Question

Obtain an expression for an intensity of electric field at a point at the end of position, i.e., the axial position of an electric dipole.

x

Answer 1

Consider an electric dipole consisting of charges –q and +q separated by a small distance 2r in free space.

Let P be a point on the axial line of the dipole at a distance x from the centre O of the dipole.

(i.e OP = x)

Electric field intensity at point P due to +q charge

`E_1 = 1/(4piepsilon_0).q/(AP^2)`  (direction A to P)

`e_1 = 1/(4piepsilon_0).q/(x+r)^2`  (direction A to P)  ...(1)

Electric field intensity at point P due to –q charge

`E_2 = 1/(4piepsilon_0).q/(BP^2)`  (direction P to B)

`E_2 = 1/(4piepsilon_0).q/(x-r)^2`  (direction P to B)  ...(2)

Since E₂ > E₁ and they act in opposite directions, resultant field intensity is given by :

E = E₂ - E₁  (direction P to B)

`= 1/(4piepsilon_0).q/(x-r)^2 - 1/(4piepsilon_0) .q/(x+r)^2`

`= 1/(4piepsilon_0).q[1/(x-r^2) - 1/(x+r)^2]`

`= 1/(4piepsilon_0).q[((x+r)^2 -(x-r)^2)/((x-r)^2 (x+r)^2)]`

`=1/4piepsilon_0.q[((x+r+x-r)(x+r-x+r))/(xx^2-r^2)^2]`

`= 1/4piepsilon_0.q [(2x.2r)/(x^2-r^2)^2]`

`E= 1/(4piepsilon_0).  (2xp)/(x^2-r^2)^2 `[∵ p = 2r.q]

If the dipole is short (i.e., r << x) then r²  may be neglected as compared to x²

`E = 1/(4piepsilon_0) . (2p)/x^3`

The direction of resultant electric field E is along the dipole axis i.e., from –q charge to +q charge.