Electric fields and Coulomb's lawEdexcel A-Level Physics: Revision notes

An electric field is a region in which a charged particle experiences a force. The field is a property of the space around a charge or a charged object and is a vector quantity.

The electric field strength $E$ at a point is the force per unit positive charge placed at that point:

$E = \frac{F}{Q}$

The unit is N C⁻¹, which is equivalent to V m⁻¹. The direction of $E$ is the direction of the force on a positive test charge, so a negative charge experiences a force in the opposite direction. Rearranged, the force on a charge $Q$ in a field $E$ is $F = QE$.

The electrostatic force between two point charges $Q_1$ and $Q_2$ a distance $r$ apart in a vacuum is

$F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}$

where $\varepsilon_0 = 8.85 \times 10^{-12}$ F m⁻¹ is the permittivity of free space, so $1/4\pi\varepsilon_0 = 8.99 \times 10^{9}$ N m² C⁻².

• Force is proportional to the product of the charges and to $1/r^2$, an inverse square law
• Like charges repel and unlike charges attract
• The two forces form a Newton's third law pair, equal in magnitude and opposite in direction
• Doubling the separation divides the force by four

The field strength at distance $r$ from a point charge $Q$ is found by combining $E = F/Q_{\text{test}}$ with Coulomb's law:

$E = \frac{Q}{4\pi\varepsilon_0 r^2}$

The field is radial, directed away from a positive charge and towards a negative charge, and it falls as $1/r^2$. This also applies outside a charged sphere, as though all the charge were concentrated at the centre.

Where more than one charge contributes, the resultant field is the vector sum of the individual fields, so fields in opposite directions subtract.

A point charge of +6.0 nC is fixed in a vacuum. Find the force on a proton 0.15 m from it.

Field: $E = \dfrac{6.0\times10^{-9}}{4\pi \times 8.85\times10^{-12} \times 0.15^2} = 2.4\times10^{3}$ N C⁻¹, directed away from the charge.

Force on the proton: $F = QE = 1.60\times10^{-19} \times 2.4\times10^{3} = 3.8\times10^{-16}$ N, also directed away from the charge, because the proton is positive.

The acceleration is $a = F/m = 3.8\times10^{-16} / 1.67\times10^{-27} = 2.3\times10^{11}$ m s⁻².