Fields and their consequences: A-level Physics AQA revision notes

3.7.1

Fields

Types of field 3.7.1

Definitions

Field: a region of space in which a suitable object experiences a non-contact force.

Key results

Gravitational field strength: $g = \frac{F}{m}$ (units $N kg^{- 1}$ ), direction: force on a test mass, always toward the source mass.
Electric field strength: $E = \frac{F}{q}$ (units $N C^{- 1}$ ), direction: force on a positive test charge.
Magnetic field, $B$ (units tesla, T): direction defined by the force on a moving charge, which depends on both the charge's velocity and its sign — not a simple 'toward/away from source' rule the way gravitational and electric fields are.

Notes

Gravitational fields are sourced by mass and are always attractive; electric fields are sourced by charge and can be attractive or repulsive; magnetic fields are sourced by moving charge (current) and exert force only on OTHER moving charges, not on stationary ones.
Field lines represent a vector field at every point in space — they are not necessarily the actual paths a particle released into the field would follow (that depends on the particle's initial velocity too, especially for magnetic fields).

Gravitation and electrostatics compared 3.7.1

Key results

Newton's law of gravitation: $F = \frac{G m _{1} m _{2}}{r ^{2}}$ ( $G \approx 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ ).
Coulomb's law: $F = \frac{∣ q _{1} q _{2} ∣}{4 π ε _{0} r ^{2}}$ .
Both are inverse-square laws: doubling separation reduces force to one quarter; tripling separation reduces it to one ninth.

Notes

Masses always attract; like charges repel, unlike charges attract — gravity has only one 'sign' of source, electric charge has two.
Each pair (whichever force) exerts equal and opposite forces on the two different bodies involved — a direct instance of Newton's third law, not a separate rule for fields.
For a proton and an electron, the electric force between them is about $2.3 \times 1 0^{39}$ times stronger than the gravitational force — gravity is completely negligible at the scale of individual particles, and only becomes dominant for large, overall-neutral masses like planets, where electric forces cancel out but gravity does not.

Field lines and equipotential surfaces 3.7.1

Definitions

Equipotential surface: a surface joining all points at the same potential — no work is done moving along one.

Key results

Field lines are always perpendicular to equipotential surfaces, and never cross one another.
Greater field-line density indicates a stronger field; the field vector at any point is tangent to the field line through it.

Notes

Gravitational potential is potential energy per unit mass; electric potential is potential energy per unit charge — the same equipotential concept applies to both.
Moving along an equipotential surface: $Δ V = 0$ , so the field does zero work on whatever is moved — a direct, general consequence of $V$ being constant there, true for gravitational or electrostatic fields alike.
For a uniform field (e.g. between parallel plates), equipotentials are evenly spaced parallel planes; for a radial field (around a point source), they are concentric spheres — in both cases, closer equipotential spacing marks a stronger field, exactly mirroring field-line density.

Field direction and force direction 3.7.1

Key results

Electric force on a charge: $F = qE$ — a positive charge feels force in the direction of $E$ ; a negative charge feels force opposite to $E$ .
Magnetic force acts only on a MOVING charge, and only has a component perpendicular to both velocity and field — a stationary charge feels no magnetic force at all, whatever the field.

Notes

For two test charges placed in the same source field, $F = qE$ shows the force on each is proportional to its own charge (both magnitude and sign) — a useful way to separate 'what the field looks like' ( $E$ , from the source) from 'what happens to a specific test charge' ( $F = qE$ ).
Reversing a moving charge's sign (at the same velocity, same field) reverses the direction of the magnetic force it feels — the field itself hasn't changed, only how a given charge responds to it.

Worked examples

Worked example 3.7.1 · 4 marks

Compare a uniform gravitational field and a uniform electric field in terms of

(a) the direction a small test mass or test positive charge experiences a force, relative to the field lines, and

(b) whether the field can ever exert a force that pushes the test object away from the source.

Explain your answer to

(b) with reference to the sign of the source's mass or charge.

Show worked solution

(a) In both cases, the force on a small test object (a test mass, or a small positive test charge) acts along the direction of the field lines at that point — field lines are, by definition, drawn in the direction of the force on a positive test charge (or, for gravity, on a test mass).

(b) A gravitational field can never push a test mass away from its source, because mass is always positive, so the gravitational force between any two masses is always attractive.

An electric field CAN push a test charge away from its source, because electric charge can be positive or negative: two like charges (both positive, or both negative) repel, giving a field that pushes a positive test charge away, whereas gravity has no equivalent 'like charges repel' case.

