SCSA WACE 2022-style practice: An electron enters midway between two horizontal parallel plates 4.0\ cm long and 1.2\ cm apart, moving horizontally at 3.010^7\ m s^-1. The plates have a potential difference of 180\ 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.

A 7 mark response rewards the field, a projectile-style deflection calculation and a comparison with the available gap. **(a) Field.** E=Vd=1801.210^-2=1.510^4\ V m^-1. **(b) Deflection.** Time inside the plates: t=Lv_x=4.010^-23.010^7=1.3310^-9\ s. Vertical acceleration: a=qEm=(1.610^-19)(1.510^4)9.1110^-31=2.6310^15\ m s^-2. Then $y=12at^2=12(2.6310^15)(1.3310^-9)^2=2.310^-3\ m.$ **(c) Does it hit?** The electron starts midway, so it has 0.6\ cm=6.010^-3\ m of gap before reaching a plate. The deflection 2.310^-3\ m is less than 6.010^-3\ m, so the electron clears the plates. Markers reward E=V/d, the transit time, a=qE/m, y=12at^2 and a valid comparison with half the gap.

§-Syllabus dot point

WAPhysicsSyllabus dot point

How does a uniform electric field accelerate and deflect a charged particle?

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

A focused answer to the WACE Year 12 Physics Unit 3 content point on charged particles in uniform fields. The field between parallel plates, the constant force on a charge, energy gained through a potential difference, and parabolic deflection like a projectile.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksAn 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 20205 marksExplain 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.

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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.