question_answer

The total electrical flux leaving a spherical surface of radius r, in enclosing an electric dipole of charge q is

A)  zero                                     

B)  \[\frac{q}{{{\varepsilon }_{0}}}\]

C)  \[\frac{8\pi {{r}^{2}}q}{{{\varepsilon }_{0}}}\]                   

D)  \[\frac{2q}{{{\varepsilon }_{0}}}\]

E)  \[\frac{4\pi {{r}^{2}}q}{{{\varepsilon }_{0}}}\]

Correct Answer: A

Solution :

                The centre O of the cube is the point of intersection, of the diagonals AG and BH. \[G{{B}^{2}}={{a}^{2}}+{{a}^{2}}=2{{a}^{2}}\] \[H{{B}^{2}}={{a}^{2}}+2{{a}^{2}}=3{{a}^{2}}\] \[\therefore \]  \[HB=\sqrt{3}a,\]\[HO=\frac{\sqrt{3}}{2}a\] From the symmetry of the eight charges with respect to the centre of the cube, it follows that the field at the centre due to two opposite charges cancel in pairs. Hence, total electric flux \[{{\phi }_{E}}=EA=0.\]