Question

With the help of a labelled diagram, obtain an expression for the electric field intensity 'E' at a point P in broad side position (i.e., equatorial plane) of an electric dipole.

Answer 1

Let E₁ and E₂ be the magnitudes of the intensities of the electric field at P due to the charges + q and - q of the dipole respectively. The distance of P from each charge is \(\sqrt{r^ 2+l^ 2}\).

Therefore,

\(E_1=\frac{1}{4\pi \epsilon_0}\,\frac{q}{(r^2+l^2)},\) away from + q

and \(E_2=\frac{1}{4\pi\epsilon_0}\,\frac{q}{(r^ 2+l^ 2)},\) towards - q

The magnitudes of E₁ and E₂ are equal (but directions are different). On resolving E₁ and E₂ into two components parallel and perpendicular to AB, the components perpendicular to AB. (E₁ sin θ and E₂ sin θ) cancel each other (because they are equal and opposite) while the components parallel to AB (E₁ cos θ and E₂ cos θ), being in the same direction, add up (Fig. b).

Hence the resultant intensity of electric field at the point P is

E = E₁ cosθ + E₂ cos θ

\(=\frac{1}{4\pi \epsilon _0}\,\frac{q}{(r^ 2+l^2)}\,cos\theta+\frac{1}{4\pi \epsilon _0}\,\frac{q}{(r^2+l^2)}\,cos \theta\)

\(=\frac{1}{4\pi \epsilon _0}\,\frac{q}{(r^2+l^2)}2\,cos\theta\)

Now, from figure (a),

\(cos\theta=\frac{BO}{BP}=\frac{l}{(r^2+l^2)^{\frac{1}{2}}}\)

\(E=\frac{1}{4\pi \epsilon_0}\,\frac{q}{(r^2+l^2)}\,\frac{2l}{(r^2+l^2)^{\frac{1}{2}}}\)

\(=\frac{1}{4\pi \epsilon_0} \,\frac{2ql}{(r^2+l^2)^{\frac{3}{2}}}\)

But 2ql = p (moment of electric dipole).

 \(\therefore\,E =\frac{1}{4\pi \epsilon _0}\,\frac{p}{(r^2+l^2)^{\frac{3}{2}}}\)

The direction of electric field E is 'antiparallel' to the dipole axis. 

If r is very large compared to l (r > > l). then l² may be neglected in comparison to r².

Then, the electric field intensity at the point P due to the dipole is

\(E =\frac{1}{4\pi \epsilon_0}\,\frac{p}{r^3}\) newton/coulomb