Electric Fields — OCR A Level Physics Revision
Module 6 · Particles and Medical Physics

Electric Fields

Electric fields let us describe how charges influence the space around them. For OCR A, you need to move confidently between field lines, Coulomb’s law, point-charge field strength, uniform electric fields, and the motion of charged particles in those fields.

Specification: OCR A H556
Section: 6.2 Electric fields
Teaching time: ~12–15 hours
Focus: maths + QWC + investigations
By the end of this topic you should be able to…
Part 1 · 6.2.1

What is an electric field?

An electric field exists in a region where a charge would experience a force. Electric fields are caused by charges, just as gravitational fields are caused by masses. The field tells you what would happen to a small positive test charge placed at a point.

Definition of electric field strength\[ E = \frac{F}{Q} \]

Electric field strength has units of newtons per coulomb, \(\mathrm{N\,C^{-1}}\). Because it comes from force per unit charge, it is a vector quantity. Its direction is the direction of the force on a small positive test charge.

Key OCR pointField lines always point away from positive charges and towards negative charges. They never cross, because the field at one point cannot have two different directions.

Reading field lines properly

  • closer lines = stronger field
  • line direction shows force on a positive test charge
  • field lines around a point charge are radial
  • between parallel plates the field is approximately uniform

Language that earns marks

Good: “The field strength decreases with the square of distance from the charge.”

Weak: “The force gets weaker as you go away.”

Good: “The force acts in the direction of the electric field on a positive charge, but opposite to the field for a negative charge.”

Electric field patterns Negative point charge − Uniform field between parallel plates + − dashed lines suggest equal-potential regions
Reference diagram sources selected from Wikimedia Commons for field-line provenance: point charge and parallel-plate field files by Sjlegg (public domain). The local image-to-SVG conversion service was unavailable during this run, so this page uses an in-page teaching diagram and records the intended source files in the run evidence.
Common misconceptions1) The field is not “where electrons move”; it exists whether or not a charge is moving. 2) Electric field lines do not show the path every charge must follow. 3) A negative charge does not move in the direction of the field; it accelerates opposite to the field direction.
Part 2 · 6.2.2

Coulomb’s law and point-charge fields

Two point charges exert forces on each other. The magnitude of that electrostatic force is given by Coulomb’s law. The form is mathematically identical to Newton’s inverse-square gravitation law, but the interaction can be attractive or repulsive depending on the signs of the charges.

Coulomb’s law\[ F = \frac{Qq}{4\pi\varepsilon_0 r^2} \]

Here, \(\varepsilon_0\) is the permittivity of free space and \(r\) is the separation between the centres of the charges. If you divide the force by the test charge \(q\), you obtain the field strength around a point charge:

Field due to a point charge\[ E = \frac{Q}{4\pi\varepsilon_0 r^2} \]

This inverse-square behaviour is crucial. If the distance from the charge doubles, the field strength falls to one quarter. If the charge doubles at the same point, the field strength doubles.

Repulsion

Like charges repel. A positive test charge near a positive source charge is pushed away.

Attraction

Unlike charges attract. A positive test charge near a negative source charge is pulled towards it.

Direction

Quote the force direction in words, not just the number. OCR will reward clear vector language.

Worked example · field strength 0.20 m from a point charge

A charge of \(+3.0\times10^{-9}\,\mathrm{C}\) is in air. Find the electric field strength at a point \(0.20\,\mathrm{m}\) away.

\[ E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{3.0\times10^{-9}}{4\pi(8.85\times10^{-12})(0.20)^2} \]
\[ E = 6.74\times10^2\,\mathrm{N\,C^{-1}} \]
Answer: \(6.7\times10^2\,\mathrm{N\,C^{-1}}\), directed radially away from the positive charge.
Worked example · force on a charge placed in a field

A particle of charge \(-2.5\times10^{-6}\,\mathrm{C}\) is placed in a uniform field of magnitude \(4.0\times10^4\,\mathrm{N\,C^{-1}}\). Find the force.

\[ F = EQ = (4.0\times10^4)(-2.5\times10^{-6}) = -1.0\times10^{-1}\,\mathrm{N} \]
Magnitude = \(0.10\,\mathrm{N}\). Because the charge is negative, the force is opposite to the field direction.
Maths shortcutIf a question only compares values, use proportional reasoning first: \(E \propto Q/r^2\). That can save time and reduce calculator slips.
Part 3 · 6.2.3(a)

Uniform electric fields and potential difference

Between two large parallel plates with a potential difference across them, the electric field is approximately uniform in the central region. In a uniform field, the field strength is constant, so a charged particle experiences a constant force.

