An electric dipole of dipole moment `oversetto p ` consists of point charges +q and -q separated by a distance '2a' apart. Deduce the expression for the electric field `oversetto E ` due to the dipole on its axial line in term of the dipole moment `oversetto p`. Hence show that in the limit ` r gt gt a, oversetto E = (2 oversetto p)/( 4 pi in _0 r^(3)) `
(b) Given the electric field in the region `oversetto E = 2x hati ` Find the net electric flux through the cube and the charge enclosed by it.

Explanation

As electric field is given by ` overset to E = 2x hati ` , it is obvious that electric flux `phi_in = int oversetto E . overset to (ds) ` is finite for only two surface marked 1 and 2 in which lie in Y-Z plane and for all remaining four surface of cube flux is zero .
Here area of each surface `s= a^(2) `
For face 1,x =0 and `oversetto E = 0 ` . Therefore electric flux ` phi_1 =0.`
For face 2,x = a and `oversetto E =2 a hati` , Therefore electric flux
` phi _2 = overset to E . oversetto s= (2 a hati).( a^(2) hatl ) = 2a^(3) `
` therefore ` Net electric flux through the cube `phi_in =phi_1+phi_2 =0 =2a^(3) =2a^(3) `
and the charge enclosed by the cube `q= in_0 phi_in =in_0 2a^(3) =2 in_0 a^(3) `
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