SACE Board SACE 2023-style practice: A proton enters the uniform field between two parallel plates separated by 0.0200\ m, perpendicular to the field, midway between the plates, with horizontal speed 4.5010^6\ m s^-1. The potential difference is 204\ V. (i) Show the field is 1.0210^4\ V m^-1. (ii) Show the acceleration is 9.7710^11\ m s^-2. (iii) Determine the time for the proton to strike a plate. Use q = 1.6010^-19\ C, m = 1.6710^-27\ kg.

(i) E = Vd = 2040.0200 = 1.0210^4\ V m^-1. (1 mark) (ii) a = Fm = qEm = (1.6010^-19)(1.0210^4)1.6710^-27 = 9.7710^11\ m s^-2. (1 mark) (iii) The proton starts midway, so the distance to a plate is s = 0.02002 = 0.0100\ m. There is no initial vertical velocity, so s = 12at^2, giving $t = 2sa = 2(0.0100)9.7710^11 = 2.0510^-14 = 1.4310^-7\ s.$ (3 marks) Award the time marks for using half the plate separation, the kinematic equation with zero initial vertical speed, and the value of about 1.4310^-7\ s.

§-Syllabus dot point

SAPhysicsSyllabus dot point

How does a charged particle move when it is released in the uniform electric field between two parallel plates?

Analyse the motion of charged particles in the uniform electric field between parallel plates, using field strength, force and energy.

The uniform field between parallel plates, the force and acceleration on a charge, and the energy gained across a potential difference, with worked examples.

Generated by Claude Opus 4.89 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
The uniform field between parallel plates
Force, acceleration and motion
Energy and potential difference
The electronvolt
How SACE assesses this

What this dot point is asking

You need to analyse the motion and energy of a charged particle in the uniform field between parallel plates.

Accelerating an electron

An electron is accelerated from rest through 500 V

Find its final speed. Electron mass $m = 9.11\times10^{-31}\ \text{kg}$ , charge magnitude $e = 1.60\times10^{-19}\ \text{C}$ .

Energy gained:

W = qV = (1.60\times10^{-19})(500) = 8.0\times10^{-17}\ \text{J}.

Final speed (work becomes kinetic energy):

qV = \tfrac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2(8.0\times10^{-17})}{9.11\times10^{-31}}} = \sqrt{1.76\times10^{14}} \approx 1.3\times10^7\ \text{m s}^{-1}.

The uniform field between parallel plates

Two flat, parallel, oppositely charged plates produce an (almost) uniform electric field in the gap - the field lines are evenly spaced, straight, and run from the positive to the negative plate.

Because the field is uniform, a charge in the gap feels the same force everywhere, so it has constant acceleration, directly analogous to projectile motion under gravity.

Force, acceleration and motion

The force on a charge $q$ in the field is constant:

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

and the acceleration is

a = F/m = qE/m

. A particle released from rest accelerates uniformly toward the oppositely charged plate, so the constant-acceleration kinematics equations apply.

If a charge enters the field sideways (perpendicular to the field), it follows a parabolic path, exactly like a projectile: constant velocity along the plates, constant acceleration across them. The horizontal and vertical motions share only the time. This is the basis of the cathode-ray oscilloscope and the deflection plates in older displays.

Energy and potential difference

Moving a charge through a potential difference transfers energy.

This gives a direct route to the final speed of an accelerated particle without needing the field or distance.

The electronvolt

Because $W = qV$ , an electron moved through 1 volt gains $1.60\times10^{-19}\ \text{J}$ . This convenient unit is the electronvolt (eV): $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ . It is widely used in atomic and particle physics, where joules are awkwardly small.

How SACE assesses this

SACE Stage 2 parallel-plate questions are typically multi-part: find the field from $E = \dfrac{V}{d}$ , find the acceleration from $a = \dfrac{qE}{m}$ , then use the constant-acceleration kinematics to find a time or speed. A frequent geometry is a charge entering midway between the plates, so the distance to a plate is half the separation; this is the single most common slip. Because the field is uniform, the motion is exactly like projectile motion, with constant velocity along the plates and constant acceleration across them, sharing only the time. The energy route, $qV = \tfrac{1}{2}mv^2$ , gives the final speed of a charge accelerated from rest without needing the field or distance. Quote each equation before substituting, and use the correct mass for the named particle.

Exam-style practice questions

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

SACE 20235 marksA proton enters the uniform field between two parallel plates separated by

0.0200\ \text{m}

, perpendicular to the field, midway between the plates, with horizontal speed

4.50\times10^6\ \text{m s}^{-1}

. The potential difference is

204\ \text{V}

. (i) Show the field is

1.02\times10^4\ \text{V m}^{-1}

. (ii) Show the acceleration is

9.77\times10^{11}\ \text{m s}^{-2}

. (iii) Determine the time for the proton to strike a plate. Use

q = 1.60\times10^{-19}\ \text{C}

m = 1.67\times10^{-27}\ \text{kg}

Show worked answer →

(i) $E = \dfrac{V}{d} = \dfrac{204}{0.0200} = 1.02\times10^4\ \text{V m}^{-1}$ . (1 mark)

(ii) $a = \dfrac{F}{m} = \dfrac{qE}{m} = \dfrac{(1.60\times10^{-19})(1.02\times10^4)}{1.67\times10^{-27}} = 9.77\times10^{11}\ \text{m s}^{-2}$ . (1 mark)

(iii) The proton starts midway, so the distance to a plate is $s = \dfrac{0.0200}{2} = 0.0100\ \text{m}$ . There is no initial vertical velocity, so $s = \tfrac{1}{2}at^2$ , giving

t = \sqrt{\dfrac{2s}{a}} = \sqrt{\dfrac{2(0.0100)}{9.77\times10^{11}}} = \sqrt{2.05\times10^{-14}} = 1.43\times10^{-7}\ \text{s}.

(3 marks)
Award the time marks for using half the plate separation, the kinematic equation with zero initial vertical speed, and the value of about

1.43\times10^{-7}\ \text{s}

SACE 20243 marksAn electron is accelerated from rest through a potential difference of

500\ \text{V}

. Calculate its final speed. Use

m = 9.11\times10^{-31}\ \text{kg}

and

e = 1.60\times10^{-19}\ \text{C}

Show worked answer →

The work done by the field becomes kinetic energy: $qV = \tfrac{1}{2}mv^2$ .

Energy gained: $W = qV = (1.60\times10^{-19})(500) = 8.0\times10^{-17}\ \text{J}$ .

Rearrange for speed:

v = \sqrt{\dfrac{2qV}{m}} = \sqrt{\dfrac{2(8.0\times10^{-17})}{9.11\times10^{-31}}} = \sqrt{1.76\times10^{14}} = 1.3\times10^7\ \text{m s}^{-1}.

1 mark for

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

, 1 mark for the energy, 1 mark for the answer of about

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