Uniform electric field relation\[ E = \frac{V}{d} \]

This is often the quickest route when the question gives plate voltage and separation. Remember the unit link:

Equivalent units\[ 1\,\mathrm{N\,C^{-1}} = 1\,\mathrm{V\,m^{-1}} \]

So if the plates are \(0.040\,\mathrm{m}\) apart and the potential difference is \(2400\,\mathrm{V}\), then \(E = 2400/0.040 = 6.0\times10^4\,\mathrm{V\,m^{-1}}\).

What makes the field uniform?

Equal spacing of field lines means constant field strength. In reality, the field curves near the plate edges. This is called the edge effect. OCR usually expects the idealised central region unless stated otherwise.

Practical interpretation

The potential difference is the work done per unit charge in moving charge from one plate to the other. Larger \(V\) or smaller \(d\) gives a stronger field.

Worked example · field between plates

Two parallel plates are \(6.0\,\mathrm{mm}\) apart and maintained at a potential difference of \(1800\,\mathrm{V}\). Calculate the field strength.

\[ d = 6.0\,\mathrm{mm} = 6.0\times10^{-3}\,\mathrm{m} \] \[ E = \frac{V}{d} = \frac{1800}{6.0\times10^{-3}} = 3.0\times10^5\,\mathrm{V\,m^{-1}} \]
Answer: \(3.0\times10^5\,\mathrm{V\,m^{-1}}\) or \(3.0\times10^5\,\mathrm{N\,C^{-1}}\).
Part 4 · 6.2.3(c)

Motion of charged particles in a uniform electric field

If a charged particle enters a uniform electric field, it experiences a constant force \(F = EQ\). Constant force means constant acceleration, so you can combine electric-field ideas with SUVAT or projectile-motion-style reasoning.

Force and acceleration in a uniform field\[ F = EQ \qquad a = \frac{F}{m} = \frac{EQ}{m} \]

If the particle enters the region with horizontal speed while the field is vertical, the motion becomes a parabola: constant horizontal velocity plus vertical acceleration. This is exactly analogous to projectile motion, except the acceleration is electric rather than gravitational.

Exam tipFor an electron, the charge is negative, so the acceleration is opposite to the electric field direction. This is one of the easiest sign errors on the topic.

Horizontal motion

No horizontal force means horizontal velocity stays constant.

Vertical motion

Use \(a = EQ/m\), then SUVAT: \(s = ut + \tfrac12at^2\).

Overall path

Combine the components to explain why the path curves.

Practice build-up

Worked examples, step by step

Worked example · acceleration of an electron between plates

An electron enters a region where the field strength is \(2.4\times10^4\,\mathrm{N\,C^{-1}}\). Find its acceleration. Use \(e = 1.60\times10^{-19}\,\mathrm{C}\) and \(m_e = 9.11\times10^{-31}\,\mathrm{kg}\).

\[ F = EQ = (2.4\times10^4)(1.60\times10^{-19}) = 3.84\times10^{-15}\,\mathrm{N} \]
\[ a = \frac{F}{m} = \frac{3.84\times10^{-15}}{9.11\times10^{-31}} = 4.21\times10^{15}\,\mathrm{m\,s^{-2}} \]
Answer: \(4.2\times10^{15}\,\mathrm{m\,s^{-2}}\), opposite to the field direction because the electron is negatively charged.
Worked example · time in field and vertical deflection

An electron travels horizontally at \(3.0\times10^7\,\mathrm{m\,s^{-1}}\) through plates of length \(0.050\,\mathrm{m}\). The field in the plate region is \(1.8\times10^4\,\mathrm{N\,C^{-1}}\). Find the time inside the field and its vertical displacement.

