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 4.77 min answerUpdated 2026-05-29

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.

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.

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