Mark scheme · 4 marks

States the force on a test mass or test positive charge acts along the field line direction at that point, in both cases 1 mark
States a gravitational field can never repel a test mass 1 mark
States mass is always positive, so gravitational force is always attractive 1 mark
States an electric field CAN repel a test charge, since charge can be positive or negative, unlike mass 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.1 · 3 marks

A student draws the electric field lines around an isolated positive point charge as straight lines radiating outward, evenly spaced close to the charge and becoming further apart at greater distances.

Explain what the increasing spacing of the field lines with distance represents physically.

Show worked solution

Field line spacing represents field strength: field lines drawn closer together indicate a stronger field, and lines drawn further apart indicate a weaker field.

The increasing spacing with distance from the point charge correctly represents the fact that the field strength falls off with distance from a point charge (as $1/ r^{2}$ , following the inverse-square form of Coulomb's law), so the field becomes weaker further from the source.

Mark scheme · 3 marks

States that field line spacing represents field strength: closer lines mean a stronger field 1 mark
States the increasing spacing with distance shows the field getting weaker further from the charge 1 mark
Links this to the inverse-square fall-off of field strength with distance from a point charge 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.1 · 4 marks

A positive point charge sits at the centre of a set of concentric circular equipotential lines drawn around it.

A small positive test charge is moved from one equipotential line to another, further out, along a path that is NOT a straight radial line.

Explain whether any work is done against the electric field during this move, and explain why the shape of the path taken does not affect your answer.

Show worked solution

Work IS done against the field, because the two equipotential lines are at different potentials (the outer one is at lower potential, since potential falls with distance from a positive charge), and moving between two different equipotentials always requires energy transfer regardless of path — only movement ALONG a single equipotential (constant potential) involves zero work.

The path's shape does not affect the amount of work done because work done depends only on the potential difference between the start and end points (a property of the field alone), not on the route taken to get there — this is a defining property of a conservative field, which both gravitational and electric fields are.

Mark scheme · 4 marks

States that work is done, since the two equipotentials are at different potentials 1 mark
States that moving along a single equipotential does zero work, contrasted with moving between two different ones 1 mark
States work done depends only on the potential difference between start and end points, not the path taken 1 mark
Identifies this path-independence as a property of a conservative field 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.7.2

Gravitational fields

Universal gravitation 3.7.2.1–3.7.2.2

Definitions

Gravitational field strength, $g = \frac{F}{m}$ : the force per unit mass at a point in space, independent of the test mass used to measure it.

Key results

Newton's law of gravitation: $F = \frac{GM m}{r ^{2}}$ (point masses, or outside spherically symmetric bodies; $r$ measured from the source's centre).
Gravitational field strength (point mass): $g = \frac{GM}{r ^{2}}$ . At a planet's surface, $r = R$ ; at height $h$ above it, $r = R + h$ .

Notes

$g$ is independent of the test mass placed at that point — it is a property of the source and the location alone, exactly parallel to how $E$ works for electric fields.
Gravity is always attractive and acts between every pair of masses, however small — it is only significant for very large masses (planets, stars) because it is so much weaker than the electric force at small scales.

Gravitational potential and work 3.7.2.3

Definitions

Gravitational potential, $V_{g}$ : the gravitational potential energy per unit mass at a point, defined as zero at infinity and negative everywhere else, $V_{g} = - \frac{GM}{r}$ .

Key results

Work done moving a mass (slowly, no kinetic-energy change) between two points: $W_{ext} = m Δ V_{g} = Δ E_{p}$ ; work done BY the gravitational field: $W_{gravity} = - m Δ V_{g}$ .

Notes

Gravitational potential is negative because gravity is always attractive, so work must be done against it to move a mass to infinity — the mass sits in an energy 'well' at any finite distance.
Moving outwards raises $V_{g}$ (makes it less negative) and raises $E_{p}$ — external work is needed. Moving inwards lowers both, and the field itself does positive work.
Equipotential surfaces around a point mass are spheres centred on it, and gravitational field lines meet these spheres at right angles everywhere.

Field strength and potential with distance 3.7.2.3

Key results

Field strength is the negative gradient of potential: $g_{r} = - \frac{d V _{g}}{d r} = - g$ (outward taken as positive; the field itself points inward, hence the sign).
The (unsigned) area under a graph of field strength $g$ against $r$ , between two radii, gives the magnitude of the potential difference between them, $Δ V_{g}$ .

