(a)An electric dipole of dipole moment `vecp` consists of point charges +q and -q separated by a distance 2a apart. Deduce the expression for the electric field due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment Hence show that in the limit:
r > > a, `vecE to (2vecp)/(4piepsilon_0r^3)`
(b) Given the electric field in the region `vecE=-2xhati`, find the net electric flux through the cube and the charge enclosed by it.

Explanation

The electric field is along X-axis , thus electric flux through all the surfaces except the faces ABCD and face EFGH as shown in the figure will be zero.
Magnitude of electric field at the face EFGH of the cube will be zero as x=0
Magnitude of electric field at the face ABCD of the cube is given by E=2x=2a
Total electric flux through the cube is given by :
`phi=EA cos 0^@ =2axx(a)^2 xx(1)=2a^3 N m^2 C^(-1)`