Electric Fields — A-Level Physics Study Guide

For: A-Level Physics candidates sitting A-Level Physics.

Covers: Coulomb’s law, electric field around point charges, uniform fields between parallel plates, electric potential and potential energy, and comparison of electric and gravitational fields, aligned to A-Level Physics syllabus requirements.

You should already know: IGCSE Physics, basic algebra and trigonometry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Physics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Electric Fields?

An electric field is a region of space where a stationary electric charge experiences a non-contact force, caused by other nearby charged particles or charged surfaces. It is a vector quantity, often shortened to E-field in syllabus materials, and appears in Section 17 of the A-Level Physics specification, tested in both multiple-choice (Paper 1) and structured written papers (Papers 2 and 4). Field lines are used to visualise electric fields: they point in the direction a positive test charge would move, with density proportional to field strength.

2. Coulomb's law — $F=k{q}_1{q}_2/r^2$

Coulomb’s law describes the magnitude of the electrostatic force between two stationary point charges in a vacuum, derived experimentally by Charles-Augustin de Coulomb in the 18th century.

Formula breakdown

$$F=\frac{k{q}_1{q}_2}{r^2}$$ Where:

• $F$ = electrostatic force in newtons (N), attractive if charges have opposite signs, repulsive if same sign
• $k$ = Coulomb’s constant, equal to $\frac{1}{4\pi {ϵ}_0}=8.99\times 10^9{\text{ N m}}^2{\text{C}}^{-2}$ (given in the exam data booklet)
• $q_1,q_2$ = magnitudes of the two point charges in coulombs (C)
• $r$ = perpendicular distance between the centres of the two charges in meters (m)

Examiners frequently test that you recognise Coulomb’s law follows Newton’s third law: the force on $q_1$ due to $q_2$ is equal in magnitude and opposite in direction to the force on $q_2$ due to $q_1$.

Worked example

Two point charges, $+2.0\mu \text{C}$ and $-3.0\mu \text{C}$, are placed 0.50 m apart in a vacuum. Calculate the magnitude of the electrostatic force between them, and state if it is attractive or repulsive.

1. Convert units to SI: $2.0\mu \text{C}=2.0\times 10^{-6}\text{ C}$, $3.0\mu \text{C}=3.0\times 10^{-6}\text{ C}$
2. Substitute into formula: $$F=\frac{8.99\times 10^9\times 2.0\times 10^{-6}\times 3.0\times 10^{-6}}{(0.50{)}^2}=\frac{0.05394}{0.25}=0.22\text{ N (2 sig figs)}$$
3. The force is attractive, as the charges have opposite signs.

3. Electric field $E=F/q$ around a point charge

Electric field strength $E$ at a point is defined as the force per unit positive test charge placed at that point. This is a general definition that applies to all electric fields, regardless of their source.

Formula breakdown

$$E=\frac{F}{q}$$ Where:

• $E$ = electric field strength, units of newtons per coulomb (${\text{N C}}^{-1}$) or equivalent volts per meter (${\text{V m}}^{-1}$)
• $F$ = force on test charge $q$ in N
• $q$ = magnitude of positive test charge in C

For a point charge $Q$, substitute Coulomb’s law into the general $E$ definition to get the field strength at distance $r$ from $Q$: $$E=\frac{kQ}{r^2}$$ $E$ points away from positive charges and towards negative charges, matching the direction of force on a positive test charge.

Worked example

Calculate the electric field strength 0.20 m from a $+5.0\mu \text{C}$ point charge, and state its direction.

1. Convert charge to SI: $5.0\mu \text{C}=5.0\times 10^{-6}\text{ C}$
2. Substitute into point charge E formula: $$E=\frac{8.99\times 10^9\times 5.0\times 10^{-6}}{(0.20{)}^2}=\frac{44950}{0.04}=1.1\times 10^6{\text{ N C}}^{-1}\text{ (2 sig figs)}$$
3. Direction: away from the positive charge.

4. Uniform fields between parallel plates

When two parallel conducting plates are connected to a voltage supply, one plate gains positive charge and the other negative. The field between the plates (ignoring edge effects where lines curve at the plate edges) is uniform, meaning $E$ has the same magnitude and direction at all points between the plates.

Formula breakdown

$$E=\frac{V}{d}$$ Where:

• $V$ = potential difference between the two plates in volts (V)
• $d$ = perpendicular separation between the plates in meters (m)
• $E$ = uniform field strength, units ${\text{V m}}^{-1}$ (equivalent to ${\text{N C}}^{-1}$, an equivalence examiners often test for 1 mark)

Charged particles entering a uniform electric field experience constant acceleration parallel to the field lines, leading to projectile-style motion that is a common high-mark exam question.

