recall and use Coulomb’s law F = Q1Q2 / (4πε0 r 2) for the force between two point charges in free space

Uniform Electric Fields

In a uniform electric field the electric field vector \$\mathbf{E}\$ has the same magnitude and direction at every point in the region of interest. This idealised situation is useful for understanding the behaviour of charges and for solving many A‑Level problems.

Key Concepts

• The electric field \$\mathbf{E}\$ is defined as the force per unit positive test charge:

  \$\mathbf{E} = \frac{\mathbf{F}}{q}\$

• For a uniform field the lines of force are parallel and equally spaced.
• The potential difference \$V\$ between two points separated by a distance \$d\$ along the field direction is:

  \$\Delta V = -E d\$

• The force on a charge \$q\$ placed in the field is:

  \$\mathbf{F}=q\mathbf{E}\$

Coulomb’s Law

Coulomb’s law gives the magnitude of the electrostatic force between two point charges \$Q1\$ and \$Q2\$ separated by a distance \$r\$ in free space:

\$F = \frac{Q1 Q2}{4\pi\varepsilon_0 r^{2}}\$

where \$\varepsilon_0 = 8.85\times10^{-12}\,\text{F m}^{-1}\$ is the permittivity of free space.

From Coulomb’s Law to a Uniform Field

Consider a large, flat, charged plate of surface charge density \$\sigma\$ (C m⁻²). At a point close to the plate (distance much smaller than the plate dimensions) the field can be approximated as uniform:

\$E = \frac{\sigma}{2\varepsilon_0}\$

Two parallel plates with opposite charge densities produce a uniform field between them:

\$E = \frac{\sigma}{\varepsilon_0} = \frac{V}{d}\$

where \$V\$ is the potential difference between the plates and \$d\$ is the separation.

Worked Example

Problem: Two point charges, \$Q1 = +5.0\,\mu\text{C}\$ and \$Q2 = -3.0\,\mu\text{C}\$, are 0.20 m apart in free space. Calculate the magnitude and direction of the force on each charge.

1. Convert microcoulombs to coulombs:

   \$Q1 = 5.0\times10^{-6}\,\text{C},\qquad Q2 = -3.0\times10^{-6}\,\text{C}\$

2. Apply Coulomb’s law:

   \$\$F = \frac{|Q1 Q2|}{4\pi\varepsilon_0 r^{2}}

   = \frac{(5.0\times10^{-6})(3.0\times10^{-6})}{4\pi(8.85\times10^{-12})(0.20)^{2}}\$\$

3. Calculate:

   \$F \approx 1.35\,\text{N}\$

4. Direction: The force on \$Q1\$ is attractive (towards \$Q2\$) and the force on \$Q2\$ is attractive (towards \$Q1\$). Thus the forces are equal in magnitude and opposite in direction, consistent with Newton’s third law.

Practice Questions

1. Two identical point charges \$+2.0\,\mu\text{C}\$ are placed 0.10 m apart. Calculate the force between them.
2. A parallel‑plate capacitor has plates 5.0 cm apart and a potential difference of 200 V. Determine the magnitude of the uniform electric field between the plates.
3. A proton (charge \$+e\$) is released from rest in a uniform electric field of magnitude \$5.0\times10^{5}\,\text{N C}^{-1}\$. Find the force on the proton and its initial acceleration (mass of proton \$=1.67\times10^{-27}\,\text{kg}\$).

Summary Table

Key Points to Remember

• Coulomb’s law applies to point charges in vacuum; superposition can be used for multiple charges.
• A uniform electric field can be approximated by large parallel plates or by being close to an infinite charged sheet.
• The direction of the electric field is defined as the direction of the force on a positive test charge.
• When using Coulomb’s law, ensure all quantities are in SI units.