Charged particles in uniform fields: WACE Year 12 Physics Unit 3

Syllabus dot point

“Analyse the motion of charged particles in uniform electric fields between parallel plates”

8 min readUpdated 2 Jun 2026

Quick answer

Between parallel plates the field is uniform, $E=V/d$ , so a charge feels a constant force $F=qE$ and a constant acceleration. A charge accelerated through a potential difference $V$ gains kinetic energy $qV$ . A charge fired sideways into the field deflects in a parabola, exactly like a projectile under gravity, because the force is constant and perpendicular to the entry velocity.

What this dot point is asking

WACE wants you to treat a charged particle in a uniform field as a constant-acceleration problem, using either energy ( $qV$ ) or force ( $qE$ ) depending on what is asked. The parallel-plate field is the standard setting for accelerators, deflection in old cathode-ray tubes and inkjet printers.

The uniform field between plates

Two oppositely charged parallel plates a distance $d$ apart with potential difference $V$ produce a uniform field of strength

E=\frac{V}{d}\quad(\text{V m}^{-1}=\text{N C}^{-1}),

directed from the positive plate to the negative plate. Uniform means the field lines are parallel and evenly spaced, so the force on a charge is the same everywhere between the plates.

Force and acceleration

A charge $q$ in this field feels a constant force

F=qE=\frac{qV}{d},

giving a constant acceleration $a=F/m=qE/m$ . Because the force is constant, the constant-acceleration kinematics equations apply directly.

Energy gained across a potential difference

When a charge moves through a potential difference $V$ , the work done on it is $W=qV$ , which becomes kinetic energy if it starts from rest:

qV=\tfrac{1}{2}mv^2\;\Rightarrow\;v=\sqrt{\frac{2qV}{m}}.

This energy method is the quickest route to a final speed and avoids needing the plate separation. The electronvolt is defined here: one electronvolt is the energy an electron gains crossing a potential difference of one volt, $1\ \text{eV}=1.6\times10^{-19}\ \text{J}$ .

Parabolic deflection

If a charge enters the field moving parallel to the plates, the field exerts a force perpendicular to that motion. The component along the plates stays constant while the perpendicular component accelerates, so the path is a parabola, identical in form to projectile motion. The horizontal travel time inside the plates sets how far it deflects sideways before leaving.

Speed of an accelerated electron

An electron accelerated through 250 V

An electron ( $q=1.6\times10^{-19}\ \text{C}$ , $m=9.11\times10^{-31}\ \text{kg}$ ) starts from rest and is accelerated through a potential difference of $250\ \text{V}$ . Find its final speed.

Energy gained equals kinetic energy:

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

Solve for $v$ :

v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6\times10^{-19})(250)}{9.11\times10^{-31}}}.

v=\sqrt{8.78\times10^{13}}=9.4\times10^{6}\ \text{m s}^{-1}.

So the electron leaves at about $9.4\times10^6\ \text{m s}^{-1}$ , a few percent of the speed of light.

Working a deflection problem step by step

Deflection questions are the most common application, so it pays to have a fixed method. First find the field $E=V/d$ and hence the perpendicular acceleration $a=qE/m$ . Treat the motion in two independent directions: along the plates the velocity is constant, so the time spent between the plates is $t=L/v_x$ where $L$ is the plate length. Across the plates the particle starts with zero perpendicular velocity and accelerates, so the sideways deflection while still inside the field is $y=\tfrac{1}{2}at^2$ . If the question also asks for the deflection on a screen some distance beyond the plates, the particle then travels in a straight line at the exit velocity, whose perpendicular component is $v_y=at$ ; the extra deflection is $v_y$ multiplied by the time to reach the screen.

Comparing electric and gravitational deflection

A neat exam point is that the parabolic path here is governed by the same kinematics as projectile motion, but the cause is the electric force $qE$ rather than the weight $mg$ . The crucial difference is sign: a positive and a negative charge in the same field deflect in opposite directions, whereas every mass falls the same way under gravity. This is exactly why a parallel-plate region can be used to separate or select charges by sign, and why reversing the plate polarity flips the deflection.

Choosing energy or force

If the question gives a potential difference and asks for a final speed, use energy ( $qV=\tfrac{1}{2}mv^2$ ). If it gives plate separation and asks for acceleration or deflection, use force ( $F=qE=qV/d$ ). Picking the right route saves several lines of algebra.

Including gravity on a tiny charged particle

Students sometimes add the weight of an electron or proton to its free-body diagram between the plates. For subatomic particles the gravitational force is negligible compared with the electric force (by many orders of magnitude), so ignore weight unless the particle is a charged oil drop or similar where the question makes gravity relevant.

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-style7 marks

An electron enters midway between two horizontal parallel plates

4.0\ \text{cm}

long and

1.2\ \text{cm}

apart, moving horizontally at

3.0\times10^{7}\ \text{m s}^{-1}

. The plates have a potential difference of

180\ \text{V}

. (a) Calculate the electric field between the plates. (b) Calculate the vertical deflection of the electron as it leaves the plates. (c) State, with a reason, whether the electron strikes a plate.

Show worked answer →

A 7 mark response rewards the field, a projectile-style deflection calculation and a comparison with the available gap.

(a) Field: $E=\dfrac{V}{d}=\dfrac{180}{1.2\times10^{-2}}=1.5\times10^{4}\ \text{V m}^{-1}$ .
(b) Deflection: Time inside the plates: $t=\dfrac{L}{v_x}=\dfrac{4.0\times10^{-2}}{3.0\times10^{7}}=1.33\times10^{-9}\ \text{s}$ . Vertical acceleration: $a=\dfrac{qE}{m}=\dfrac{(1.6\times10^{-19})(1.5\times10^{4})}{9.11\times10^{-31}}=2.63\times10^{15}\ \text{m s}^{-2}$ . Then
$y=\tfrac{1}{2}at^2=\tfrac{1}{2}(2.63\times10^{15})(1.33\times10^{-9})^2=2.3\times10^{-3}\ \text{m}.$
(c) Does it hit: The electron starts midway, so it has $0.6\ \text{cm}=6.0\times10^{-3}\ \text{m}$ of gap before reaching a plate. The deflection $2.3\times10^{-3}\ \text{m}$ is less than $6.0\times10^{-3}\ \text{m}$ , so the electron clears the plates.

Markers reward $E=V/d$ , the transit time, $a=qE/m$ , $y=\tfrac{1}{2}at^2$ and a valid comparison with half the gap.

WACE 2020-style5 marks

Explain why a charged particle moving through a uniform electric field between parallel plates follows a parabolic path, and explain why the motion is analogous to projectile motion under gravity.

Show worked answer →

A 5 mark explanation needs the constant-force argument and the two-component decomposition.

Constant perpendicular force: Between the plates the field is uniform, so a charge feels a constant force $F=qE$ directed across the plates, giving a constant acceleration perpendicular to its entry velocity.
Two independent components: The velocity component parallel to the plates is unchanged (no force in that direction), while the perpendicular component grows linearly with time. Combining a constant horizontal velocity with a uniformly increasing perpendicular velocity produces a parabolic path, since $y\propto t^2$ while $x\propto t$ .
Analogy with gravity: This is mathematically identical to a projectile, where gravity supplies a constant downward acceleration while horizontal velocity stays constant. The electric field simply plays the role gravity does for a thrown object.

Markers reward the constant force, the independence of the two components and the explicit $y\propto t^2$ , $x\propto t$ parabola 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