Electric potential on an equatorial plane of an electric dipole is taken as zero. This is based on
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Superposition principle followed for electric charge potentials.
The potential at a point due to the dipole is the algebraic sum of the potentials of the two charges, by the superposition principle. Every point on the equatorial plane is at the same distance r from \( +q \) and from \( -q \), so the two potentials are equal in size and opposite in sign.
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} + \frac{1}{4\pi\varepsilon_0}\frac{(-q)}{r} = 0 \)
Why the other options are incorrect
Option (2) Net charge on electric dipole is zero — Incorrect. The net charge is zero at every point, yet the potential is zero only on the equatorial plane; on the axis it is \( \pm\frac{p}{4\pi\varepsilon_0 r^{2}} \), which is not zero. Zero net charge is not the reason.
Option (3) Electric potential for electric dipole exists along axial line only — Incorrect. The potential exists at every point in space.
Option (4) Electric potential due to a dipole has symmetrical distribution — Incorrect. The potential \( V = \frac{p\cos\theta}{4\pi\varepsilon_0 r^{2}} \) depends on the angle \( \theta \), so it is not symmetrical about the dipole.
NCERT (2026-27):
Potential Due to an Electric Dipole – “… Now we know that the electric field obeys the superposition principle. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_1} - \frac{q}{r_2}\right) \qquad (2.9) \)”
Potential Due to an Electric Dipole – “… The potential in the equatorial plane (\( \theta = \pi/2 \)) is zero.”