Coulomb's lawAQA A-Level Physics: Revision notes

The electrostatic force between two point charges in a vacuum is

$F = \dfrac{Q_1Q_2}{4\pi\varepsilon_0 r^2}$

where $Q_1$ and $Q_2$ are the charges, $r$ is the separation and $\varepsilon_0$ is the permittivity of free space, $8.85\times10^{-12}$ F m⁻¹. The force acts along the line joining the charges. Like charges repel and unlike charges attract. The forces on the two charges are equal and opposite (Newton's third law).

It is an inverse square law: doubling $r$ divides $F$ by 4, and the force is proportional to each charge.

Charges of $+3.0$ nC and $-5.0$ nC are 4.0 cm apart in air:

$F = \dfrac{3.0\times10^{-9}\times5.0\times10^{-9}}{4\pi\times8.85\times10^{-12}\times0.040^2} = 8.4\times10^{-5}$ N, attractive.

Convert units first: nC to C ($10^{-9}$), cm to m ($10^{-2}$).

Air can be treated as a vacuum when calculating the force between charges, because the permittivity of air is almost the same as $\varepsilon_0$.

For a charged sphere the charge may be considered to be at the centre, so $r$ is the distance between centres. This is why, for example, an 8.0 nC sphere with a ball 10.0 cm from its centre gives the same force as a 2.0 nC sphere with the ball at 5.0 cm.

Forces are vectors: the resultant on a charge is the vector sum of the forces from each other charge. Draw the directions of attraction and repulsion first, then resolve components. Where components in one direction are equal and opposite, they cancel.

Similarities: inverse square laws for point sources; field lines; potential and equipotential surfaces.

Differences: masses only attract, charges can attract or repel. Between subatomic particles the electrostatic force is vastly larger. For a proton and electron, $F_e/F_g = e^2/(4\pi\varepsilon_0Gm_em_p) \approx 2\times10^{39}$, independent of separation because both are inverse square laws.

On large scales planets are almost neutral, so electrostatic forces cancel and gravity dominates.