Two charges ±q are placed 5 mm apart. Determine the electric field at a point 15 cm away from O on a line passing through O and normal to the axis of dipole as shown in diagram (b). [Given: Electric field at a point P on the axis of the dipole 15 cm away from its centre O on the side of positive charge is \( 2.66\times10^{5}\,\mathrm{N/C} \) along AB] (diagram a)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.33\times10^{5}\,\mathrm{N/C} \) along BA
The dipole is short (5 mm compared with 15 cm), so its field at distance r is \( E_{\mathrm{axial}} = \frac{2p}{4\pi\varepsilon_0 r^{3}} \) on the axis and \( E_{\mathrm{equatorial}} = \frac{p}{4\pi\varepsilon_0 r^{3}} \) on the perpendicular bisector. The equatorial field is half the axial field at the same distance, and it is opposite in direction to the dipole moment \( \vec{p} \), which points from \( -q \) (A) to \( +q \) (B).
Magnitude
\( E_{\mathrm{eq}} = \frac{1}{2}E_{\mathrm{ax}} = \frac{2.66\times10^{5}}{2}\,\mathrm{N/C} = 1.33\times10^{5}\,\mathrm{N/C} \)
Direction
At P, on the side of the positive charge, the axial field is along AB, that is along \( \vec{p} \). The equatorial field is opposite to \( \vec{p} \), so it points along BA.
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 – …
