}}^{{}}{E.dA}=\frac{q}{{{\varepsilon }_{0}}}\] Total flux through six faces of the cube is \[\text{o}\text{ }\!\!|\!\!\text{  = }\frac{Q\times {{10}^{-6}}}{{{\varepsilon }_{0}}}\] Since, cube has six faces, therefore flux through one face is \[\text{o }\!\!|\!\!\text{ }{{\,}_{s}}=\frac{1\times Q\times {{10}^{-6}}}{6{{\varepsilon }_{0}}}\] \[\text{o }\!\!|\!\!\text{ }{{\,}_{s}}=\frac{Q\times {{10}^{-6}}}{6{{\varepsilon }_{0}}}\frac{N{{M}^{2}}}{C}\]