Skip to main content
ExamExplained

Uniform fields and charged particles (SACE Stage 2 Physics)

Syllabus dot point

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

9 min read

Quick answer

Between two parallel charged plates the electric field is uniform: E=VdE = \dfrac{V}{d}, where VV is the voltage across the plates and dd their separation. A charge qq feels a constant force F=qEF = qE, giving constant acceleration. The work done moving a charge through the field is W=qVW = qV, which becomes the particle's kinetic energy if it starts from rest.

Jump to a section
  1. What this dot point is asking
  2. The uniform field between parallel plates
  3. Force, acceleration and motion
  4. Energy and potential difference
  5. The electronvolt
  6. How SACE assesses this
  7. Exam-style questions

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=9.11×10−31 kgm = 9.11\times10^{-31}\ \text{kg}, charge magnitude e=1.60×10−19 Ce = 1.60\times10^{-19}\ \text{C}.

Energy gained:

W=qV=(1.60×10−19)(500)=8.0×10−17 J.W = qV = (1.60\times10^{-19})(500) = 8.0\times10^{-17}\ \text{J}.

Final speed (work becomes kinetic energy):

qV=12mv2  ⟹  v=2qVm=2(8.0×10−17)9.11×10−31=1.76×1014≈1.3×107 m s−1.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.

Field between parallel plates

E=VdE = \frac{V}{d}

EE is the uniform field strength, VV the potential difference across the plates (V), and dd the plate separation (m). In this context the field can be quoted in V m−1\text{V m}^{-1}, which is equivalent to N C−1\text{N C}^{-1}.

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 qq in the field is constant:

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

and the acceleration is a=F/m=qE/ma = 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.

Work done across a potential difference

W=qVW = qV

WW is the work done (J) on a charge qq (C) moved through a potential difference VV (V). If the charge starts from rest, all this work becomes kinetic energy: qV=12mv2qV = \tfrac{1}{2}mv^2.

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

The electronvolt

Because W=qVW = qV, an electron moved through 1 volt gains 1.60×10−19 J1.60\times10^{-19}\ \text{J}. This convenient unit is the electronvolt (eV): 1 eV=1.60×10−19 J1\ \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=VdE = \dfrac{V}{d}, find the acceleration from a=qEma = \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=12mv2qV = \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.

Common errors
Confusing VV and EE
The voltage VV is across the whole gap; the field E=V/dE = V/d is per metre. They are linked by the plate separation.
Forgetting gravity is usually negligible
For electrons and ions, the electric force vastly exceeds the gravitational force, so gravity is normally ignored, unlike a charged oil drop, where it may matter.
Mismatching mass and charge
Use the correct particle: an electron and a proton have the same charge magnitude but very different masses, giving very different accelerations and speeds.
Treating the entry-velocity case as straight-line
A charge entering sideways follows a parabola, not a straight line; analyse it like a projectile.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2023-style5 marks
A proton enters the uniform field between two parallel plates separated by 0.0200 m0.0200\ \text{m}, perpendicular to the field, midway between the plates, with horizontal speed 4.50×106 m s−14.50\times10^6\ \text{m s}^{-1}. The potential difference is 204 V204\ \text{V}. (i) Show the field is 1.02×104 V m−11.02\times10^4\ \text{V m}^{-1}. (ii) Show the acceleration is 9.77×1011 m s−29.77\times10^{11}\ \text{m s}^{-2}. (iii) Determine the time for the proton to strike a plate. Use q=1.60×10−19 Cq = 1.60\times10^{-19}\ \text{C}, m=1.67×10−27 kgm = 1.67\times10^{-27}\ \text{kg}.
Show worked answer →

(i) E=Vd=2040.0200=1.02×104 V m−1E = \dfrac{V}{d} = \dfrac{204}{0.0200} = 1.02\times10^4\ \text{V m}^{-1}. (1 mark)

(ii) a=Fm=qEm=(1.60×10−19)(1.02×104)1.67×10−27=9.77×1011 m s−2a = \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=0.02002=0.0100 ms = \dfrac{0.0200}{2} = 0.0100\ \text{m}. There is no initial vertical velocity, so s=12at2s = \tfrac{1}{2}at^2, giving

t=2sa=2(0.0100)9.77×1011=2.05×10−14=1.43×10−7 s.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 1.43×10−7 s1.43\times10^{-7}\ \text{s}.

SACE 2024-style3 marks
An electron is accelerated from rest through a potential difference of 500 V500\ \text{V}. Calculate its final speed. Use m=9.11×10−31 kgm = 9.11\times10^{-31}\ \text{kg} and e=1.60×10−19 Ce = 1.60\times10^{-19}\ \text{C}.
Show worked answer →

The work done by the field becomes kinetic energy: qV=12mv2qV = \tfrac{1}{2}mv^2.

Energy gained: W=qV=(1.60×10−19)(500)=8.0×10−17 JW = qV = (1.60\times10^{-19})(500) = 8.0\times10^{-17}\ \text{J}.

Rearrange for speed:

v=2qVm=2(8.0×10−17)9.11×10−31=1.76×1014=1.3×107 m s−1.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=12mv2qV = \tfrac{1}{2}mv^2, 1 mark for the energy, 1 mark for the answer of about 1.3×107 m s−11.3\times10^7\ \text{m s}^{-1}.

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

ExamExplained