Notes

$g$ falls off as $\frac{1}{r ^{2}}$ with distance, while $V_{g}$ (always negative, approaching zero from below) falls off more slowly, as $\frac{1}{r}$ — the two graphs have genuinely different shapes, not just different signs of the same curve.
Field strength being the (negative) gradient of potential is exactly why field lines always point from high to low potential, and why the field is strongest wherever equipotential lines are drawn closest together — the same relationship holds for electric fields and potential too.

Circular orbits 3.7.2.4

Key results

For a stable circular orbit, gravitational force provides exactly the centripetal force needed: $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$ .
Orbital speed: $v = \frac{GM}{r}$ . Kepler's third law: $T^{2} = \frac{4 π ^{2} r ^{3}}{GM}$ , i.e. $T^{2} \propto r^{3}$ .

Notes

For the same central mass $M$ : orbital speed $v \propto r^{- 1/2}$ and period $T \propto r^{3/2}$ — a higher orbit is always slower and takes longer to complete, not merely longer because the path is bigger.
An orbiting satellite is in continuous free fall — gravity provides its ENTIRE centripetal force, with no other force needed to 'hold it up'. Apparent weightlessness aboard a satellite reflects this free-fall state; it does not mean gravity is somehow absent or negligible there.

Orbital energy and escape speed 3.7.2.4

Key results

For a circular orbit: $E_{p} = - \frac{GM m}{r}$ , $E_{k} = \frac{GM m}{2 r}$ , total energy $E_{total} = - \frac{GM m}{2 r}$ .
Escape speed: setting total mechanical energy to zero (arriving at infinity with zero remaining speed), $\frac{1}{2} m v_{esc}^{2} - \frac{GM m}{r} = 0$ , giving $v_{esc} = \frac{2 GM}{r}$ .

Notes

Total orbital energy is NEGATIVE for any bound circular orbit — this is exactly what 'gravitationally bound' means: escaping to infinity would require adding energy, since energy there is defined as zero.
At the same starting radius $r$ , escape speed is exactly $2$ times the circular orbital speed at that radius — a frequently tested direct comparison.
As orbital radius increases: $E_{p}$ rises (becomes less negative), $E_{k}$ falls, but total energy also rises (becomes less negative) — the total does NOT stay fixed between different possible circular orbits, since a higher orbit is a genuinely higher-energy (less bound) state.

Earth satellite orbits 3.7.2.4

Definitions

Geosynchronous orbit: an orbit whose period exactly matches Earth's own rotation period (one sidereal day, $\approx 23 h 56 min$ ).
Geostationary orbit: a geosynchronous orbit that is also circular, equatorial, and in the same direction as Earth's rotation — so the satellite stays permanently above the same point on Earth's surface.

Key results

Geostationary orbit radius $\approx 42200 km$ (height above the surface $\approx 35800 km$ ) — a single value, fixed entirely by requiring $T \approx 24 h$ in Kepler's third law, with no freedom to choose a different radius.

Notes

A geosynchronous orbit need not be geostationary — only a circular, equatorial, same-direction geosynchronous orbit stays fixed over one ground point; other geosynchronous orbits still return to the same point once per day but trace a path in the sky in between.
A geostationary orbit is the standard choice for communications and continuous weather-monitoring satellites, since a ground antenna can then point at one fixed position in the sky.
A low Earth orbit (e.g. height $\approx 400 km$ , giving orbital speed $\approx 7.67 km s^{- 1}$ and period $\approx 92 min$ ) is used for Earth imaging and crewed spacecraft — much lower altitude and much shorter period than geostationary, and often near-polar so successive orbits sweep over different longitudes as Earth rotates beneath.

Worked examples

Worked example 3.7.2 · 5 marks

A satellite orbits the Earth in a circular orbit of radius $7.20 \times 1 0^{6} m$ .

Calculate the orbital period of the satellite, in hours. (:

G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}

mass of Earth:

M = 5.97 \times 1 0^{24} kg

)

Show worked solution

Using Kepler's third law,

T^{2} = \frac{4 π ^{2} r ^{3}}{GM}

r^{3} = (7.20 \times 1 0^{6})^{3} = 3.732 \times 1 0^{20} m^{3}

T^{2} = \frac{4 π ^{2} \times 3.732 \times 1 0 ^{20}}{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}} = \frac{1.474 \times 1 0 ^{22}}{3.982 \times 1 0 ^{14}} \approx 3.702 \times 1 0^{7} s^{2}

T = 3.702 \times 1 0^{7} \approx 6085 s

which is:

\frac{6085}{3600} \approx 1.69 hours

Mark scheme · 5 marks

Selects and correctly rearranges Kepler's third law $T^{2} = \frac{4 π ^{2} r ^{3}}{GM}$ 1 mark
Calculates $r^{3} \approx 3.732 \times 1 0^{20} m^{3}$ 1 mark
Substitutes correctly to find $T^{2} \approx 3.70 \times 1 0^{7} s^{2}$ 1 mark
Calculates $T \approx 6085 s$ 1 mark
Converts to hours, $T \approx 1.69 hours$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.2 · 5 marks

Calculate the escape speed from the surface of Mars, given its mass is $6.42 \times 1 0^{23} kg$ and its radius is $3.39 \times 1 0^{6} m$ .

