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Electric Field Strength of a Point Charge

Key idea: Define electric field strength, calculate the field due to point charges, determine direction, and apply vector superposition.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Define electric field strength and calculate resultant fields due to point charges.

1. Definitions (must know)

Electric field strength, vector E, at a point is the electric force per unit positive charge on a small stationary test charge placed at that point:

vector E = (vector F)/q

Electric field strength is a vector. Its direction is the direction of the force on a positive test charge. Its units are N C⁻¹, equivalent to V m⁻¹.

For a point source charge Q in free space or air, the field magnitude at distance r is

E = (1/4πε₀)|Q|/r² = k|Q|/r²

The field points away from + Q and towards -Q.

Electric field lines and equipotentials around point chargesTwo panels compare a positive and negative point charge. Solid radial field lines point outwards from the positive charge and inwards towards the negative charge. Dashed concentric circles are equipotentials.Positive source chargeNegative source charge+Q−Qfield lineequipotentialE is tangent to field linesand points towards lower V
Field lines point away from a positive source charge and towards a negative source charge. Dashed equipotentials are perpendicular to the field lines; wider spacing farther away represents a smaller potential gradient and weaker field.

2. Key ideas

  • The source charge Q creates the field; a small test charge reveals it without significantly disturbing it.
  • E ∝ |Q| and E ∝ 1/r² for a point charge.
  • Use the sign of Q to decide field direction, not to make the field magnitude negative.
  • The force on a separate charge q is vector F = q vector E; a negative q experiences force opposite to vector E.
  • Fields obey superposition: add the individual field vectors at the point.
Field direction is not particle motion

A field line shows the force direction for a positive test charge. A negative charge accelerates opposite to the field, and any initial sideways velocity can make the particle follow a curved path.

3. Detailed reasoning

A. From Coulomb force to field strength

Place a positive test charge q a distance r from source charge Q. Coulomb’s law gives

F = k|Qq|/r²

Dividing by the test charge magnitude gives

E = F/|q| = k|Q|/r²

The test charge cancels: the field is a property of the source configuration and position, not of the particular test charge used.

B. Superposition

For several source charges,

vector Eᵣₑₛᵤₗₜₐₙₜ = vector E₁ + vector E₂ + …

On a straight line, choose a positive direction and add signed components. In two dimensions, resolve each contribution into perpendicular components before adding.

4. Common mistakes

  • Confusing E = kQ/r² with the potential relationship V = kQ/r.
  • Giving only a magnitude when direction is required.
  • Adding field magnitudes when the contributions point in opposite directions.
  • Using the sign of a test charge to decide the field direction.
  • Measuring r from the surface rather than from the point charge or centre of a spherical source model.

5. Exam tips

  • Mark the point where E is required and draw a separate arrow for each source contribution.
  • Write E = k|Q|/r² for magnitude, then state direction in words or with a signed component.
  • Check inverse-square scaling: doubling r reduces E to one quarter.
  • State the point-charge or spherically symmetric-source assumption when it matters.

6. Worked examples

Modelled example 1

Field due to one point charge

Core

Problem

A charge Q = -6.0 nC is in free space. Find the electric field strength 0.20 m away, including direction.
Study the worked solution

  1. Calculate the magnitude

    Method

    Use the source-charge magnitude in E = k|Q|/r².

    Reason

    Field magnitude is non-negative; source sign is used for direction.

    Working

    E = ((8.99 × 10⁹)(6.0 × 10⁻⁹))/(0.20)² = 1.35 × 10³ N C⁻¹

  2. State direction

    Method

    Point the field towards the negative charge.

    Reason

    Field direction is the force direction on a positive test charge.

    Working

    vector E is radially inward towards Q.

Guided practice 2

Resultant field between like charges

About 4 min

Problem

Two identical positive charges are fixed at opposite ends of a line. What is the resultant electric field at the midpoint?

Try this before viewing the solution

Resultant field

Hints

Hint 1: draw both field arrows
A positive source produces field away from itself.
View solution step by step

  1. Compare magnitudes

    Method

    Recognise equal fields.

    Reason

    The source charges and midpoint distances are identical.

    Working

    E_L = E_R.

  2. Add vectors

    Method

    Cancel the opposing directions.

    Reason

    Each positive charge’s field points away from that charge.

    Working

    vector Eₙₑₜ = vector E_L + vector E_R = 0.

Common misconception 3

From field to force

Find and correct the mistake

Learner claim

An electron is in a 4.0 × 10⁴ N C⁻¹ field directed east. A learner says its force is east because force always follows the field. Diagnose the claim and find the force.

Try this before viewing the solution

Electron-force direction

View solution step by step

  1. Find force magnitude

    Method

    Use |F| = |q|E.

    Reason

    Magnitude uses the elementary-charge magnitude.

    Working

    |F| = (1.60 × 10⁻¹⁹)(4.0 × 10⁴) = 6.4 × 10⁻¹⁵ N

  2. Apply the charge sign

    Method

    Direct the force west.

    Reason

    vector F = q vector E reverses direction when q < 0.

    Working

    vector F is opposite the eastward field.

Examiner practice 4

Distance for a given point-charge field

3 marks

Examination question

A + 4.0 nC point charge produces electric field magnitude 400 N C⁻¹ at a point in free space. Find the distance from the charge and state the field direction. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Rearrange for distance

    1 mark

    Method

    Use r = square root of (k|Q|/E).

    Reason

    Point-charge field follows an inverse square.

    Working

    r = square root of (((8.99 × 10⁹)(4.0 × 10⁻⁹))/400)

  2. Evaluate

    1 mark

    Method

    Obtain 0.300 m.

    Reason

    Distance is the positive square root.

    Working

    r = 0.300 m.

  3. State direction

    1 mark

    Method

    Point radially away from the charge.

    Reason

    The source is positive.

    Working

    Direction: away from + 4.0 nC.

Challenge 5

Resultant field between opposite charges

Minimal support

Independent transfer

Charges + 8.0 nC and -2.0 nC are 0.30 m apart. Find the resultant field at the midpoint, including direction.

Try this before viewing the solution

Hints

Hint 1: decide directions before combining
At the midpoint, field points away from the positive charge and towards the negative charge.
View solution step by step

  1. Use the midpoint distance

    Method

    Set r = 0.15 m for both sources.

    Reason

    The point is halfway across the 0.30 m separation.

    Working

    r = 0.15 m.

  2. Calculate both fields

    Method

    Obtain 3.20 × 10³ and 7.99 × 10² N C⁻¹.

    Reason

    Use E = k|Q|/r² for each source.

    Working

    E₊ = 3.20 × 10³; E₋ = 7.99 × 10² N C⁻¹.

  3. Add aligned vectors

    Method

    Obtain 4.00 × 10³ N C⁻¹ towards the negative charge.

    Reason

    Both midpoint fields point from positive to negative.

    Working

    Eₙₑₜ = 3.20 × 10³ + 0.799 × 10³ = 4.00 × 10³ N C⁻¹

7. Mind stretchers

Mind stretcher 1: Zero field does not imply zero potentialExtension

At the midpoint between two identical positive charges, E = 0. Must the electric potential also be zero?

Show Answer

No. Field contributions are vectors and cancel, but potential contributions are scalars and add. Both potentials are positive, so the total potential is positive even though the resultant field is zero.

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