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Electric Fields · 11.1

Electric Field Strength & Force

An electric field tells you the force per unit positive charge at each point. The source creates the field; a second charge responds to it.

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01

Build the model

Predict direction before calculating magnitude.

Electric field strength is a vector property of a point in space. Its direction is the force direction on a small positive test charge. Once E⃗ is known, multiply by the actual signed charge q: a positive charge accelerates along E⃗ and a negative charge accelerates opposite E⃗.

Simple definition
Electric field strength is the force per unit positive charge at a point, directed as the force on a positive test charge.
Example
In an eastward field, a positive charge feels force east while an electron feels force west.
Field definitionE⃗ = F⃗/q₀

The electric field at a point is the force a small positive test charge would feel there, per unit of its charge.

q₀ is positive and small enough not to disturb the sources

Force in a fieldF⃗ = qE⃗

Put a charge in a field and it feels field strength times its charge — negative charges are pushed the opposite way to the arrows.

A negative q reverses the vector direction

Equivalent units1 N/C = 1 V/m

Two ways of stating field strength: force per charge, or how quickly the voltage changes with distance.

Both units describe electric field strength

01

Separate source and probe

Source charges establish E⃗. A test charge is an imagined measuring probe; changing that probe does not change the pre-existing field in the ideal model.

02

Define one direction

At each point, E⃗ points in the direction a positive test charge would initially accelerate. State that direction before using the sign of q.

03

Keep the vectors

E⃗ and F⃗ are vectors. Magnitudes alone cannot tell you whether a negative charge moves east or west, so draw an arrow or use signed components.

02

Change one variable at a time

Interrogate the field with a probe.

Build E⃗, then place q

Set q = 0: the force disappears, but the source-created electric field remains.

Field directioneast

Force directionwest

Force magnitude5.00e-1 N

RelationshipF⃗ opposite E⃗

03

Catch the common trap

Say what the arrows mean.

The electric field points east. Which way is the electric force on an electron?

Choose an answer, then explain the direction in your own words.

04

Worked examples

Sketch, calculate, then restore direction.

EasyA +2.0 μC charge feels a 0.50 N force at a point. Find the field strength there.
  1. E = F/q = 0.50 ÷ 2.0 × 10⁻⁶.
  2. E = 2.5 × 10⁵ N/C, along the force.

AnswerE = 2.5 × 10⁵ N/C

MediumA uniform electric field of 2.0 × 10⁴ N/C points east. Find the electric force on an electron.
  1. Write the signed charge: q = −1.60 × 10⁻¹⁹ C.
  2. Magnitude: |F| = |q|E = (1.60 × 10⁻¹⁹)(2.0 × 10⁴) = 3.2 × 10⁻¹⁵ N.
  3. Because q is negative, the force is opposite the eastward field.

AnswerF = 3.2 × 10⁻¹⁵ N west

HardAn oil droplet of mass 3.3 × 10⁻¹⁵ kg floats motionless between plates producing a downward field of 2.0 × 10⁵ N/C. Find the droplet's charge, including sign.
  1. Balance: qE = mg → |q| = 3.3 × 10⁻¹⁵ × 9.81 ÷ 2.0 × 10⁵ ≈ 1.6 × 10⁻¹⁹ C.
  2. The electric force must point UP against a DOWNWARD field, so the charge is negative.
  3. |q| = e — a single excess electron: Millikan's experiment.

Answerq = −1.6 × 10⁻¹⁹ C — one electron

ChallengingIn a thundercloud field of 3.0 × 10⁵ N/C, compare the acceleration of a free electron with that of a nitrogen ion (m ≈ 4.7 × 10⁻²⁶ kg, charge +e). What does this imply about who does the ionising in a spark?
  1. a(e) = eE/mₑ = 1.6 × 10⁻¹⁹ × 3.0 × 10⁵ ÷ 9.11 × 10⁻³¹ ≈ 5.3 × 10¹⁶ m/s².
  2. a(ion) = 4.8 × 10⁻¹⁴ ÷ 4.7 × 10⁻²⁶ ≈ 1.0 × 10¹² m/s² — fifty thousand times smaller.
  3. Electrons reach ionising speeds almost instantly; the avalanche of a spark is an electron story.

