SQA SQA AH style-style practice: Two point charges of +3.0\ μC and +5.0\ μC are 0.20\ m apart in a vacuum. Calculate the electrostatic force between them. Take the permittivity of free space _0 = 8.85 × 10^-12\ F m^-1.

Use Coulomb's law, F = Q_1 Q_24π_0 r^2. Substitute (charges in coulombs): F = 3.0 × 10^-6 × 5.0 × 10^-64π × 8.85 × 10^-12 × 0.20^2. Numerator: 1.5 × 10^-11. Denominator: 4π × 8.85 × 10^-12 × 0.040 = 4.45 × 10^-12. So F = 1.5 × 10^-114.45 × 10^-12 = 3.4\ N, repulsive (both charges positive). Markers reward Coulomb's law, converting microcoulombs to coulombs, squaring the separation, and noting the force is repulsive.

SQA SQA AH style-style practice: An electron is accelerated from rest through a potential difference of 250\ V. Calculate the kinetic energy and the speed it gains. Take e = 1.6 × 10^-19\ C and m_e = 9.11 × 10^-31\ kg.

The work done on the charge equals the energy gained: E_k = QV = eV. E_k = 1.6 × 10^-19 × 250 = 4.0 × 10^-17\ J. Equate to kinetic energy: 12mv^2 = 4.0 × 10^-17, so v = 2 × 4.0 × 10^-179.11 × 10^-31. v = 8.8 × 10^13 = 9.4 × 10^6\ m s^-1. Markers reward using E_k = QV, then equating to 12mv^2 to find the speed, both with units.

ScotlandPhysicsSyllabus dot point

How do electric fields and potential describe the force on charges?

Electric charge and Coulomb's law, electric field strength, electric potential, and the motion of a charged particle in an electric field.

An SQA Advanced Higher Physics answer on electric fields, covering electric charge and Coulomb's law, electric field strength, electric potential, and the motion of a charged particle accelerated through a potential difference or deflected in a uniform field.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-14

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

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Quick answer

An electric field is the region in which a charge experiences a force. Coulomb's law gives the force between two point charges, $F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}$ , an inverse-square law like gravitation but able to attract or repel. The electric field strength is the force per unit positive charge, $E = \frac{F}{Q}$ , equal to $\frac{Q}{4\pi\varepsilon_0 r^2}$ around a point charge and $\frac{V}{d}$ in a uniform field. The electric potential is the work done per unit charge bringing a charge from infinity, $V = \frac{Q}{4\pi\varepsilon_0 r}$ . A charge accelerated through a potential difference gains kinetic energy $E_k = QV$ , and a charge crossing a uniform field is deflected like a projectile.

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What this key area is asking
Coulomb's law
Electric field strength
Electric potential
Motion of a charged particle
Examples in context
Try this

What this key area is asking

The SQA wants you to use Coulomb's law for the force between point charges, define and use electric field strength, define and use electric potential, and analyse the motion of a charged particle accelerated through a potential difference or deflected in a uniform field.

Coulomb's law

Coulomb's law is the electrostatic analogue of Newton's law of gravitation: both are inverse-square laws, but the electric force can be either attractive or repulsive and is enormously stronger. The constant $\frac{1}{4\pi\varepsilon_0}$ replaces $G$ . Doubling the separation quarters the force, exactly as in gravitation.

Electric field strength

Around a point charge the field strength is $E = \frac{Q}{4\pi\varepsilon_0 r^2}$ , pointing away from a positive charge. Between two parallel plates the field is uniform and given by $E = \frac{V}{d}$ , where $V$ is the potential difference and $d$ the plate separation. Field lines run from positive to negative, and their spacing shows the field strength. The uniform-field result is the one used for parallel-plate and deflection problems.

Electric potential

Potential describes the energy landscape: a charge $q$ at potential $V$ has potential energy $qV$ . Unlike gravitational potential, electric potential can be positive (near a positive charge) or negative (near a negative charge). The work done moving a charge between two points is $W = q\Delta V$ , which is the key to accelerating charges.

Motion of a charged particle

Accelerating a charge through a potential difference is how electron guns work, and equating $QV$ to $\tfrac{1}{2}mv^2$ gives the final speed. Deflecting a charge in a uniform field is treated exactly like projectile motion: the field provides a constant transverse acceleration $a = \frac{QE}{m}$ , while the forward velocity is unchanged, so the path is parabolic.

