Coulomb's law (A-level only)

AQA·Physics·3.7.3.1

Coulomb's law

Two electric charges exert forces on each other. Coulomb's law gives the magnitude of this electrostatic force for two point charges in a vacuum:

F=14πε0∣Q1Q2∣r2F=\frac{1}{4\pi\varepsilon_0}\frac{|Q_1Q_2|}{r^2}

Here, $$F$$ is the force magnitude in newtons, $$Q_1$$ and $$Q_2$$ are the charges in coulombs, and $$r$$ is their separation in metres. The constant $$\varepsilon_0$$ is the permittivity of free space. The modulus signs in $$|Q_1Q_2|$$ mean that the equation gives a positive force magnitude, regardless of the signs of the charges.

Direction of the electrostatic force

The force acts along the line joining the two charges. Like charges repel, so each charge is pushed away from the other. Unlike charges attract, so each is pulled towards the other. The forces on the two charges have equal magnitudes and opposite directions.

Effect of charge and separation

The force magnitude is proportional to the product of the charge magnitudes. For example, doubling the magnitude of one charge doubles the force, provided the other charge and the separation stay the same.

The force also follows an inverse-square relationship with separation. Doubling the separation reduces the force to one-quarter of its original magnitude; halving the separation makes the force four times as large, provided the charges stay the same.

Applying Coulomb's law

Air can be treated as a vacuum when calculating the force. For a point outside a spherically symmetric charged sphere, the sphere's charge can be treated as though it were concentrated at its centre. The separation $$r$$ is then measured from the centre of the sphere to the point charge.

Comparing gravitational and electrostatic forces between particles

Two charged subatomic particles exert both gravitational and electrostatic forces on each other. To compare the magnitudes of these forces, use the particles’ masses, charges and separation.

Fg=Gm1m2r2,Fe=∣Q1Q2∣4πε0r2F_g = \frac{Gm_1m_2}{r^2}, \qquad F_e = \frac{|Q_1Q_2|}{4\pi\varepsilon_0r^2}

Here, $$m_1$$ and $$m_2$$ are the particle masses, $$Q_1$$ and $$Q_2$$ are their charges, and $$r$$ is their separation. $$G$$ is the gravitational constant and $$\varepsilon_0$$ is the permittivity of free space. These expressions treat the particles as point particles, with the electrostatic expression applying in a vacuum. The absolute value of $$Q_1Q_2$$ gives the electrostatic force magnitude regardless of the charges’ signs.

Comparing the force magnitudes

Dividing the electrostatic force by the gravitational force gives:

FeFg=∣Q1Q2∣4πε0Gm1m2\frac{F_e}{F_g} = \frac{|Q_1Q_2|}{4\pi\varepsilon_0Gm_1m_2}

The separation cancels because both forces are proportional to $$1/r^2$$. Changing the separation therefore changes both force magnitudes by the same factor, leaving their ratio unchanged. For charged subatomic particles, the electrostatic force is typically much greater than the gravitational force.

Comparing the force directions

Gravity between the particles is always attractive. The electrostatic force is attractive if the charges have opposite signs and repulsive if they have the same sign. The ratio above compares magnitudes only; it does not show whether the forces act in the same or opposite directions.