Explain, without further calculation, whether the escape speed of a more massive object launched from the same point on Mars would be larger, smaller, or the same. (:

G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}

)

Show worked solution

v_{esc} = \frac{2 GM}{r} = \frac{2 \times 6.67 \times 1 0 ^{- 11} \times 6.42 \times 1 0 ^{23}}{3.39 \times 1 0 ^{6}} = \frac{8.566 \times 1 0 ^{13}}{3.39 \times 1 0 ^{6}} = 2.527 \times 1 0^{7} \approx 5.03 \times 1 0^{3} m s^{- 1}

Since the escape speed formula does not contain the mass of the escaping object at all, a more massive object launched from the same point would have exactly the same escape speed.

Mark scheme · 5 marks

Selects and uses $v_{esc} = 2 GM / r$ with the planet's own mass and radius 1 mark
Substitutes correctly to find $\frac{2 GM}{r} \approx 2.53 \times 1 0^{7}$ 1 mark
Calculates $v_{esc} \approx 5.03 \times 1 0^{3} m s^{- 1}$ 1 mark
States a more massive object would have the same escape speed 1 mark
Explains this is because the escaping object's own mass does not appear in the escape-speed formula 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.2 · 4 marks

A geostationary satellite orbits the Earth once every 23 h 56 min (one sidereal day).

A second satellite, in a different circular orbit, has an orbital radius exactly $4$ times that of the geostationary satellite.

Calculate the orbital period of the second satellite, in days, without recalculating $GM$ .

Show worked solution

From Kepler's third law, $T^{2} \propto r^{3}$ , so:

\frac{T _{2}^{2}}{T _{1}^{2}} = (\frac{r _{2}}{r _{1}})^{3} = 4^{3} = 64

So:

T_{2} = T_{1} 64 = T_{1} \times 8

Since $T_{1} \approx 1$ day (a sidereal day), $T_{2} \approx 8$ days.

Mark scheme · 4 marks

States $T^{2} \propto r^{3}$ from Kepler's third law 1 mark
Calculates $(r_{2} / r_{1})^{3} = 4^{3} = 64$ 1 mark
Takes the square root to find $T_{2} / T_{1} = 8$ 1 mark
Calculates $T_{2} \approx 8$ days, using $T_{1} \approx 1$ day for the geostationary orbit 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.7.3

Electric fields

Coulomb's law 3.7.3.1

Key results

Coulomb's law: $F = \frac{∣ Q _{1} Q _{2} ∣}{4 π ε _{0} r ^{2}} = \frac{k ∣ Q _{1} Q _{2} ∣}{r ^{2}}$ , with $k = \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9} N m^{2} C^{- 2}$ and $ε_{0} \approx 8.85 \times 1 0^{- 12} F m^{- 1}$ (permittivity of free space).

Notes

Coulomb's law is mathematically identical in form to Newton's law of gravitation — an inverse-square law — but electric charge can be positive or negative, so electric force can be attractive OR repulsive, unlike gravity, which is always attractive.
For a proton and an electron, the electric force between them is about $2.3 \times 1 0^{39}$ times the gravitational force between them — both vary as $1/ r^{2}$ , but their relative strength never changes with distance, since that ratio is fixed by the particles' charge and mass alone.

Electric field strength 3.7.3.2

Definitions

Electric field strength, $E = \frac{F}{q}$ : force per unit positive test charge at a point, units $N C^{- 1}$ (equivalently $V m^{- 1}$ ).

Key results

Electric field strength (point charge): $E = \frac{k ∣ Q ∣}{r ^{2}}$ .

Notes

An isolated positive charge's field points radially OUTWARD (a positive test charge would be repelled); an isolated negative charge's field points radially INWARD (a positive test charge would be attracted) — field direction is always defined by the force on a POSITIVE test charge, by convention.

A uniform field between parallel plates 3.7.3.2

Key results

Uniform field between parallel plates: $E = \frac{U}{d}$ (plate separation $d$ , potential difference $U$ between them) — derived from $W_{field} = q U = F d$ and $F = qE$ .