Worked example

Two parallel plates are separated by 2.0 cm, connected to a 12 V battery. Calculate the uniform electric field strength between the plates, and the force on an electron (charge $-1.60\times 10^{-19}\text{ C}$) placed in the field.

1. Convert d to SI: $2.0\text{ cm}=0.020\text{ m}$
2. Calculate E: $$E=\frac{12}{0.020}=600{\text{ V m}}^{-1}=600{\text{ N C}}^{-1}$$
3. Calculate force on electron: $$F=E\times |q|=600\times 1.60\times 10^{-19}=9.6\times 10^{-17}\text{ N}$$
4. Direction: towards the positive plate, opposite to the direction of the electric field.

5. Electric potential and potential energy

Electric potential $V$ at a point in an electric field is the work done per unit positive charge to move a test charge from infinity (where potential is defined as 0) to that point. It is a scalar quantity, so only has magnitude and sign, no direction.

Key formulas

1. General potential definition: $V=\frac{W}{q}$, units volts (V = ${\text{J C}}^{-1}$), where $W$ is work done in joules
2. Potential around a point charge: $V=\frac{kQ}{r}$
3. Potential change in a uniform field: $\Delta V=E\times \Delta d$, where $\Delta d$ is distance moved parallel to field lines
4. Electric potential energy $E_p$ of a charge $q$ in a field: $E_p=qV$, so for two point charges: $E_p=\frac{k{q}_1{q}_2}{r}$

A negative potential energy value means the system is bound: you have to do external work to separate the charges.

Worked example

Calculate the electric potential 0.10 m from a $-4.0\mu \text{C}$ point charge, then find the potential energy of a $+1.0\mu \text{C}$ test charge placed at that point.

1. Calculate potential: $$V=\frac{8.99\times 10^9\times (-4.0\times 10^{-6})}{0.10}=-3.6\times 10^5\text{ V (2 sig figs)}$$
2. Calculate potential energy: $$E_p=qV=1.0\times 10^{-6}\times (-3.6\times 10^5)=-0.36\text{ J}$$

6. Comparison with gravitational field

Electric and gravitational fields are both non-contact, inverse-square force fields, but have key differences that examiners frequently ask you to state for 4-6 mark questions.

Similarities

1. Both follow an inverse-square law for force between point quantities: $F=\frac{G{m}_1{m}_2}{r^2}$ for gravity, $F=\frac{k{q}_1{q}_2}{r^2}$ for electrostatics
2. Field strength is defined as force per unit quantity: $g=F/m$ for gravity, $E=F/q$ for electric fields
3. Potential is defined as work done per unit quantity to move a test mass/charge from infinity to the point
4. Both can act through a vacuum, no medium required

Differences

Worked comparison

Compare the electrostatic force and gravitational force between a proton (mass $1.67\times 10^{-27}\text{ kg}$, charge $+1.60\times 10^{-19}\text{ C}$) and electron (mass $9.11\times 10^{-31}\text{ kg}$, charge $-1.60\times 10^{-19}\text{ C}$) separated by 0.10 nm.

1. Electrostatic force magnitude: $$F_e=\frac{8.99\times 10^9\times (1.60\times 10^{-19}{)}^2}{(1.0\times 10^{-10}{)}^2}=2.3\times 10^{-8}\text{ N}$$
2. Gravitational force magnitude: $$F_g=\frac{6.67\times 10^{-11}\times 1.67\times 10^{-27}\times 9.11\times 10^{-31}}{(1.0\times 10^{-10}{)}^2}=1.0\times 10^{-47}\text{ N}$$ The electrostatic force is ~$10^{39}$ times stronger, so gravity is negligible in atomic and subatomic systems.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to convert charge units from $\mu \text{C}$/nC to C, or distance from cm/mm to m before calculations. Why: Students rush through questions and skip unit checks. Correct move: Write all unit conversions as your first step in any calculation, cross out non-SI units to confirm you have converted correctly.
• Wrong move: Treating electric potential as a vector, summing magnitudes instead of signed values. Why: Confusion with electric field strength, which is a vector. Correct move: Remember V is scalar, always include the sign of the charge when calculating potential, and add signed values for multiple charges, no direction required.
• Wrong move: Using $E=\frac{kQ}{r^2}$ for uniform parallel plate fields. Why: Mixing point charge and uniform field formulas. Correct move: Only use $E=\frac{kQ}{r^2}$ for point charges or spherical charge distributions, use $E=V/d$ exclusively for uniform parallel plate fields.
• Wrong move: Stating electric field lines point in the direction a negative charge moves. Why: Mixing up the test charge convention. Correct move: Electric field direction is defined as the direction of force on a positive test charge, so negative charges move opposite to field lines.
• Wrong move: Applying Coulomb’s law to extended charged objects. Why: Forgetting Coulomb’s law only applies to point charges. Correct move: Only use Coulomb’s law if the question specifies point charges, or the separation between charges is at least 10x larger than the size of the charged objects.

