The magnitudes of electric field and of the electric potential on the axis of an electric dipole at a distance 'r' from its centre are E and V, respectively. The magnitude of the electric field and potential at the same distance 'r' on its equatorial line will be: (given r>>a, where 2a is dipole length)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{E}{2},\ 0 \)
For a short dipole of moment p (\( r\gg a \)), the field and potential at a distance r are:
On the axis
\( \displaystyle E = \frac{2p}{4\pi\varepsilon_0r^{3}},\qquad V = \frac{p}{4\pi\varepsilon_0r^{2}} \)
On the equatorial line (the perpendicular bisector)
\( \displaystyle E_{\mathrm{eq}} = \frac{p}{4\pi\varepsilon_0r^{3}} = \frac{E}{2},\qquad V_{\mathrm{eq}} = 0 \)
The field at an equatorial point is half the axial field at the same distance. The potential is zero there because every point on the equatorial line is equidistant from the two equal and opposite charges, so their potentials cancel.
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 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
\( \displaystyle \mathbf{E} = -\frac{\mathbf{p}}{4\pi\varepsilon_0 r^3} \qquad (r \gg a) \qquad (1.21) \)”
Potential Due to an Electric Dipole – “… From Eq. (2.15), potential on the dipole axis (\( \theta = 0, \pi \)) is given by
\( \displaystyle V = \pm\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2} \qquad (2.16) \)
(Positive sign for \( \theta = 0 \), negative sign for \( \theta = \pi \).) The potential in the equatorial plane (\( \theta = \pi/2 \)) is zero.”