question_answer

A charge q is located at the centre of a cube. The electric flux through any face is

A) \[\frac{4\pi q}{6(4\pi {{\varepsilon }_{0}})}\]                  

B)  \[\frac{4\pi q}{6(4\pi {{\varepsilon }_{0}})}\]

C)  \[\frac{q}{6(4\pi {{\varepsilon }_{0}})}\]                       

D)  \[\frac{2\pi q}{6(4\pi {{\varepsilon }_{0}})}\]

Correct Answer: A

Solution :

This problem can be solved by symmetry consideration and Gauss Law. The total flux passing through the close cube \[\phi =\frac{q}{{{\varepsilon }_{0}}}.\] All the six surfaces are symmetrical with respect  to charge, hence they will have equal contribution of the flux. So, flux through any one face: \[\phi '=\frac{\phi }{6}=\frac{q}{6{{\varepsilon }_{0}}}.\] \[{{\phi }_{face}}=\frac{q}{6{{\varepsilon }_{0}}}=\frac{4\pi q}{6(4\pi {{\varepsilon }_{0}})}\]