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Electric Potential Energy

Key idea: Relate electric potential energy to potential using U = qV, use U = kQq/r for point charges, and apply work–energy links for moving charges (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Relate electric potential, potential energy and work for systems of point charges.

1. Definitions (Must Know)

A. Electric potential energy, U

Electric potential energy, U, is the energy associated with the position of a charge in an electric field (relative to a chosen reference).

With the standard reference V = 0 at infinity:

U = qV

where q is the charge placed at the point and V is the electric potential at that point.

B. Two point charges

For two point charges Q and q separated by r in free space / air:

U = (1/4πε₀)Qq/r

Sign matters:

  • If Qq > 0, then U > 0 (repulsive pair).
  • If Qq < 0, then U < 0 (attractive, bound pair).

C. Change in potential energy

Moving a charge q from A to B:

Δ U = qΔ V = q(V_B-V_A)

2. Key Ideas (What Earns Marks)

  • Use U = qV to convert between potential and potential energy.
  • For point charges: U = kQq/r with k = 1/4πε₀.
  • Work done by the electric field is the negative change in potential energy: W_field = -Δ U = -qΔ V
  • A positive charge “falls” towards lower potential; a negative charge “falls” towards higher potential.
Keep track of signs

Always write U = kQq/r with the sign of Qq. Don’t use absolute values unless the question asks for a magnitude.

3. Detailed Explanations

A. Why U = qV is so useful

V is energy per unit charge (J C⁻¹). Multiply by charge to recover energy:

U = qV

This is often faster than working directly with forces.

B. Energy interpretation for attraction/repulsion

  • Like charges: you must do positive work to bring them closer (they repel), so U increases and is positive.
  • Unlike charges: the field does the work as they move closer (they attract), so U decreases and is negative.

C. Linking energy changes to motion

If a charge moves freely under the electric force:

Δ K = -Δ U

So:

  • if Δ U < 0, kinetic energy increases,
  • if Δ U > 0, kinetic energy decreases.

4. Common Mistakes

  • Forgetting the sign of q in U = qV (especially for electrons).
  • Using U = k|Qq|/r when the question needs direction/energy change.
  • Mixing up V (volts) and U (joules).

5. Exam Tips

  • Write Δ U = qΔ V early. It keeps the sign logic clear.
  • For two charges, check whether the pair is attractive (U < 0) or repulsive (U > 0).
  • If the question says “energy gained by the charge”, that is usually Δ K; if it says “work done against the field”, that is Δ U.

6. Worked Examples

Modelled example 1

Potential energy of two charges

Core

Problem

Charges Q = +3.0 μC and q = -2.0 μC are 0.50 m apart. Find the system’s potential energy relative to infinity.
Study the worked solution

  1. Use signed charges

    Method

    Apply U = kQq/r without absolute values.

    Reason

    The sign of potential energy distinguishes attractive and repulsive configurations.

    Working

    U = ((8.99 × 10⁹)(3.0 × 10⁻⁶)(-2.0 × 10⁻⁶))/0.50

  2. Evaluate and interpret

    Method

    Obtain -0.108 J.

    Reason

    Unlike charges form an attractive pair with negative energy relative to infinite separation.

    Working

    U = -0.108 J

Guided practice 2

Energy change moving through a potential difference

About 5 min

Problem

An electron moves from V_A = +200 V to V_B = 0 V. Find Δ U.

Try this before viewing the solution

Unit: J

Hints

Hint 1: two signs matter
Δ V = 0-200 V and q = -e.
View solution step by step

  1. Find potential change

    Method

    Use final minus initial.

    Reason

    Energy change is defined between ordered endpoints.

    Working

    Δ V = 0-200 = -200 V

  2. Multiply by signed charge

    Method

    Obtain a positive energy change.

    Reason

    The product of negative charge and negative potential change is positive.

    Working

    Δ U = (-1.60 × 10⁻¹⁹)(-200) = 3.2 × 10⁻¹⁷ J

Common misconception 3

Linking to kinetic energy

Find and correct the mistake

Learner claim

A positive charge released from rest moves to lower potential. A learner says its kinetic energy falls because the potential value falls. Diagnose the claim.

Try this before viewing the solution

Kinetic-energy change

View solution step by step

  1. Find the potential-energy sign

    Method

    Use Δ U = qΔ V < 0.

    Reason

    The charge is positive and moves to lower potential.

    Working

    q > 0, Δ V < 0 ⇒ Δ U < 0.

  2. Apply energy conservation

    Method

    Conclude that kinetic energy increases.

    Reason

    With the electric field doing work, Δ K = -Δ U.

    Working

    Δ K > 0.

Examiner practice 4

Energy gained in electronvolts (accelerating voltage)

3 marks

Examination question

An electron is accelerated through 800 V. Find its kinetic-energy gain in joules and electronvolts. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Use charge magnitude

    1 mark

    Method

    Apply Δ K = e|Δ V|.

    Reason

    The question asks for energy gained, a positive magnitude.

    Working

    Δ K = (1.60 × 10⁻¹⁹)(800)

  2. State joule energy

    1 mark

    Method

    Obtain 1.28 × 10⁻¹⁶ J.

    Reason

    Coulomb-volts are joules.

    Working

    Δ K = 1.28 × 10⁻¹⁶ J.

  3. State electronvolt energy

    1 mark

    Method

    Obtain 800 eV.

    Reason

    One electron gains one electronvolt per volt traversed.

    Working

    Δ K = 800 eV.

Challenge 5

Using U = qV to find potential

Minimal support

Independent transfer

At a point, an electron has U = -3.2 × 10⁻¹⁷ J relative to infinity. Find the electric potential and explain its sign.

Try this before viewing the solution

Hints

Hint 1: retain both negative signs
Use V = U/q with q = -1.60 × 10⁻¹⁹ C.
View solution step by step

  1. Rearrange

    Method

    Divide potential energy by charge.

    Reason

    U = qV relates a particle’s energy to the field property at that point.

    Working

    V = U/q

  2. Evaluate and interpret

    Method

    Obtain + 200 V.

    Reason

    Negative energy divided by negative electron charge gives positive potential.

    Working

    V = (-3.2 × 10⁻¹⁷)/(-1.60 × 10⁻¹⁹) = +200 V

7. Mind Stretchers

Mind stretcher 1: Work done and sign of UExtension

Two identical positive charges are initially far apart (effectively infinity). Work is done to bring them to separation r. Explain the sign of the work done and the sign of U.

Show Answer

Like charges repel, so you must do positive work against the repulsive force to bring them closer. Since U is defined as the work done by an external force from infinity, U is positive.

Mind stretcher 2: Negative charge moving to higher potentialExtension

An electron moves to a region of higher potential. State the sign of Δ U and whether the electric field does positive or negative work on it.

Show Answer

For an electron, q < 0. If potential increases, Δ V > 0, so Δ U = qΔ V < 0.

Then W_field = -Δ U > 0: the field does positive work on the electron.

Mind stretcher 3: Optional (Enrichment)Extension

A. Electronvolt (eV) as an energy unit

An electronvolt, eV, is the energy gained by a charge of magnitude e when moved through a potential difference of 1 V:

1 eV = (1.60 × 10⁻¹⁹ C)(1 V) = 1.60 × 10⁻¹⁹ J

8. Practice (Quiz)

Practice (Quiz)

Practice U = qV, Δ U = qΔ V, and sign questions:

A Level Electric Fields Quiz

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027