Notes

A uniform field's strength is constant in magnitude and direction throughout the gap (away from the edges), unlike the field around a point charge, which weakens with distance.
Equipotentials in this region are planes parallel to the plates, evenly spaced — potential falls steadily in the field's own direction, from the higher-potential plate to the lower.
Edge effects (the field bulging outward near the plates' physical edges) are deliberately excluded from this ideal model — it holds well only when the plate dimensions are large compared with their separation.

Charged-particle motion in a uniform field 3.7.3.2

Key results

A charged particle entering a uniform field at right angles to it: horizontal motion is unaffected ( $a_{x} = 0$ , $v_{x} = u$ constant); vertical acceleration $a_{y} = \frac{qE}{m}$ — giving a parabolic path within the field, $y = \frac{qE x ^{2}}{2 m u ^{2}}$ .
Using energy instead of force: for motion under the electric force alone, $Δ E_{k} = - q Δ V$ .

Notes

This is structurally identical to projectile motion under gravity — a constant force perpendicular to the initial velocity produces a parabolic path, whether that force is $m g$ (gravity) or $qE$ (a uniform electric field).
Once the particle leaves the field region (e.g. passes beyond the plates), no further force acts on it (in this idealised model), so its path becomes a straight line again, continuing at the velocity it had on exit.
For an electron ( $q = - e$ , negative), the force is directed opposite to $E$ — toward the positive plate — even though the standard deflection formula is written in terms of the general (signed) charge $q$ .

Electric potential and equipotentials 3.7.3.3

Definitions

Electric potential, $V = \frac{E _{p}}{q} = \frac{k Q}{r}$ : the electric potential energy per unit charge at a point, defined as zero at infinity.

Key results

Work done moving a charge (slowly): $W_{ext} = q Δ V = Δ E_{p}$ ; work done BY the field: $W_{field} = - q Δ V$ — use the signed value of $q$ .

Notes

Electric potential, unlike gravitational potential, can be positive (near a positive charge) or negative (near a negative charge), since charge itself can be either sign.
Field lines are always perpendicular to equipotential surfaces — around an isolated point charge, equipotentials are concentric spheres, exactly as for a gravitational point source.

Radial field and potential graphs 3.7.3.3

Key results

$E_{r} = - \frac{d V}{d r}$ — field strength is the negative gradient of potential, just as for gravitational fields.
The (signed) area under a graph of $E_{r}$ against $r$ , between two radii, gives $- Δ V$ between them.

Notes

Both $E$ and $V$ approach zero as $r \to \infty$ , but at different rates ( $E \propto 1/ r^{2}$ , $V \propto 1/ r$ ) — the same relationship, and the same graph-shape contrast, as for gravitational field strength and potential.
For a negative source charge, $V$ is negative and rises (becomes less negative) with distance, while the field points inward — the mathematics mirrors the positive-charge case exactly, with every sign flipped.

Worked examples

Worked example 3.7.3 · 6 marks

In a cathode ray tube, electrons are accelerated from rest through a potential difference of $2.5 kV$ , then enter a region between two parallel plates of length $4.0 cm$ and separation $2.0 cm$ , across which a potential difference of $300 V$ is applied.

Calculate the vertical deflection of the electrons as they leave the plates. (:

e = 1.60 \times 1 0^{- 19} C

m_{e} = 9.11 \times 1 0^{- 31} kg

)

Show worked solution

Horizontal (entry) speed from the accelerating p.d.:

e V_{1} = \frac{1}{2} m_{e} v_{x}^{2} \Rightarrow v_{x} = \frac{2 e V _{1}}{m _{e}} = \frac{2 \times 1.60 \times 1 0 ^{- 19} \times 2500}{9.11 \times 1 0 ^{- 31}} \approx 2.965 \times 1 0^{7} m s^{- 1}

Time between the plates:

t = \frac{L}{v _{x}} = \frac{0.040}{2.965 \times 1 0 ^{7}} \approx 1.349 \times 1 0^{- 9} s

Electric field between the plates:

E = \frac{V _{2}}{d} = \frac{300}{0.020} = 1.5 \times 1 0^{4} V m^{- 1}

Vertical acceleration:

a = \frac{e E}{m _{e}} = \frac{1.60 \times 1 0 ^{- 19} \times 1.5 \times 1 0 ^{4}}{9.11 \times 1 0 ^{- 31}} \approx 2.635 \times 1 0^{15} m s^{- 2}

Vertical deflection:

y = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2.635 \times 1 0^{15} \times (1.349 \times 1 0^{- 9})^{2} \approx 2.4 \times 1 0^{- 3} m

(2.4 mm).

