Electric fields and potential: WACE Year 12 Physics Unit 3

Syllabus dot point

“Describe electric fields and potential and apply them to point charges and uniform fields between parallel plates”

8 min readUpdated 2 Jun 2026

Quick answer

A charge sets up an electric field that exerts a force on other charges. For a point charge the field follows an inverse-square law $E=k\frac{Q}{r^2}$ ; between parallel plates the field is uniform with $E=\frac{V}{d}$ . Moving a charge through a potential difference does work $W=qV$ , which lets you find the speed a charge gains when accelerated.

What this dot point is asking

WACE treats electric fields as a direct parallel to gravitational fields, with the key differences that charge comes in two signs and the force can attract or repel. You need to handle both the point-charge case and the uniform-field-between-plates case, and connect field, force, work and potential difference.

Coulomb's law and the field of a point charge

The force between two point charges is

F=k\frac{q_1q_2}{r^2},\qquad k=8.99\times10^{9}\ \text{N m}^2\,\text{C}^{-2}.

Like charges repel; opposite charges attract. The electric field strength is the force per unit positive charge,

E=\frac{F}{q}=k\frac{Q}{r^2}\quad(\text{N C}^{-1}).

Field is a vector, pointing away from positive source charges and toward negative ones. Field lines never cross, and their density represents field strength.

The uniform field between parallel plates

Two oppositely charged parallel plates produce a uniform field (except near the edges). Its magnitude is

E=\frac{V}{d}

where $V$ is the potential difference across the plates and $d$ their separation. Here the units $\text{V m}^{-1}$ and $\text{N C}^{-1}$ are equivalent. The field points from the positive plate to the negative plate, and the force $F=qE$ on a charge is constant throughout the gap.

Work and potential difference

Potential difference is the work done per unit charge moved between two points,

V=\frac{W}{q}\quad\Rightarrow\quad W=qV.

When a charge is accelerated from rest through a potential difference, that work becomes kinetic energy,

qV=\tfrac{1}{2}mv^2,

so $v=\sqrt{\dfrac{2qV}{m}}$ . This is the standard electron-gun calculation.

Motion of a charge in a uniform field

A charged particle entering a uniform field experiences a constant force, so its path is a parabola in exactly the same way a projectile is. The component along the field accelerates uniformly while the perpendicular component stays constant. This deflection of charges is the basis of cathode-ray tubes and is a favourite exam scenario.

Accelerating an electron and finding the field

Electron accelerated through 250 V between plates

An electron starts from rest and is accelerated through a potential difference of $250\ \text{V}$ . Take $q=1.60\times10^{-19}\ \text{C}$ and $m=9.11\times10^{-31}\ \text{kg}$ .

Work done on the electron:

W=qV=(1.60\times10^{-19})(250)=4.00\times10^{-17}\ \text{J}.

This equals its kinetic energy, so

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(4.00\times10^{-17})}{9.11\times10^{-31}}}=\sqrt{8.78\times10^{13}}=9.4\times10^{6}\ \text{m s}^{-1}.

If the plates are $5.0\ \text{cm}$ apart, the field between them is

E=\frac{V}{d}=\frac{250}{0.050}=5.0\times10^{3}\ \text{V m}^{-1}.

Field as a parallel to gravity

WACE deliberately mirrors electric fields with gravitational fields so you can transfer reasoning. Gravitational field strength $g=GM/r^2$ is force per unit mass; electric field strength $E=kQ/r^2$ is force per unit charge. Both are inverse-square radial fields, both are vectors, and both give a constant uniform field in special geometries (near a planet's surface for gravity, between parallel plates for electricity). The decisive difference is that mass is single-signed so gravity only attracts, whereas charge has two signs so the electric field can point toward or away from a source and forces can repel. Carrying this analogy explicitly into an answer earns marks and helps you remember which formula has which constant.

Field lines and equipotentials

Field lines start on positive charges and end on negative charges, and they are always perpendicular to surfaces of equal potential (equipotentials). Between parallel plates the field lines are parallel and evenly spaced, so the equipotentials are flat planes parallel to the plates. Around a point charge the lines are radial and the equipotentials are concentric spheres. Because no work is done moving a charge along an equipotential (there is no potential difference between its points), the force has no component along it, which is exactly why field lines and equipotentials meet at right angles.

Linking the field and potential pictures

A graph of $E$ against $r$ for a point charge falls as $1/r^2$ , while a graph for parallel plates is flat. The work done moving a charge equals $q$ times the change in potential, which is the area interpretation you may be asked to use. Keep clear which scenario you are in before choosing a formula.

Mixing the point-charge and parallel-plate formulae

Students reach for $E=k\frac{Q}{r^2}$ inside a parallel-plate problem, or use $E=\frac{V}{d}$ for a single point charge. The point-charge formula applies to the radial, inverse-square field around an isolated charge; $E=\frac{V}{d}$ applies only to the uniform field between plates. Identify the geometry first, then pick the matching equation, and never substitute a plate separation into the inverse-square law.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks

A small charged sphere carries a charge of

+8.0\ \text{nC}

. (a) Calculate the electric field strength at a point

0.15\ \text{m}

from the centre of the sphere. (b) A point charge of

-2.0\ \text{nC}

is placed at that point. Calculate the magnitude and direction of the force it experiences.

Show worked answer →

A 6 mark calculation rewards the point-charge field, the force from it and the correct direction.

(a) Field strength. Use $E=\dfrac{kQ}{r^2}$ :

E=\frac{(8.99\times10^9)(8.0\times10^{-9})}{(0.15)^2}=\frac{71.9}{0.0225}=3.2\times10^{3}\ \text{N C}^{-1},

directed radially away from the positive sphere.

(b) Force on the test charge. $F=qE=(2.0\times10^{-9})(3.2\times10^{3})=6.4\times10^{-6}\ \text{N}$ . The charge is negative, so the force is opposite to the field, that is directed toward the sphere (attraction).

Markers reward $E=kQ/r^2$ , the field value near $3.2\times10^{3}\ \text{N C}^{-1}$ , $F=qE$ and the attractive direction because the test charge is negative.

WACE 2020-style5 marks

An electron gun accelerates electrons from rest through a potential difference of

1.5\ \text{kV}

. (a) Calculate the speed of the electrons as they leave the gun. (b) Explain why the accelerating field does the same work on every electron regardless of the path it takes between the electrodes.

Show worked answer →

A 5 mark answer needs the energy calculation and a statement about the field being conservative.

(a) Exit speed. Work done equals kinetic energy gained: $qV=\tfrac{1}{2}mv^2$ , so

v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6\times10^{-19})(1500)}{9.11\times10^{-31}}}=2.3\times10^{7}\ \text{m s}^{-1}.

(b) Path independence. The work done moving a charge between two points depends only on the potential difference between them, $W=qV$ , not on the route taken. The electric field is conservative, so any path between the same start and end electrodes gives the same energy gain and therefore the same final speed.

Markers reward $qV=\tfrac{1}{2}mv^2$ , the value near $2.3\times10^{7}\ \text{m s}^{-1}$ and the conservative-field path-independence argument.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics ATAR course: syllabus and support materials — SCSA
Physics: past ATAR course examinations — SCSA