Field of an electric dipole of length '2a' and moment |p⃗| in its equatorial plane at a distance x from the centre would be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( E \cong \displaystyle\frac{-\mathbf{p}}{4\pi\epsilon_0 x^{3}} \) for \( x \gt\gt a \)
On the equatorial plane, the point is at the same distance \( \sqrt{x^{2}+a^{2}} \) from both charges, so the two fields have equal magnitude. Their components perpendicular to the dipole axis cancel, and the components along the axis add, pointing from +q towards −q, that is opposite to the dipole moment.
\( \displaystyle E_{\pm q} = \frac{q}{4\pi\varepsilon_0 (x^{2}+a^{2})} \)
\( \displaystyle E = 2E_{\pm q}\cos\theta = \frac{2q}{4\pi\varepsilon_0 (x^{2}+a^{2})} \times \frac{a}{\sqrt{x^{2}+a^{2}}} \)
\( \displaystyle \vec{E} = \frac{-\vec{p}}{4\pi\varepsilon_0 (x^{2}+a^{2})^{3/2}} \qquad (p = 2qa) \)
For a point far away, \( x \gt\gt a \), the term \( a^{2} \) can be neglected:
\( \displaystyle \vec{E} \cong \frac{-\vec{p}}{4\pi\varepsilon_0 x^{3}} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
The Field of an Electric Dipole – “… The total electric field is opposite to \( \hat{\mathbf{p}} \).”
The Field of an Electric Dipole – “… In terms of p, the electric field of a dipole at large distances takes simple forms:
At a point on the equatorial plane
\( \displaystyle \mathbf{E} = -\frac{\mathbf{p}}{4\pi\varepsilon_0 r^3} \qquad (r \gg a) \qquad (1.21) \)”