p = q × 2a is the dipole moment and r ≫ a is the distance from the centre along the axis.
Derivation
- At distance r on the axis, the +q end is r − a away and the −q end is r + a away.
- E = kq/(r − a)² − kq/(r + a)² = kq × 4ar/(r² − a²)².
- For r ≫ a: E = 4kqa/r³ = 2kp/r³, pointing along p.
Points to remember
- Axial field is twice the equatorial field at the same distance.
- A dipole field falls as 1/r³, faster than a point charge.
All field formulas
| Field due to | Formula |
|---|---|
| Electric field due to a point charge | E = kq/r² = q/(4πε₀r²) |
| Electric field due to an electric dipole on its axis | E = 2kp/r³ |
| Electric field due to an electric dipole on its equatorial line | E = kp/r³ |
| Electric field due to a charged ring on its axis | E = kQx/(x² + R²)^(3/2) |
| Electric field due to an infinite line charge | E = λ/(2πε₀r) = 2kλ/r |
| Electric field due to an infinite charged sheet | E = σ/(2ε₀) |
| Electric field due to a charged conducting sphere (or spherical shell) | E = kQ/r² outside, 0 inside |
| Electric field due to a uniformly charged non-conducting sphere | E = kQr/R³ inside, kQ/r² outside |
| Magnetic field due to a long straight wire | B = μ₀I/(2πr) |
| Magnetic field due to a finite straight wire | B = (μ₀I/4πd)(sin α + sin β) |
| Magnetic field due to a circular loop at its centre | B = μ₀I/(2R) |
| Magnetic field due to a circular loop on its axis | B = μ₀IR²/(2(R² + x²)^(3/2)) |
| Magnetic field due to a long solenoid | B = μ₀nI |
| Magnetic field due to a toroid | B = μ₀NI/(2πr) |
Questions students ask
What is the electric field due to an electric dipole on its axis?
E = 2kp/r³. p = q × 2a is the dipole moment and r ≫ a is the distance from the centre along the axis.
How do you derive the electric field due to an electric dipole on its axis?
At distance r on the axis, the +q end is r − a away and the −q end is r + a away. E = kq/(r − a)² − kq/(r + a)² = kq × 4ar/(r² − a²)². For r ≫ a: E = 4kqa/r³ = 2kp/r³, pointing along p.