Horizontal motion gives the time: \[ t = \frac{x}{v_x} = \frac{0.050}{3.0\times10^7} = 1.67\times10^{-9}\,\mathrm{s} \]
Vertical acceleration: \[ a = \frac{EQ}{m} = \frac{(1.8\times10^4)(1.60\times10^{-19})}{9.11\times10^{-31}} = 3.16\times10^{15}\,\mathrm{m\,s^{-2}} \]
Vertical displacement, with initial vertical velocity 0: \[ y = \frac12at^2 = \frac12(3.16\times10^{15})(1.67\times10^{-9})^2 = 4.4\times10^{-3}\,\mathrm{m} \]
Answer: time \(=1.67\times10^{-9}\,\mathrm{s}\); vertical displacement \(=4.4\,\mathrm{mm}\).
Worked example · comparing fields without a calculator

Point A is \(r\) from a charge \(Q\). Point B is \(3r\) from a charge \(2Q\). Compare field strengths.

\[ E \propto \frac{Q}{r^2} \] So \[ \frac{E_B}{E_A} = \frac{2Q/(3r)^2}{Q/r^2} = \frac{2}{9} \]
Field strength at B is \(2/9\) of the field strength at A.
Check your understanding

Knowledge Check

1
Explain what is meant by electric field strength at a point and state the direction convention used when drawing electric field lines.
3 marks
  • electric field strength is force per unit positive charge / \(E = F/Q\)
  • unit is \(\mathrm{N\,C^{-1}}\) or \(\mathrm{V\,m^{-1}}\)
  • field lines point in the direction of force on a positive test charge
2
A point is moved from \(0.10\,\mathrm{m}\) to \(0.20\,\mathrm{m}\) from an isolated point charge. How does the field strength change?
2 marks
  • field strength follows an inverse-square law
  • doubling distance makes the field strength one quarter
Interactive investigation 1

Point-charge field explorer

Use the controls to investigate how charge size and distance affect field strength. This is built to make the inverse-square pattern visible, not just memorised.

Point-charge field explorer
Field strength —
Force on probe —
Scaling idea —

Investigation prompts

  1. Keep \(Q\) constant and double the distance. What factor change do you observe in \(E\)?
  2. Keep distance constant and double \(Q\). What happens to field strength and force?
  3. Change the probe charge \(q\). Why does the force change while the field strength at the point does not?
  4. Write a one-sentence conclusion using the phrase inverse-square law.
Interactive investigation 2

Parallel-plate deflection explorer

Now investigate the motion of an electron passing through a uniform electric field. This models the classic charged-particle deflection setup used throughout school and early university physics.

Electron deflection in a uniform field
Field strength —
Electron acceleration —
Exit deflection —

Investigation prompts

  1. Keep the plate gap fixed. What happens to the deflection when the voltage doubles? Explain using \(E = V/d\).
  2. Keep voltage fixed but increase the plate gap. Why does the electron deflect less?
  3. Increase the entry speed. Why does the deflection reduce even though the field is unchanged?
  4. Write a short method for how you could test the relationship between deflection and voltage in a school-lab style simulation study.
Extended writing

QWC practice

QWC prompt 1

Explain, in a clear sequence, why the path of an electron through a uniform electric field is curved when it enters with horizontal velocity.

  • define the field direction
  • state the force on the electron is opposite to the field
  • link constant force to constant vertical acceleration
  • explain why horizontal velocity is unchanged
  • combine the motions to justify a parabolic path

QWC prompt 2

Describe the similarities and differences between gravitational fields and electric fields.

  • both are fields that exert non-contact forces
  • both can obey inverse-square forms around point/spherical sources
  • gravity is always attractive, electric forces may attract or repel
  • electric field acts on charge; gravitational field acts on mass
Exam Practice