Mark scheme · 6 marks

Uses $e V_{1} = \frac{1}{2} m_{e} v_{x}^{2}$ to find the horizontal entry speed $v_{x} \approx 2.97 \times 1 0^{7} m s^{- 1}$ 1 mark
Calculates the time spent between the plates, $t = L / v_{x} \approx 1.35 \times 1 0^{- 9} s$ 1 mark
Calculates the field between the plates, $E = V_{2} / d = 1.5 \times 1 0^{4} V m^{- 1}$ 1 mark
Calculates the vertical acceleration, $a = e E / m_{e} \approx 2.64 \times 1 0^{15} m s^{- 2}$ 1 mark
Uses $y = \frac{1}{2} a t^{2}$ to combine the acceleration and time 1 mark
Calculates the vertical deflection $y \approx 2.4 \times 1 0^{- 3} m$ (2.4 mm) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.3 · 4 marks

Two point charges, $+ 6.0 nC$ and $- 4.0 nC$ , are fixed $0.30 m$ apart.

Calculate the net electric force on the negative charge, stating its direction. (:

ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}

)

Show worked solution

F = \frac{Qq}{4 π ε _{0} r ^{2}} = \frac{6.0 \times 1 0 ^{- 9} \times 4.0 \times 1 0 ^{- 9}}{4 π \times 8.85 \times 1 0 ^{- 12} \times 0.3 0 ^{2}} = \frac{2.4 \times 1 0 ^{- 17}}{1.001 \times 1 0 ^{- 11}} \approx 2.40 \times 1 0^{- 6} N

(2.4 μN).

Since the two charges have opposite signs, the force between them is attractive, so the force on the negative charge is directed towards the positive charge.

Mark scheme · 4 marks

Selects and uses Coulomb's law $F = \frac{Qq}{4 π ε _{0} r ^{2}}$ 1 mark
Substitutes correctly (using the magnitudes of both charges) 1 mark
Calculates $F \approx 2.4 \times 1 0^{- 6} N$ 1 mark
States the force is attractive, directed towards the positive charge, since the charges have opposite signs 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.7.4

Capacitance

Charge separation in a capacitor 3.7.4.1

Definitions

Capacitance, $C = \frac{Q}{V}$ : how much charge a component stores per unit potential difference across it, measured in farads ( $1 F = 1 C V^{- 1}$ ).

Key results

Parallel-plate capacitor (vacuum/air gap): $C_{0} = \frac{ε _{0} A}{d}$ (overlapping plate area $A$ , separation $d$ ).

Notes

Larger plate area or smaller separation both give larger capacitance — for fixed geometry with a linear dielectric, $C$ itself is independent of $Q$ and $V$ , a property of the capacitor's construction, not of how much it happens to be charged at any moment.
The two plates carry equal and opposite charge ( $+ Q$ and $- Q$ ), so the capacitor's NET charge is always zero — $Q$ in $C = Q / V$ refers to the magnitude of charge on either plate alone, not a sum.

How a dielectric increases capacitance 3.7.4.2

Definitions

Relative permittivity, $ε_{r}$ : the factor by which inserting a dielectric between a capacitor's plates increases its capacitance compared with a vacuum, $ε_{r} = \frac{C}{C _{0}}$ .

Key results

Capacitance with a dielectric fully filling the gap: $C = ε_{r} ε_{0} \frac{A}{d} = ε_{r} C_{0}$ .

Notes

An insulating dielectric's molecules polarise in the plates' field — in a polar dielectric, molecules rotate so their positive ends turn toward the negative plate; this partial alignment produces its own field that opposes the field from the plates' free charge.
For an isolated capacitor (charge $Q$ fixed, disconnected from any supply): inserting a dielectric leaves $Q$ unchanged but reduces the resultant field and so reduces $V$ — since $C = Q / V$ , capacitance rises.
For a capacitor connected to a fixed-voltage supply: $V$ (and so $E$ ) stays fixed, and extra charge flows onto the plates from the supply to maintain that voltage against the dielectric's opposing field — capacitance again rises, but this time because $Q$ increases at fixed $V$ .

Energy stored in the electric field 3.7.4.3

Key results

Capacitor energy: $E = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C}$ — the area under a graph of charge against potential difference, for an ideal linear capacitor.

Notes

The factor of $\frac{1}{2}$ arises because potential difference across the capacitor rises from zero as it charges — the average potential difference during charging is half the final value, not the final value itself, so work done is half of $Q V$ , not all of it.
At fixed capacitance, doubling $V$ quadruples the stored energy (since energy $\propto V^{2}$ ) — a proportionality worth remembering directly, not just a formula to substitute into.

