18.3 Electric force between point charges
- Syllabus
- 9702–2028–2029
- Topic
- 18.3
- Level
- A2
Learning objectives
18.3.1That, for a point outside a spherical conductor, the charge on the sphere• understand that, for a point outside a spherical conductor, the charge on the sphere may be considered to be a point charge at its centre18.3.2Coulomb’s law F = Q1Q2 / (4πε0 r 2) for the force between two point charges• recall and use Coulomb’s law F = Q1Q2 / (4πε0 r 2) for the force between two point charges in free space
Outside a charged spherical conductor, its field is equivalent to a point charge at the centre
For a spherical conductor in electrostatic equilibrium, excess charge resides on the surface and the external field acts as if total charge Q were at the centre.
Use centre distance r and remember the field inside the conductor is zero in electrostatic equilibrium.
A charged metal sphere produces an external field decreasing as 1/r², even though the charge is spread over its surface.
The point-charge equivalence does not describe the field inside the conductor or an irregular charged object without symmetry.
Use Coulomb's inverse-square law for two point charges
Coulomb's law states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation.
forcemagnitudeF=∣Q1Q2∣/(4πε0r2),whereε0=8.85×10−12Fm−1
| Charges | Direction of the two equal and opposite forces |
|---|---|
| same sign | repel along the joining line |
| opposite signs | attract along the joining line |
For +3.0 nC and -5.0 nC separated by 0.040 m, F=(3.0×10⁻⁹)(5.0×10⁻⁹)/[4π(8.85×10⁻¹²)(0.040)²]=8.43×10⁻⁵ N. The force is attractive.
Qdoubled→Fdoubled;rdoubled→Fdividedby4
Use centre-to-centre separation in metres and square it. Force magnitude is positive; charge signs choose attraction or repulsion. With several charges, calculate each force vector and add vectors.