Question

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 `vecE` due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment `vecp`. Hence show that in the limit x>> a, `vecE->2vecp"/"(4piepsilon_0x^3)`

Answer 1

Electric Field on Axial Line of an Electric Dipole

Let P be at distance r from the centre of the dipole on the side of charge −q. Then, the electric field at point P due to charge −q of the dipole is given by

`vecE_-q=-q/(4piepsilon_0(r+a)^2)hatp`

Where `hatp `is the unit vector along the dipole axis (from − q to q).

Also, the electric field at point P due to charge +q of the dipole is given by

`vecE_+q=q/(4piepsilon_0(r-a)^2)hatp`

The total field at P is

`vecE=vecE_+q+vecE_-q=q/(4piepsilon_0)[1/(r-a)^2-1/(r+a)^2]hatp`

`=>vecE=q/(4piepsilon_0)(4ar)/(r^2-a^2)^2hatp `

Given:

r = x

`vecE=q/(4piepsilon_0)(4ax)/(x^2-a^2)^2hatp`

For x >> a, 

`vecE=(4qa)/(4piepsilon_0x^3)hatp`

`vecE=(2vecp)/(4piepsilon_0x^3)" "[.:vecp=(qxx2a)hatp]`