Charging and discharging through a resistor 3.7.4.4

Definitions

Time constant, $τ = RC$ : governs the exponential charging/discharging rate of a capacitor through a resistor.

Key results

Charging (source connected): $V_{0} = I R + V_{C}$ . Discharging (source disconnected, resistor remains in the loop): $0 = I R + V_{C}$ .

Notes

A changeover switch models this cleanly: position A connects the source to charge the capacitor through $R$ ; position B disconnects the source and lets the capacitor discharge back through the same $R$ .
By this circuit's own sign convention, positive current flows toward the capacitor's upper plate: charging current is positive, discharging current is negative — the magnitude of current is what actually decays exponentially in both cases, not its sign, which stays fixed by which phase you're in.

Exponential charging and discharging 3.7.4.4

Key results

Discharging: $Q = Q_{0} e^{- t / RC}$ , $V_{C} = V_{0} e^{- t / RC}$ , $I = - I_{0} e^{- t / RC}$ — after one time constant, each has fallen to $\frac{1}{e} \approx 36.8%$ of its initial value.
Charging: $Q = C V_{0} (1 - e^{- t / RC})$ , $V_{C} = V_{0} (1 - e^{- t / RC})$ — after one time constant, each has risen to $1 - \frac{1}{e} \approx 63.2%$ of its final value.

Notes

The gradient of a $Q$ - $t$ graph gives current directly (since $I = d Q / d t$ ); the signed area under an $I$ - $t$ graph gives the change in charge $Δ Q$ — the same graphical relationships used throughout mechanics, reapplied here.
The $63.2%$ / $36.8%$ figures after one time constant hold regardless of the actual starting charge, current or voltage — a genuinely universal property of exponential growth/decay, not specific to any one circuit.

Time constant and practical analysis 3.7.4.4

Key results

Time constant: $τ = RC$ (seconds, with $R$ in ohms and $C$ in farads).
Half-life of the decay: $t_{1/2} = RC ln 2 \approx 0.693 RC$ — the constant time for $Q$ , $V$ or $∣ I ∣$ to halve during discharge, exactly analogous to radioactive half-life.

Method

Required practical 9: discharge a charged capacitor through a known resistor, recording $V_{C}$ at regular time intervals; plot $ln (V_{C} / V_{0})$ against $t$ — the graph is a straight line of gradient $- \frac{1}{RC}$ , so $C = - \frac{1}{R \times gradient}$ .

In practiceFor a capacitor discharging through 47 kΩ, the gradient of $ln (V / V_{0})$ against $t$ is $- 0.105$ s $^{- 1}$ . Find $C$ .

$ln \frac{V}{V _{0}} = - \frac{t}{RC}$ a straight line through the origin
$RC = \frac{1}{0.105} = 9.52 s$ the time constant
$C = \frac{9.52}{47 \times 1 0 ^{3}} = 2.0 \times 1 0^{- 4} F = 200 μ F$

Notes

Taking logarithms turns the exponential decay into a straight line (the same linearisation technique used for radioactive decay) — this is what makes $τ$ (and so $C$ , given a known $R$ ) readable directly from a graph gradient, rather than needing to fit a curve.
Worked example: $R = 100 k Ω$ , $C = 470 μ F$ gives $τ = 47 s$ and $t_{1/2} \approx 32.6 s$ .
After 5 time constants, a capacitor is about $99.3%$ charged (or, equivalently, only about $0.67%$ of the initial charge remains during discharge) — in practice, 'fully' charged or discharged usually means about five time constants have passed.

Worked examples

Worked example 3.7.4 · 5 marks

A $470 μ F$ capacitor is charged to a p.d. of $9.0 V$ , then discharged through a $2.2 k Ω$ resistor.

Calculate

(a) the time constant of the discharge,

(b) the p.d. across the capacitor 3.0 s after discharge begins.

Show worked solution

(a):

τ = RC = 2.2 \times 1 0^{3} \times 470 \times 1 0^{- 6} \approx 1.034 s

(b):

V = V_{0} e^{- t / τ} = 9.0 \times e^{- 3.0/1.034} = 9.0 \times e^{- 2.902} \approx 9.0 \times 0.0548 \approx 0.49 V

Mark scheme · 5 marks

Calculates the time constant $τ = RC \approx 1.03 s$ 1 mark
Selects and uses $V = V_{0} e^{- t / τ}$ 1 mark
Calculates the exponent $t / τ \approx 2.90$ 1 mark
Evaluates $e^{- 2.90} \approx 0.055$ 1 mark
Calculates $V \approx 0.49 V$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.4 · 5 marks

A $2200 μ F$ capacitor is charged to $12.0 V$ by a battery of negligible internal resistance, through a $1.0 k Ω$ resistor.

