Physics · Electrostatic Potential And Capacitance · NEET
An equatorial point is the same distance from the +q charge and the -q charge. The +q charge gives a potential +kq/r and the -q charge gives -kq/r. Being equal in size and opposite in sign, they add to zero. In the formula V = kp·cosθ/r², the equatorial line means θ = 90°, and cos 90° = 0, so V = 0. This is exact, not approximate.
On the axial line θ = 0°, so cosθ = 1, which is the largest possible value of cosθ. That makes V = kp/r², the maximum magnitude for a given distance r. On the side of the +q charge it is positive; on the side of the -q charge it is negative. Any other direction has cosθ < 1, so the potential is smaller.
No. Potential zero does NOT mean field zero. On the equatorial line the electric field is E = kp/r³ (for a short dipole), pointing anti-parallel to the dipole moment. Potential is a scalar sum that cancels, but the field is a vector and the two charges' fields add up in a fixed direction. This is a very common NEET trap.
For a dipole the potential falls as 1/r² (V ∝ 1/r²), which is faster than a single point charge where V ∝ 1/r. This is because the two opposite charges partly cancel each other at large distance. Remember: point charge → 1/r; short dipole potential → 1/r²; short dipole field → 1/r³.
A short dipole means the distance r of the point is much larger than the dipole length 2a (r >> a). Only under this condition do the clean results V = kp cosθ/r² (axial kp/r², equatorial 0) hold. NEET questions almost always say 'short dipole' so you can use these formulas directly.
A short electric dipole has a dipole moment of 16 × 10⁻⁹ C·m. The electric potential due to the dipole at a point 0.6 m from the centre, on a line making 60° with the dipole axis, is (1/4πε₀ = 9 × 10⁹ N·m²/C²):
Assertion (A): The potential at an axial point 2 m from the centre of a dipole of moment 4 × 10⁻⁶ C·m is ±9 × 10³ V (1/4πε₀ = 9 × 10⁹ SI). Reason (R): V = ±2P/(4πε₀ r²).
Try the real previous-year questions from this chapter — each with the answer and a full solution.
V = kp·cosθ / r² = (1/4πε₀)·(p cosθ)/r², where p is the dipole moment, r is the distance from the centre, and θ is the angle measured from the dipole axis (from +q side).
On the axial line θ = 0° or 180°, so V = ±kp/r². This is the maximum magnitude: positive on the +q side, negative on the -q side.
On the equatorial line θ = 90°, cos 90° = 0, so V = 0 at every equatorial point, regardless of distance.
A point charge is symmetric, so its potential kq/r depends only on distance. A dipole has a direction (from -q to +q), so the two charges' contributions depend on which side you are on — this brings in the cosθ factor.
Dipole potential falls as 1/r² while the dipole electric field falls as 1/r³. Both are faster than a single point charge (V ∝ 1/r, E ∝ 1/r²) because the opposite charges partly cancel.
