Investigate and quantitatively derive and analyse the interaction between charged particles and uniform electric fields, including: electric field between parallel charged plates E = V/d, acceleration of charged particles by the electric field F_net = ma, F = qE, work done on the charge W = qV, W = qEd, K = (1/2)mv^2

What this dot point is asking

NESA wants you to treat a uniform electric field (the field between parallel charged plates) like a uniform gravitational field for projectile work, then quantify the force on a charged particle ($F = qE$), the work done on it ($W = qV = qEd$), and its final kinetic energy ($K = \frac{1}{2} m v^2$). You should be able to derive a final speed from a potential difference, or a deflection from a transverse field.

The answer

The uniform field between parallel plates

Two parallel conducting plates held at a potential difference $V$ and separated by a distance $d$ produce a nearly uniform electric field in the region between them:

The field points from the positive plate to the negative plate. Units: volts per metre (V/m), equivalent to newtons per coulomb (N/C). Outside the plates (the fringing region) the field is weaker and non-uniform, but for HSC-level problems treat the inter-plate region as perfectly uniform.

Force and acceleration

A particle of charge $q$ in the field experiences a force:

For a positive charge, the force is along the field; for a negative charge (such as an electron), the force is opposite to $E$. Newton's second law gives the acceleration:

This acceleration is constant in a uniform field, so once you have $a$ the motion reduces to SUVAT (or to projectile-style two-axis kinematics if the particle has an initial transverse velocity).

Work done by the field

If the charge moves a distance $d$ in the direction of the field (or, equivalently, through a potential difference $V$ between its start and end positions):

The two forms $W = qEd$ and $W = qV$ are equivalent because $V = Ed$ for a uniform field. Use $W = qV$ whenever the potential difference is given; use $W = qEd$ when only the field strength and distance are given.

Kinetic energy and final speed

By the work-energy theorem, the work done by the net force equals the change in kinetic energy:

For a particle accelerated from rest:

This is the standard electron-gun result: knowing the accelerating voltage fixes the final speed, regardless of plate geometry.

Two motion regimes you must distinguish

Parallel acceleration. The particle enters along the field direction (or starts at rest). Motion is one-dimensional, constant acceleration. Use $v^2 = u^2 + 2as$ with $s = d$, or use energy: $qV = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.

Transverse deflection (the projectile analogue). The particle enters horizontally between the plates with speed $v_x$, and the field is vertical. The horizontal speed is constant, the vertical motion has constant acceleration $a = qE/m$. After time $t = L / v_x$ in plates of length $L$, the vertical deflection is $y = \frac{1}{2} a t^2$. This is the same maths as projectile motion under gravity, with $a$ replacing $g$.

Worked example with numbers

A proton ($m = 1.67 \times 10^{-27}$ kg, $q = 1.60 \times 10^{-19}$ C) enters horizontally at $v_x = 3.0 \times 10^5$ m/s between two horizontal plates 8.0 cm long, separated by $d = 2.0$ cm, with a potential difference $V = 200$ V (top plate positive).

Field: $E = V/d = 200 / 0.020 = 1.0 \times 10^4$ V/m, pointing down.

Force on the proton: $F = qE = 1.60 \times 10^{-19} \times 1.0 \times 10^4 = 1.6 \times 10^{-15}$ N, downward.

Acceleration: $a = F/m = 1.6 \times 10^{-15} / 1.67 \times 10^{-27} = 9.58 \times 10^{11}$ m/s$^2$.

Time in plates: $t = L / v_x = 0.080 / 3.0 \times 10^5 = 2.67 \times 10^{-7}$ s.

Vertical deflection: $y = \frac{1}{2} a t^2 = 0.5 \times 9.58 \times 10^{11} \times (2.67 \times 10^{-7})^2 = 3.4 \times 10^{-2}$ m.

That deflection (3.4 cm) exceeds the half-gap (1.0 cm), so the proton would actually strike the bottom plate before exiting. A good answer flags this physical check.

Try it: Electric field calculator for converting between $V$, $d$ and $E$ between parallel plates.

Common traps

Forgetting that electrons accelerate opposite to $E$. The force on a negative charge is $-|q|E$, so an electron near a negative plate is pushed toward the positive plate.

Mixing centimetres and metres. $E = V/d$ requires $d$ in metres. A 5 cm gap is 0.05 m, not 5.

Using $W = Fd$ for the wrong $d$. The $d$ in $W = qEd$ is the distance moved along the field. Transverse motion does no work; only the component of displacement along $E$ counts.

Including gravity. For electrons and protons in the electric fields of standard HSC problems, the gravitational force is many orders of magnitude smaller than the electric force and is usually neglected. State this assumption explicitly if asked.

Quoting $W = qV$ with the wrong sign. $V$ is the potential difference through which the charge moves; a positive charge gains kinetic energy when moving from high to low potential, a negative charge gains kinetic energy when moving from low to high potential.

In one sentence

A uniform electric field $E = V/d$ between parallel plates exerts a constant force $F = qE$ on a charge, doing work $W = qV = qEd$ that becomes kinetic energy $\frac{1}{2} m v^2$, so a particle from rest reaches $v = \sqrt{2qV/m}$.