Deflection in a uniform field

An electron travelling at $2.0 \times 10^{7}\ \text{m s}^{-1}$ enters a uniform field of $3.0 \times 10^{4}\ \text{V m}^{-1}$ at right angles, between plates of length $0.040\ \text{m}$ . Find the vertical deflection on leaving the plates. Take $e = 1.6 \times 10^{-19}\ \text{C}$ , $m_e = 9.11 \times 10^{-31}\ \text{kg}$ .

step 1 Find the transverse acceleration

$a = \dfrac{QE}{m} = \dfrac{1.6 \times 10^{-19} \times 3.0 \times 10^{4}}{9.11 \times 10^{-31}} = 5.27 \times 10^{15}\ \text{m s}^{-2}$ .

step 2 Find the time in the field

$t = \dfrac{\text{length}}{v} = \dfrac{0.040}{2.0 \times 10^{7}} = 2.0 \times 10^{-9}\ \text{s}$ .

step 3 Find the deflection

$s = \tfrac{1}{2}at^2 = \tfrac{1}{2}(5.27 \times 10^{15})(2.0 \times 10^{-9})^2 = 1.1 \times 10^{-2}\ \text{m}$ .

Examples in context

The electron gun in older displays accelerates electrons through a potential difference and deflects them with fields to scan an image, exactly this key area's physics. Inkjet printers charge and deflect droplets in a field to steer them onto the page. Mass spectrometers accelerate ions through a known potential difference to give them a known energy. The van de Graaff generator demonstrates large potentials and the inverse-square field around charged spheres.

Try this

Q1. State the quantity defined as the force per unit positive charge. [1 mark]

Cue. The electric field strength, $E$ .

Q2. Write the relationship for the field strength between two parallel plates a distance $d$ apart at potential difference $V$ . [1 mark]

Cue. $E = \frac{V}{d}$ .

Q3. State the kinetic energy gained by a charge $Q$ accelerated through a potential difference $V$ . [1 mark]

Cue. $E_k = QV$ .

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 AH style5 marksTwo point charges of

+3.0\ \mu\text{C}

and

+5.0\ \mu\text{C}

are

0.20\ \text{m}

apart in a vacuum. Calculate the electrostatic force between them. Take the permittivity of free space

\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}

Show worked answer →

Use Coulomb's law, $F = \dfrac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}$ .

Substitute (charges in coulombs): $F = \dfrac{3.0 \times 10^{-6} \times 5.0 \times 10^{-6}}{4\pi \times 8.85 \times 10^{-12} \times 0.20^2}$ .

Numerator: $1.5 \times 10^{-11}$ . Denominator: $4\pi \times 8.85 \times 10^{-12} \times 0.040 = 4.45 \times 10^{-12}$ .

So $F = \dfrac{1.5 \times 10^{-11}}{4.45 \times 10^{-12}} = 3.4\ \text{N}$ , repulsive (both charges positive).

Markers reward Coulomb's law, converting microcoulombs to coulombs, squaring the separation, and noting the force is repulsive.

SQA AH style4 marksAn electron is accelerated from rest through a potential difference of

250\ \text{V}

. Calculate the kinetic energy and the speed it gains. Take

e = 1.6 \times 10^{-19}\ \text{C}

and

m_e = 9.11 \times 10^{-31}\ \text{kg}

Show worked answer →

The work done on the charge equals the energy gained: $E_k = QV = eV$ .

$E_k = 1.6 \times 10^{-19} \times 250 = 4.0 \times 10^{-17}\ \text{J}$ .

Equate to kinetic energy: $\tfrac{1}{2}mv^2 = 4.0 \times 10^{-17}$ , so $v = \sqrt{\dfrac{2 \times 4.0 \times 10^{-17}}{9.11 \times 10^{-31}}}$ .

$v = \sqrt{8.8 \times 10^{13}} = 9.4 \times 10^{6}\ \text{m s}^{-1}$ .

Markers reward using $E_k = QV$ , then equating to $\tfrac{1}{2}mv^2$ to find the speed, both with units.

Related dot points

Sources & how we know this

SQA Advanced Higher Physics Course Specification — SQA (2019)