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 $g$.

Examples in context

Example 1. Electron gun in a heritage CRT at the Powerhouse Museum. A retired Sydney TV studio CRT accelerates electrons across a potential difference of $V = 18{,}000 \text{ V}$. Energy conservation gives $\tfrac{1}{2} m v^2 = e V$, so $v = \sqrt{2 e V / m} = \sqrt{2 \times 1.60 \times 10^{-19} \times 18{,}000 / 9.11 \times 10^{-31}} = 7.95 \times 10^7 \text{ m/s}$, about $0.27 c$. The relativistic correction $\gamma = 1.04$ adds only $\sim 4\%$, so the non-relativistic estimate is acceptable here. Each electron carries $eV = 1.60 \times 10^{-19} \times 18{,}000 = 2.88 \times 10^{-15} \text{ J} = 18 \text{ keV}$ of kinetic energy when it strikes the phosphor screen, exciting electrons in the phosphor to emit visible light.

Example 2. Deflecting electrons in a Lucas Heights linear accelerator beam line. At ANSTO's Lucas Heights, a $200 \text{ keV}$ electron passes through deflection plates of length $L = 5.0 \text{ cm}$ separated by $d = 1.0 \text{ cm}$ at $V = 80 \text{ V}$. Field strength $E = V/d = 8000 \text{ V/m}$. Transit speed $v = 2.66 \times 10^8 \text{ m/s}$ (mildly relativistic; we use classical estimate). Transit time $t = L/v = 1.88 \times 10^{-10} \text{ s}$. Transverse acceleration $a = e E / m = 1.41 \times 10^{15} \text{ m/s}^2$. Deflection $y = \tfrac{1}{2} a t^2 = 2.49 \times 10^{-5} \text{ m} = 25\ \mu\text{m}$, enough to steer the beam onto a downstream target slot.

Try this

Q1. Define the electric field strength between two parallel plates and write the equation relating it to potential difference and separation. [2 marks]

• Cue. Field per unit positive test charge; $E = V/d$ for uniform field between parallel plates.

Q2. A proton is accelerated from rest through $V = 1500 \text{ V}$. Calculate its final speed. ($m_p = 1.67 \times 10^{-27} \text{ kg}$.) [3 marks]

• Cue. $\tfrac{1}{2} m v^2 = e V$, so $v = \sqrt{2 e V / m_p} = \sqrt{2 \times 1.6 \times 10^{-19} \times 1500 / 1.67 \times 10^{-27}} = 5.36 \times 10^5 \text{ m/s}$.

Q3. An electron enters a $200 \text{ V}$, $2.0 \text{ cm}$-separation parallel plate region horizontally at $v_0 = 3.0 \times 10^7 \text{ m/s}$. (a) Find the field strength. (b) Find the transverse acceleration. (c) Calculate the vertical deflection after traversing $4.0 \text{ cm}$ of plate length. [2+2+2 marks]

• Cue. (a) $E = 200/0.02 = 1.0 \times 10^4 \text{ V/m}$. (b) $a = eE/m = 1.76 \times 10^{15} \text{ m/s}^2$. (c) $t = L/v_0 = 1.33 \times 10^{-9} \text{ s}$; $y = \tfrac{1}{2} a t^2 = 1.56 \times 10^{-3} \text{ m} = 1.6 \text{ mm}$.