Physics · Electrostatic Potential And Capacitance · NEET
A single point charge has potential V = kq/r, falling as 1/r. A dipole is +q and -q very close together. Far away their potentials almost cancel, and only the tiny difference (because +q and -q are at slightly different distances) survives. This leftover difference falls off faster, as 1/r². So the dipole potential dies out quicker than a point charge because the two opposite charges nearly cancel.
θ is the angle between the dipole axis (direction of the dipole moment vector p, which points from -q to +q) and the line joining the centre of the dipole to the point P where you measure potential. When θ = 0° the point is on the axis (nearer the +q side), when θ = 90° it is on the equatorial line.
On the equatorial line θ = 90°, so cosθ = 0, giving V = 0. Physically, the point is equally far from +q and -q, so the positive potential from +q exactly cancels the negative potential from -q. The net potential is zero even though the electric field there is not zero.
Potential is always a scalar. You do NOT add potentials like vectors. You simply add the arithmetic values V₊ = +kq/r₊ and V₋ = -kq/r₋. Only the direction of the point (through cosθ) matters, not vector addition. This is why dipole potential problems are usually easier than dipole field problems.
The exact potential involves the two distances r₊ and r₋ separately. When the point P is far compared to the dipole size (r >> 2a, where 2a is the separation), we approximate r₊ ≈ r - a cosθ and r₋ ≈ r + a cosθ. This lets the two terms combine neatly into V = kp cosθ / r². For NEET, almost every dipole is treated as 'short'.
A short electric dipole has a dipole moment of 16 × 10⁻⁹ C·m. The electric potential due to the dipole at a point at distance 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 at 2 m from the centre of a dipole of moment 4 × 10⁻⁶ C·m is ±9 × 10³ V (1/4πε₀ = 9 × 10⁹ SI units). Reason R: V = ±2P/(4πε₀ r²) for any axial point at 2 m from the centre.
A dipole of moment 5 × 10⁻⁶ C·m is aligned with a uniform field of 4 × 10⁵ N/C. It is then rotated by 60° from the field. The change in potential energy of the dipole is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
V = kp cosθ / r², where k = 1/(4πε₀) = 9×10⁹ SI units, p is the dipole moment, r is the distance from the centre, and θ is the angle between the dipole axis and the line to the point.
On the axis θ = 0°, so cosθ = 1 and V = kp / r². It is positive on the +q side and negative on the -q side. There is no factor of 2 (unlike the axial field).
On the equatorial line θ = 90°, cosθ = 0, so V = 0. The point is equally distant from +q and -q, so their potentials cancel exactly.
A point charge gives V = kq/r (falls as 1/r) and does not depend on direction. A dipole gives V = kp cosθ/r² (falls as 1/r², faster) and depends on the angle θ.
No. Potential is a scalar. We add the numbers V₊ = +kq/r₊ and V₋ = -kq/r₋ arithmetically. This is why dipole potential is simpler to compute than dipole field.
