Fields and potential energy

Key idea: A field assigns a force tendency to each point in space. Potential energy belongs to the interacting system, and it falls when the field does positive work.

  • GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 2 of 3

Check your understanding

By the end of this lesson, you should be able to

  • Define gravitational and electric field strength and read field-line patterns.
  • Relate work by a field to potential-energy change.
  • Distinguish gravitational, electric and elastic potential energy and use force–extension area.

Learn the idea

Big question: How do field direction and energy change describe the same interaction from two viewpoints?

Read and draw field patterns

Gravitational field strength is force per unit mass. Electric field strength is force per unit positive charge, so the arrow direction is not necessarily the force direction on a negative charge.

Parallel equally spaced lines show a uniform field. Radial lines show a field whose direction and strength vary with position; closer line spacing represents a stronger field qualitatively. Field lines never cross because a field cannot have two directions at one point.

Check your understanding: Which way is the force on an electron in an electric field?

Opposite to the electric field direction because the electron has negative charge.

Link work and potential energy

When a field does positive work, its potential-energy store decreases: Wfield = −ΔEp. Work done by an external agent against the field can increase that store.

Gravitational potential energy belongs to an interacting mass system such as object and Earth; electric potential energy belongs to interacting charges; elastic potential energy belongs to a deformed material. Naming the interaction prevents these stores from being treated as interchangeable labels.

Elastic potential energy is the area under a force–extension graph. For a Hooke's-law straight line this is ½Fx; for a curved graph, use the actual area rather than a triangle formula.

Check your understanding: A positive charge moves spontaneously along an electric field. What happens to electric potential energy?

It decreases because the electric field does positive work.

Work and potential-energy change in gravitational and electric fieldsA mass moves downward along uniform gravitational field lines and a positive charge moves right along uniform electric field lines. In both cases the field does positive work and potential energy decreases. Equipotential lines are perpendicular to the field lines.Gravitational fieldElectric fieldmmotionfield and motion downward+qmotionfield and positive-charge motion rightWfield > 0, so ΔU < 0
Scroll diagram horizontally to read all labels.
For motion along either field, positive work done by the field corresponds to a negative potential-energy change: Wfield = −ΔU.

Key ideas

  • Gravitational potential energy belongs to the mass–Earth system.
  • Electric force reverses for negative charge, although E keeps its defined direction.
  • Area under an F–x graph has unit N m = J.

Relationships to know

  • g = F/m
  • E = F/Q for a positive test charge
  • Wfield = −ΔEp
  • Eelastic = area under F–x graph

Follow the reasoning

Worked example

Track force, work and energy for a negative charge

Question: An electron moves 0.080 m opposite to a uniform electric field of 3.0 × 10⁴ N C⁻¹. Find the work done by the field and the change in electric potential energy. Use electron charge −1.60 × 10⁻¹⁹ C.

  1. Step 1: Find the force direction

    Why: Negative charge reverses the direction given by the field arrow.

    Working: F = qE, so the force is opposite E with magnitude 1.60 × 10⁻¹⁹(3.0 × 10⁴) = 4.8 × 10⁻¹⁵ N.

  2. Step 2: Compare force and displacement

    Why: The electron moves opposite E, which is along its force.

    Working: Wfield = Fs = (4.8 × 10⁻¹⁵)(0.080) = 3.84 × 10⁻¹⁶ J.

  3. Step 3: Convert work to store change

    Why: Positive work by the field reduces the associated potential-energy store.

    Working: ΔEp = −Wfield = −3.84 × 10⁻¹⁶ J.

Answer: The field does +3.8 × 10⁻¹⁶ J of work and the electric potential energy decreases by 3.8 × 10⁻¹⁶ J.

Check: The electron accelerates along its force, so a potential-energy decrease can become kinetic energy.

Now try it with support

Practise with support

A +3.0 μC charge in a uniform 2.0 × 10⁴ N C⁻¹ field moves 0.40 m along the field. Find the force and change in electric potential energy.

Hints

  1. Use F = QE.
  2. Work by the field is positive for displacement along the force.
View the guided answer

F = (3.0 × 10⁻⁶)(2.0 × 10⁴) = 0.060 N. Wfield = Fs = 0.024 J, so ΔEelectric = −0.024 J.

Your turn

Practise independently

A spring force rises linearly from 0 to 80 N over 0.20 m. Find the stored elastic energy and state the work done by the spring during release.

Check your answer

The stored elastic energy is the triangular area ½(80)(0.20) = 8.0 J. During complete release, the spring’s elastic store decreases by 8.0 J and the spring can do +8.0 J of work on the attached body if losses are negligible.

Common mistakes and exam guidance

Watch out for

  • Saying electric field direction is the force direction on an electron.
  • Using the final spring force times the full extension for a linearly increasing force.

In an exam

  • State whether work is done by the field or against the field before assigning a sign.
  • On field diagrams, arrows show direction and spacing shows relative strength.

Put the ideas together

Exam-style practice [6 marks]

A spring has force 20 N at 0.10 m extension and 50 N at 0.20 m. Its force–extension graph consists of straight lines from (0,0) to these points. Find the energy stored at 0.20 m and explain how the graph shows that the spring does not obey Hooke's law throughout.

Plan before you answer

  • Split the area under the graph into simple shapes.
  • Use gradient, not the fact that the graph is made of straight segments, to test Hooke's law.
View the marking points and model answer

Marking points

  1. Uses area under the F–x graph.
  2. Finds first triangular area 1.0 J.
  3. Finds second trapezium area 3.5 J.
  4. Obtains total 4.5 J.
  5. Compares gradients 200 N m⁻¹ and 300 N m⁻¹.
  6. Concludes F is not proportional to x over the full range.

Model answer

The first area is ½(0.10)(20) = 1.0 J. From 0.10 m to 0.20 m the trapezium area is ½(20 + 50)(0.10) = 3.5 J, so the total stored energy is 4.5 J. The gradient changes from 200 N m⁻¹ to 300 N m⁻¹, so F/x is not constant and Hooke's law is not obeyed throughout.

Finish from memory

Three-question recap

  1. Define gravitational field strength.

    Check

    Force per unit mass on a small test mass.

  2. What does the area under a force–extension graph represent?

    Check

    Elastic potential energy stored, or work done in stretching.

  3. When is Wfield positive?

    Check

    When the field force has a component along the displacement.

Try this next

Compare uniform and radial fields, then practise one elastic-energy question from a non-linear graph.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H1 Physics
Edition
GCE A-Level H1 Physics 2027