How does the magnitude of the electric field on the equatorial plane of an electric dipole vary with the distance r from its centre at large distances \( (r\gg a) \), where 2a is the length of the dipole?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Proportional to \( 1/r^{3} \)
The field of a dipole is the sum of the fields of \( +q \) and \( -q \). Far from the dipole these two fields nearly balance, so the dipole field falls off faster than the \( 1/r^{2} \) field of a single charge. On the equatorial plane, for \( r\gg a \), it is
\( \displaystyle \vec{E}=-\frac{\vec{p}}{4\pi\varepsilon_0r^{3}} \)
where the dipole moment \( p=q\times2a \) is fixed for a given dipole. Hence \( E\propto1/r^{3} \).
Option (1) Proportional to \( r^{3} \) — Incorrect. The field weakens as the point moves away from the dipole. A field growing as \( r^{3} \) would grow without limit far away.
Option (2) Proportional to \( 1/r^{3} \) — Correct. With \( p \) fixed, the magnitude \( p/4\pi\varepsilon_0r^{3} \) falls as the cube of the distance. Doubling \( r \) makes the field one-eighth.
Option (3) Proportional to \( r^{2} \) — Incorrect. The dipole field decreases as the distance increases. So it cannot grow as \( r^{2} \).
Option (4) Proportional to \( 1/r \) — Incorrect. A \( 1/r \) law would fall off even more slowly than the \( 1/r^{2} \) field of a single charge. The dipole field falls off faster than that.
NCERT (2026-27):
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
\( \displaystyle \mathbf{E}=-\frac{\mathbf{p}}{4\pi\varepsilon_0r^{3}} \)\( \displaystyle \quad(r\gg a) \qquad (1.21) \)”
The Field of an Electric Dipole – “Notice the important point that the dipole field at large distances falls off not as 1/r2 but as 1/r3.”