σ is the charge per unit area.
Derivation
- Take a cylindrical Gaussian 'pillbox' that pokes through the sheet, with end faces of area A.
- Flux leaves through both ends: 2EA.
- Gauss's law: 2EA = σA/ε₀, so E = σ/2ε₀. It doesn't depend on distance.
Points to remember
- Uniform field on both sides.
- Just outside a charged conductor, E = σ/ε₀, because all the flux goes out one side.
- Between two oppositely charged sheets (a capacitor): σ/ε₀; outside: 0.
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 infinite charged sheet?
E = σ/(2ε₀). σ is the charge per unit area.
How do you derive the electric field due to an infinite charged sheet?
Take a cylindrical Gaussian 'pillbox' that pokes through the sheet, with end faces of area A. Flux leaves through both ends: 2EA. Gauss's law: 2EA = σA/ε₀, so E = σ/2ε₀. It doesn't depend on distance.