The electric field \( (\mathbf{E}) \) due to an electric dipole of size \( 2a \) and dipole moment \( \mathbf{p} \) on its equatorial plane at a distance \( r\ (r \gt\gt a) \) from its center is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \mathbf{E} = \displaystyle\frac{-\mathbf{p}}{4\pi\varepsilon_{0}r^{3}} \)
For a short dipole (r much greater than a), the field at a point on the equatorial plane is set by the components of the fields of +q and −q parallel to the dipole axis; the components along the line joining the point to the centre cancel. The resultant points from +q to −q, that is, opposite to the dipole moment p.
\( \displaystyle \mathbf{E} = -\frac{2q\,a}{4\pi\varepsilon_{0}(r^{2} + a^{2})^{3/2}}\,\hat{\mathbf{p}} \)
For r much greater than a, \( (r^{2} + a^{2})^{3/2} \approx r^{3} \), and \( 2qa\,\hat{\mathbf{p}} = \mathbf{p} \):
\( \displaystyle \mathbf{E} = \frac{-\mathbf{p}}{4\pi\varepsilon_{0}r^{3}} \)
NCERT (2026-27):
The Field of an Electric Dipole – “… In terms of \( \mathbf{p} \), the electric field of a dipole at large distances takes simple forms:
At a point on the dipole axis
\( \displaystyle \mathbf{E} = \frac{2\mathbf{p}}{4\pi\varepsilon_0 r^{3}} \qquad (r \gg a) \qquad (1.20) \)
At a point on the equatorial plane
\( \displaystyle \mathbf{E} = -\frac{\mathbf{p}}{4\pi\varepsilon_0 r^{3}} \qquad (r \gg a) \qquad (1.21) \)
Notice the important point that the dipole field at large distances falls off not as \( 1/r^{2} \) but as \( 1/r^{3} \).”