Question

Show that the intensity of electric field at a point in end on position i.e., axial position of an electric dipole is given by:

`E = (1/(4πε_0))(2pr)/(r^2 - l^2)^2`

where the terms have their usual meaning.

Answer 1

Consider an electric dipole consisting of two charges, −q and +q, separated by a distance 2l. Let O be the centre of the dipole.

• The distance of point P (the observation point) from the centre O is r.
• The distance of P from charge +q is (r − l).
• The distance of P from charge −q is (r + l).

Electric fields due to charges

`E = (1/(4πε_0))q/r^2`

Field due to +q (E₁): Directed away from the charge.

`E_1 = (1/(4πε_0))q/(r - l)^2`

Field due to −q (E₂): Directed towards the charge.

`E_2 = (1/(4πε_0))q/(r + l)^2`

Since E₁ and E₂ act along the same line but in opposite directions, the net intensity E is the difference between them (E₁ > E₂):

E = E₁ − E₂

E = `1/(4πε_0)[q/(r - l)^2 - q/(r + l)^2]`

= `q/(4πε_0)[1/(r - l)^2 - 1/(r + l)^2]`

= `q/(4πε_0)[((r + l)^2 - (r - l)^2)/((r - l)^2(r + l)^2)]`

= `q/(4πε_0)[((r + l)^2 - (r - l)^2)/(r^2 - l^2)^2]`

= `q/(4πε_0)[((r^2 + 2rl + l^2) - (r^2 - 2rl + l^2))/(r^2 - l^2)^2]`

= `q/(4πε_0)[((r^2 + 2rl + l^2 - r^2 + 2rl - l^2))/(r^2 - l^2)^2]`

= `q/(4πε_0)[(4rl)/(r^2 - l^2)^2]`

We know that the electric dipole moment is p = q × 2l. Rearranging the numerator:

= `1/(4πε_0)[((2*ql)*2r)/(r^2 - l^2)^2]`

Substituting p = q × 2l

= `(1/(4πε_0))(2pr)/(r^2 - l^2)^2`