AnswerElectron accelerates ~5 × 10⁴ times harder — sparks are electron avalanches

Exam diagrams for this topic1 figure to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

  1. 01

    InspectRead the figure comment.

  2. 02

    TraceFollow labels, arrows and axes.

  3. 03

    AnswerWork one part at a time.

  4. 04

    CheckReveal hints and marking points.

A Level

01Fig. 10.1Electric fields · CapacitanceA Level
Two parallel metal plates in a vacuum, connected to a 2.0 kV supply2.0 kV+−vacuum5.0 mm

Figure comment

Fig. 10.1Two long horizontal metal plates are drawn one directly above the other and connected by wires at their left-hand ends to a battery labelled 2.0 kV; the upper plate carries a plus sign and the lower plate a minus sign. The space between the plates is empty and labelled vacuum, and a dimension at the right-hand end gives the separation of the plates as 5.0 mm. No field lines are drawn between the plates.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

18 Electric fields19 Capacitance20 Magnetic fieldsOpen A Level question · A Level extension Q10
Guided questions 5 parts

Reading cue. 5.0 mm is the gap, not a plate length, and no field lines are drawn — the plus and minus signs on the plates are the only thing in the figure that fixes the field direction.

  1. aState State the direction of the electric field in the gap in Fig. 10.1, and the direction of the force it exerts on an electron placed there.

    recall2 marks

    Check answer 2 marks
    1. the field points vertically downwards, from the positive upper plate to the negative lower plate
    2. the force on an electron is vertically upwards, towards the positive plate

  2. bCalculate Calculate the work done on an electron that moves from the lower plate to the upper plate, and the speed with which it arrives.

    routine3 marks

    Check answer 3 marks
    1. W = eV = 1.60 × 10⁻¹⁹ × 2.0 × 10³ = 3.2 × 10⁻¹⁶ J
    2. ½mv² = 3.2 × 10⁻¹⁶ J
    3. v = 2.7 × 10⁷ m s⁻¹

  3. cDetermine A charged dust particle of weight 1.28 × 10⁻¹³ N is held at rest midway between the plates. Determine the magnitude and sign of its charge, and the number of excess electrons it carries.

    demanding4 marks

    Check answer 4 marks
    1. E = V/d = 2.0 × 10³ / 5.0 × 10⁻³ = 4.0 × 10⁵ V m⁻¹
    2. for equilibrium qE = weight, so q = 1.28 × 10⁻¹³ / 4.0 × 10⁵ = 3.2 × 10⁻¹⁹ C
    3. the electric force must act upwards while the field points downwards, so the charge is negative
    4. 3.2 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2 excess electrons

  4. dDeduce The battery p.d. is suddenly reduced to 1.0 kV while the particle is still midway between the plates. Deduce the acceleration of the particle, and calculate the time it takes to reach a plate.

    top of the paper4 marks

    Check answer 4 marks
    1. new field = 2.0 × 10⁵ V m⁻¹, so the electric force = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N upwards
    2. resultant = 1.28 × 10⁻¹³ − 6.4 × 10⁻¹⁴ = 6.4 × 10⁻¹⁴ N downwards
    3. mass = 1.28 × 10⁻¹³ / 9.81 = 1.30 × 10⁻¹⁴ kg, so a = 4.9 m s⁻² downwards, that is g/2
    4. falling the 2.5 mm to the lower plate: t = √(2 × 2.5 × 10⁻³ / 4.9) = 3.2 × 10⁻² s

Transfer challenge

An electron travelling at 2.0 × 10⁷ m s⁻¹ enters midway between two parallel plates 5.0 cm long, moving parallel to them, in a uniform field of strength 1.2 × 10⁴ V m⁻¹. Calculate the sideways deflection of the electron as it leaves the plates.

Check answer 3 marks
  1. a = eE/m = (1.60 × 10⁻¹⁹ × 1.2 × 10⁴) / 9.11 × 10⁻³¹ = 2.1 × 10¹⁵ m s⁻²
  2. time between the plates t = 0.050 / 2.0 × 10⁷ = 2.5 × 10⁻⁹ s
  3. deflection = ½at² = 6.6 × 10⁻³ m