Calculate the work done when a charge of 3.0 × 10^-6\ C is accelerated through 200\ V. [2 marks]

ScotlandPhysicsSyllabus dot point

How do electric and magnetic fields exert forces on charged particles, and how are particles accelerated?

The work done accelerating a charge through a potential difference, the force on a charge moving in a magnetic field, and the operation of particle accelerators.

An SQA Higher Physics answer on forces on charged particles, covering the work done accelerating a charge through a potential difference, the force on a charge moving in a magnetic field, and how particle accelerators work.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this key area is asking
Work done accelerating a charge
The force on a charge in a magnetic field
Particle accelerators
Examples in context
Try this

What this key area is asking

The SQA wants you to calculate the work done (and speed gained) when a charge is accelerated through a potential difference, describe the force on a charge moving in a magnetic field, and explain how particle accelerators use these ideas.

Work done accelerating a charge

An electric field does work on a charge as it moves through a potential difference. For a charge starting from rest, all of that work becomes kinetic energy, so equating $QV$ to $\tfrac{1}{2}mv^2$ gives the speed. Because the electron charge is fixed, a larger accelerating voltage gives a faster electron, which is the basis of the electron gun in an old television tube and in an electron microscope.

The force on a charge in a magnetic field

Because the magnetic force is always at right angles to the velocity, it does no work on the particle: it changes the direction of motion but not the speed. A constant-magnitude force always perpendicular to the velocity is exactly a centripetal force, so a charged particle entering a uniform magnetic field at right angles follows a circular path at constant speed. A faster or more massive particle follows a larger circle; a stronger field gives a tighter circle.

Particle accelerators

Accelerators combine the two effects. An electric field is used to speed the particles up (do work on them), and a magnetic field is used to steer and focus them.

A linear accelerator (linac) sends particles down a straight line of accelerating gaps, giving them a push at each one.
A circular accelerator (such as a cyclotron or synchrotron) uses magnetic fields to bend the particles round a loop, passing them through the same accelerating field many times so they gain energy each lap.

These machines are used to study fundamental particles (by smashing high-energy beams together), to produce medical isotopes, and in radiotherapy.

Proton accelerated then bent in a field

A proton ( $Q = 1.6 \times 10^{-19}\ \text{C}$ , $m = 1.67 \times 10^{-27}\ \text{kg}$ ) is accelerated from rest through $1.0 \times 10^{6}\ \text{V}$ . Find its kinetic energy and final speed.

step 1 Work done equals kinetic energy

$E_W = QV = 1.6 \times 10^{-19} \times 1.0 \times 10^{6} = 1.6 \times 10^{-13}\ \text{J}$ . This is the kinetic energy gained.

step 2 Final speed

$QV = \tfrac{1}{2}mv^2$ , so $v = \sqrt{\frac{2E_W}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-13}}{1.67 \times 10^{-27}}}$ .

step 3 Answer

$v = 1.4 \times 10^{7}\ \text{m s}^{-1}$ . Entering a magnetic field at right angles, this fast proton would then curve along a circular arc whose radius depends on the field strength.

Examples in context

The cathode-ray tube in an old television accelerates electrons through a few kilovolts with an electric field, then deflects them with magnetic coils to paint the picture. A mass spectrometer accelerates ions through a known voltage and bends them in a magnetic field; the radius of the circle reveals the mass-to-charge ratio, which identifies the ion. The Large Hadron Collider at CERN accelerates protons to almost the speed of light using electric fields and steers them round a 27 km ring with powerful superconducting magnets. Proton-beam radiotherapy uses an accelerator to deliver a precisely targeted dose to a tumour.

Try this

Q1. Calculate the work done when a charge of $3.0 \times 10^{-6}\ \text{C}$ is accelerated through $200\ \text{V}$ . [2 marks]

Cue. $E_W = QV = 3.0 \times 10^{-6} \times 200 = 6.0 \times 10^{-4}\ \text{J}$ .

Q2. State the direction of the magnetic force on a charge relative to its velocity and the field. [1 mark]

Cue. Perpendicular to both the velocity and the magnetic field.

Q3. Explain why the speed of a charged particle stays constant as it moves in a circle in a uniform magnetic field. [2 marks]

Cue. The force is always perpendicular to the velocity, so it does no work; with no work done the kinetic energy and therefore the speed are unchanged.

Exam-style practice questions

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

SQA Higher 20194 marksAn electron is accelerated from rest through a potential difference of 2500 V. Calculate the work done on the electron and its final speed. Take the charge on an electron as 1.6 times ten to the power minus nineteen C and its mass as 9.11 times ten to the power minus thirty-one kg.

Show worked answer →

Work done by the field equals the energy gained.

Relationship: $E_W = QV$ .

Substitution: $E_W = 1.6 \times 10^{-19} \times 2500$ .

Answer: $E_W = 4.0 \times 10^{-16}$ J.

This becomes kinetic energy: relationship $E_W = \tfrac{1}{2}mv^2$ , so $v = \sqrt{\frac{2E_W}{m}}$ . Substitution: $v = \sqrt{\frac{2 \times 4.0 \times 10^{-16}}{9.11 \times 10^{-31}}}$ . Answer: $v = 3.0 \times 10^{7}$ m per second.

Markers reward $E_W = QV$ , equating it to kinetic energy, and the final speed with unit.

SQA Higher 20213 marksState the direction of the force on a positive charge that moves at right angles to a magnetic field, and explain why a charged particle moving through a uniform magnetic field follows a circular path while its speed stays constant.

Show worked answer →

The force on a moving charge is perpendicular to both its velocity and the magnetic field; its direction is given by the right-hand rule (for a positive charge).

Because the force is always perpendicular to the velocity, it acts as a centripetal force, continually changing the direction of motion but never the speed. A constant-magnitude force always at right angles to the velocity produces uniform circular motion. No work is done on the particle (force perpendicular to displacement), so the kinetic energy and therefore the speed stay constant.

Markers reward the perpendicular direction, the centripetal role, and the no-work-done reason for constant speed.

Related dot points

Sources & how we know this

SQA Higher Physics Course Specification — SQA (2018)