Q is spread evenly through a sphere of radius R; r is the distance from the centre.
Derivation
- Outside: Gauss's law as for a point charge, E = kQ/r².
- Inside: a Gaussian sphere of radius r encloses Q × r³/R³.
- E × 4πr² = Qr³/(ε₀R³), so E = kQr/R³. It grows linearly from 0 at the centre to kQ/R² at the surface.
Points to remember
- E is largest at the surface.
- Inside E = ρr/3ε₀, where ρ is the charge density.
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 uniformly charged non-conducting sphere?
E = kQr/R³ inside, kQ/r² outside. Q is spread evenly through a sphere of radius R; r is the distance from the centre.
How do you derive the electric field due to a uniformly charged non-conducting sphere?
Outside: Gauss's law as for a point charge, E = kQ/r². Inside: a Gaussian sphere of radius r encloses Q × r³/R³. E × 4πr² = Qr³/(ε₀R³), so E = kQr/R³. It grows linearly from 0 at the centre to kQ/R² at the surface.