q is the charge and r the distance from it; k = 1/4πε₀ = 9 × 10⁹ N m² C⁻².
Derivation
- Coulomb's law gives the force on a test charge q₀: F = kqq₀/r².
- The field is force per unit test charge: E = F/q₀ = kq/r².
- It points away from a positive charge and towards a negative one.
Points to remember
- E falls as 1/r².
- Fields from several charges add as vectors (superposition).
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 a point charge?
E = kq/r² = q/(4πε₀r²). q is the charge and r the distance from it; k = 1/4πε₀ = 9 × 10⁹ N m² C⁻².
How do you derive the electric field due to a point charge?
Coulomb's law gives the force on a test charge q₀: F = kqq₀/r². The field is force per unit test charge: E = F/q₀ = kq/r². It points away from a positive charge and towards a negative one.