The ratio of the magnitude of electric field at a large distance 'r'…

The correct answer is Option (2) → 1:2

For a short dipole at a large distance r, the field on the axis is twice the field at the same distance in the equatorial plane.

    \( \displaystyle E_{\mathrm{axis}} = \frac{1}{4\pi\varepsilon_0}\,\frac{2p}{r^{3}} \)

    \( \displaystyle E_{\mathrm{eq}} = \frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^{3}} \)

    \( \displaystyle \frac{E_{\mathrm{eq}}}{E_{\mathrm{axis}}} = \frac{p/r^{3}}{2p/r^{3}} = \frac{1}{2} \)

    \( E_{\mathrm{eq}} : E_{\mathrm{axis}} = 1 : 2 \)

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 – “… 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

    \( \displaystyle \mathbf{E} = -\frac{\mathbf{p}}{4\pi\varepsilon_0 r^{3}} \qquad (r \gg a) \qquad (1.21) \)”