8. Practice Questions (A-Level Physics Style)

Question 1 (1 mark, multiple choice)

Two point charges of $+Q$ and $-2Q$ are separated by distance $d$. The magnitude of the force on $+Q$ due to $-2Q$ is $F$. What is the magnitude of the force on $-2Q$ due to $+Q$? A) $F/2$ B) $F$ C) $2F$ D) $4F$

Solution

Correct answer: B) $F$. Explanation: Coulomb’s law follows Newton’s third law, so the force on each charge is equal in magnitude and opposite in direction, regardless of the relative size of the charges. The most common wrong answer is C, where students incorrectly assume the larger charge experiences a larger force.

Question 2 (4 marks, structured)

(a) Define electric field strength at a point. (2 marks) (b) A uniform electric field is set up between two parallel plates separated by 50 mm, with a potential difference of 250 V across the plates. Calculate the acceleration of an alpha particle (charge $+3.2\times 10^{-19}\text{ C}$, mass $6.64\times 10^{-27}\text{ kg}$) placed in the field, ignoring gravitational effects. (2 marks)

Solution

(a) Award 1 mark for stating electric field strength is force per unit positive test charge placed at the point, 1 mark for noting it is a vector with direction equal to the force on a positive test charge. (b) Step 1: Convert separation to SI: $50\text{ mm}=0.050\text{ m}$ Step 2: Calculate $E=V/d=250/0.050=5000{\text{ V m}}^{-1}$ (1 mark) Step 3: Calculate force $F=Eq=5000\times 3.2\times 10^{-19}=1.6\times 10^{-15}\text{ N}$ Step 4: Acceleration $a=F/m=1.6\times 10^{-15}/6.64\times 10^{-27}=2.4\times 10^{11}{\text{ m s}}^{-2}$ (1 mark for correct final answer to 2-3 sig figs)

Question 3 (6 marks, long structured)

(a) State two similarities and two differences between electric and gravitational fields. (4 marks) (b) Calculate the electric potential at a point midway between a $+2.0\mu \text{C}$ charge and a $-3.0\mu \text{C}$ charge separated by 0.40 m. (2 marks)

Solution

(a) Similarities (any 2, 1 mark each, max 2): Both are non-contact force fields; both follow inverse square force laws for point quantities; both have field strength defined as force per unit quantity; both have potential defined as work done per unit quantity from infinity. Differences (any 2, 1 mark each, max 2): Gravitational forces are always attractive, electric forces can be attractive or repulsive; gravitational fields act on mass, electric fields act on charge; gravitational constant G is much smaller than Coulomb’s constant k; gravitational potential is always negative, electric potential can be positive or negative. (b) Step 1: Midpoint is 0.20 m from each charge. Potential is scalar, so sum signed potentials: $$V_{total}=\frac{k{Q}_1}{r}+\frac{k{Q}_2}{r}=\frac{8.99\times 10^9}{0.20}(2.0\times 10^{-6}-3.0\times 10^{-6})$$ Step 2: Calculate result: $V_{total}=4.5\times 10^{10}\times (-1.0\times 10^{-6})=-4.5\times 10^4\text{ V}$ Award 1 mark for correct use of scalar addition with signed charges, 1 mark for correct final value with sign.

9. Quick Reference Cheatsheet

10. What's Next

This electric fields topic is a foundational building block for multiple later sections of the A-Level Physics syllabus. You will apply these concepts when studying capacitance (where uniform parallel plate fields are core to capacitor operation), electromagnetism (where electric fields interact with magnetic fields to produce Lorentz force on moving charges), and even modern physics topics like particle acceleration and atomic structure, where electrostatic forces govern interactions between subatomic particles. A strong grasp of electric field calculations will also help you answer cross-topic questions that combine fields with mechanics, such as projectile motion of charged particles in uniform fields, which is a frequent high-mark structured question in Paper 2 and Paper 4.

If you are struggling with any of the concepts, formulas, or practice questions in this guide, you can ask Ollie, our AI tutor, for step-by-step explanations, additional practice problems, or clarification of any syllabus points any time. You can also find more topic-specific study guides, past paper walkthroughs, and revision quizzes for A-Level Physics on the homepage to help you prepare for your exams.

Aligned with the Cambridge International AS & A Level Physics 9702 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.