Exam-Style Questions

1
A small positively charged sphere produces a field. At a point \(0.15\,\mathrm{m}\) from the centre, the field strength is \(1.2\times10^3\,\mathrm{N\,C^{-1}}\). Calculate the force on a \(+4.0\times10^{-8}\,\mathrm{C}\) test charge placed at that point, and state its direction.
Generated, OCR-style
3 marks
  • use \(F = EQ\)
  • \(F = (1.2\times10^3)(4.0\times10^{-8}) = 4.8\times10^{-5}\,\mathrm{N}\)
  • direction away from the sphere because both charges are positive
2
Two large horizontal plates are separated by \(12\,\mathrm{mm}\) and connected to a \(1500\,\mathrm{V}\) supply. An electron enters the field midway between the plates, travelling horizontally. (a) Calculate the field strength. (b) Show that the initial vertical acceleration of the electron is about \(2.2\times10^{16}\,\mathrm{m\,s^{-2}}\).
Adapted from common PMT / textbook style
5 marks
  • \(E = V/d = 1500 / 0.012 = 1.25\times10^5\,\mathrm{V\,m^{-1}}\)
  • \(F = EQ = (1.25\times10^5)(1.60\times10^{-19}) = 2.0\times10^{-14}\,\mathrm{N}\)
  • \(a = F/m = 2.0\times10^{-14}/9.11\times10^{-31} = 2.2\times10^{16}\,\mathrm{m\,s^{-2}}\)
  • direction opposite to electric field because electron has negative charge
3
A student writes: “Doubling the distance from a point charge halves the electric field strength.” Explain why this statement is incorrect.
Generated QWC mini-question
2 marks
  • field strength follows an inverse-square law, not an inverse law
  • doubling distance makes the field strength one quarter, not one half
4
Describe how the motion of a proton differs from the motion of an electron when both enter the same uniform electric field with the same horizontal speed.
Adapted concept question
4 marks
  • proton force is in the direction of the field; electron force opposite to the field
  • both experience force magnitude \(F = EQ\) if the charge magnitudes are equal
  • electron has much larger acceleration because its mass is much smaller
  • therefore electron path curves much more strongly
5
Challenge: An electron enters horizontally midway between two parallel plates of length \(60\,\mathrm{mm}\). The plates are \(18\,\mathrm{mm}\) apart and maintained at a potential difference of \(2.4\,\mathrm{kV}\). The electron’s horizontal speed on entry is \(3.6\times10^7\,\mathrm{m\,s^{-1}}\).

(a) Calculate the electric field strength between the plates.
(b) Show that the vertical acceleration of the electron is about \(2.3\times10^{16}\,\mathrm{m\,s^{-2}}\).
(c) Determine the time the electron spends between the plates and hence its vertical deflection on leaving the field.
(d) State whether the electron hits a plate before leaving, giving a reason.
Generated “Unified Physics” style challenge
9 marks
  • \(E = V/d = 2400 / 0.018 = 1.33\times10^5\,\mathrm{V\,m^{-1}}\)
  • \(F = EQ = (1.33\times10^5)(1.60\times10^{-19}) = 2.13\times10^{-14}\,\mathrm{N}\)
  • \(a = F/m = 2.13\times10^{-14}/9.11\times10^{-31} = 2.34\times10^{16}\,\mathrm{m\,s^{-2}}\)
  • time in field: \(t = x/v_x = 0.060 / 3.6\times10^7 = 1.67\times10^{-9}\,\mathrm{s}\)
  • vertical deflection: \(y = \tfrac12at^2 = 0.5(2.34\times10^{16})(1.67\times10^{-9})^2 = 3.26\times10^{-2}\,\mathrm{m}\)
  • \(3.26\times10^{-2}\,\mathrm{m} = 32.6\,\mathrm{mm}\), which is larger than half the plate gap \((9.0\,\mathrm{mm})\)
  • therefore the electron would hit a plate before leaving the field region
  • allow equivalent reasoning based on comparing displacement with half-gap

Topic Summary

Field idea

An electric field is a region where a charge experiences a force. The field direction is the force direction on a positive test charge.

Point charge

For a point charge, \(E\) falls with \(1/r^2\). This is why field lines spread out rapidly as you move away.

Uniform field

Between parallel plates, \(E = V/d\). Charged particles accelerate at \(a = EQ/m\).

\(E = F/Q\)
\(F = Qq/4\pi\varepsilon_0 r^2\)
\(E = Q/4\pi\varepsilon_0 r^2\)
\(E = V/d\)
\(F = EQ\)
\(a = EQ/m\)