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…
  • describe electric fields using field lines and explain what they show
  • use the definition of electric field strength \(E = F/Q\)
  • apply Coulomb’s law and the point-charge form \(E = \dfrac{Q}{4\pi\varepsilon_0 r^2}\)
  • use the uniform-field relation \(E = V/d\)
  • use signed potential, potential energy and work; compare scalar potential with vector field superposition
  • derive the capacitance of an isolated sphere and use the parallel-plate dielectric model
  • predict and calculate the motion of charged particles in a uniform electric field
  • answer OCR-style maths and QWC questions about electric fields with precise scientific language
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.”

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 \(3.0\times10^5\,\mathrm{N\,C^{-1}}\).
OCR 6.2.4 · electric potential and energy

Potential, potential energy and work

Electric potential V at a point is the work done per unit positive charge by an external agent bringing a small test charge slowly from infinity to that point. Set V = 0 at infinity for an isolated charge. Potential is a scalar; field strength is a vector. A negative source charge gives negative potential.

\[V=\frac{Q}{4\pi\varepsilon_0r}=\frac{kQ}{r},\qquad U=qV=\frac{kQq}{r}\] \[\Delta U=q(V_B-V_A),\qquad W_{\rm external}=\Delta U,\qquad W_{\rm field}=-\Delta U\]

The external-work equation assumes a slow move with no change in kinetic energy. If the electric force alone accelerates a particle, its gain in kinetic energy is −ΔU. Use the sign of the moving charge q: an electron moves naturally toward higher potential while its electric potential energy decreases.

Positive charge potential decreases toward zero with distance; negative charge potential rises toward zero. Equal like charges give zero midpoint field but positive potential; opposite charges give zero midpoint potential but nonzero field.
Signed potential approaches zero at infinity. Zero field and zero potential are different conditions.

Graphs and superposition

For a point charge, field magnitude varies as 1/r²; signed potential varies as 1/r. Potential energies have the same radial form multiplied by q, so like charges have positive U and unlike charges negative U. Add potentials algebraically and fields as vectors. The area under a force–distance graph gives work, with the sign determined by the force direction and displacement. For a positive outward force Fᵣ, an outward move has positive field work and negative ΔU; the slow external agent does the opposite work. The area under a graph of force magnitude alone cannot identify the agent or sign.

Worked example · signed energy transfer

A +3.0 nC point charge is fixed in vacuum. Move a small test charge from r = 0.20 m to r = 0.40 m. Use k = 8.99 × 10⁹ N m² C⁻².

VA = 134.85 V; VB = 67.425 V; ΔV = −67.425 V. For a proton with q = +1.60 × 10⁻¹⁹ C, ΔU = −1.0788 × 10⁻¹⁷ J; the field does +1.0788 × 10⁻¹⁷ J of work. For an electron, ΔU = +1.0788 × 10⁻¹⁷ J and the slowly moving external agent must supply that energy. The signs reverse because q reverses, not because V changes.

Worked example · midpoint comparison

Two +2.0 nC charges are 0.60 m apart. At the midpoint, their fields cancel but their potentials add: E = 0 and V = 2kQ/0.30 = +120 V (3 s.f.). Replace the right charge by −2.0 nC: V = 0 while both fields point right, giving E = 2kQ/(0.30)² = 400 N C⁻¹ to the right. Neither result implies the other quantity is zero.

Capacitance: a sphere and parallel plates

For an isolated conducting sphere of radius R, all charge resides on its surface and the outside potential is that of a point charge at its centre: Vsurface = Q/(4πε₀R). Combining this with C = Q/V gives the OCR 6.2.4(c) result C = 4πε₀R. A 0.10 m sphere has C = 1.11 × 10⁻¹¹ F = 11.1 pF.

For large parallel plates, neglect fringing and use OCR 6.2.3(b): C = εA/d = εᵣε₀A/d. A dielectric that fills the gap increases capacitance by εᵣ. With A = 0.020 m² and d = 1.0 mm, C = 177 pF in vacuum; εᵣ = 4.0 gives 708 pF. At fixed charge this reduces V and stored energy; at fixed voltage the supply adds charge and stored energy increases. State which quantity is held constant.

Original practice [4 marks] For the +3.0 nC source above, calculate V at 0.20 m and 0.40 m. Find the change in potential energy of an electron moved slowly outward between these positions, and state who supplies the energy.

Original practice [3 marks] Explain why the midpoint of two equal positive charges has zero field but nonzero potential.

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
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
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. Model: fixed uniform field, no fringing, horizontal entry midway between plates, nonrelativistic motion and negligible gravity. The upper positive plate attracts the electron; the electric field points down and the electron accelerates up. The trajectory stops if the electron reaches a plate.

Electron deflection in a uniform field
Field strength —
Electron acceleration —
Hypothetical deflection at exit plane if no plate intervened —

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.
Original question written by the PhysicsUK teaching team
3 marks
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 \(2.2\times10^{16}\,\mathrm{m\,s^{-2}}\).
Original question written by the PhysicsUK teaching team.
5 marks
3
A student writes: “Doubling the distance from a point charge halves the electric field strength.” Explain why this statement is incorrect.
Written explanation task
2 marks
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
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) Calculate the hypothetical time to the exit plane and the deflection there if no plate intervened.
(d) Decide whether the electron actually exits or hits a plate first. Calculate the horizontal distance to the collision if one occurs.
Electric-field calculation challenge
9 marks

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.

Potential and energy

V = kQ/r; U = qV; ΔU = qΔV; slow external work is ΔU and field work is −ΔU. Add potentials algebraically and fields as vectors.

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 \(a = EQ/m\).

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

Written by: PhysicsUK teaching team

Expertise: Built by a UK A Level Physics teacher and examiner.

Reviewed for: OCR A Level Physics H556

Last reviewed: 2026-10-07

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