Calculate

(a) the maximum charge stored,

(b) the initial charging current, and

Show worked solution

(a):

Q_{m a x} = C V = 2200 \times 1 0^{- 6} \times 12.0 = 2.64 \times 1 0^{- 2} C

(26.4 mC).

(b) At the instant charging begins, the capacitor has zero p.d., so the full battery p.d. is across the resistor:

I_{0} = \frac{V}{R} = \frac{12.0}{1000} = 1.2 \times 1 0^{- 2} A

(12 mA). (c):

E = \frac{1}{2} C V^{2} = \frac{1}{2} \times 2200 \times 1 0^{- 6} \times 12. 0^{2} \approx 0.158 J

Mark scheme · 5 marks

Calculates $Q_{m a x} = C V \approx 2.64 \times 1 0^{- 2} C$ 1 mark
States that at $t = 0$ the capacitor has zero p.d., so the full supply p.d. is across $R$ 1 mark
Calculates $I_{0} = V / R = 1.2 \times 1 0^{- 2} A$ 1 mark
Selects and uses $E = \frac{1}{2} C V^{2}$ 1 mark
Calculates $E \approx 0.158 J$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.4 · 4 marks

A student wants to increase the maximum energy a parallel-plate capacitor can store for a fixed maximum working voltage, without changing the plate area or separation.

Explain, with reference to the relevant equation, why inserting a dielectric between the plates achieves this.

Show worked solution

Inserting a dielectric increases capacitance, since $C = ε_{r} C_{0}$ with $ε_{r} > 1$ .

Because energy stored at a fixed maximum voltage is:

E = \frac{1}{2} C V^{2}

and $V$ (the fixed maximum working voltage) is unchanged, increasing $C$ directly increases the maximum energy that can be stored at that same working voltage.

Mark scheme · 4 marks

States that a dielectric increases capacitance via $C = ε_{r} C_{0}$ with $ε_{r} > 1$ 1 mark
Selects $E = \frac{1}{2} C V^{2}$ as the relevant energy equation 1 mark
States $V$ , the fixed maximum working voltage, is unchanged 1 mark
Concludes a larger $C$ at the same fixed $V$ gives a larger maximum stored energy 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.7.5

Magnetic fields

Force on a current-carrying wire 3.7.5.1

Key results

Force on a current-carrying conductor: $F = B I l$ (wire length $l$ within the field, perpendicular to it).
Tesla defined: $1 T = 1 N A^{- 1} m^{- 1}$ — one tesla gives $1 N$ on $1 m$ of wire carrying $1 A$ at right angles to the field.

Notes

The force on a current-carrying conductor is zero when the conductor is parallel to the field and maximum when perpendicular to it — the magnitude form $F = B I l$ already assumes the perpendicular case.
Fleming's left-hand rule gives the direction for conventional current: thumb = force, first finger = field, second finger = (conventional) current — reversing either the current or the field reverses the force direction.

Charged particles in a magnetic field 3.7.5.2

Key results

Force on a moving charge: $F = B ∣ q ∣ v$ (velocity perpendicular to the field), always perpendicular to both velocity and field.
Because this force is always perpendicular to velocity, it changes direction but never speed — giving circular motion with radius $r = \frac{m v}{B ∣ q ∣}$ and period $T = \frac{2 πm}{B ∣ q ∣}$ .
Cyclotron frequency: $f = \frac{B ∣ q ∣}{2 πm}$ — independent of the particle's speed or orbit radius.

Notes

Equating $B ∣ q ∣ v = \frac{m v ^{2}}{r}$ and rearranging gives $r = \frac{m v}{B ∣ q ∣}$ — the orbital radius of a charged particle in a magnetic field, the working principle behind a mass spectrometer separating particles by their mass-to-charge ratio.
A cyclotron accelerates charged particles across a gap between two hollow 'dees' using an alternating potential difference, while a magnetic field bends their path into a circle inside each dee — because the cyclotron frequency $f = B ∣ q ∣/2 πm$ doesn't depend on speed, the same alternating frequency keeps accelerating the particle correctly even as its orbit radius (and speed) grows with each crossing.

Magnetic flux and flux linkage 3.7.5.3

Definitions

Magnetic flux, $Φ = B A cos θ$ : the magnetic flux through one turn of a coil of area $A$ , with $θ$ measured between the field and the coil's normal.
Flux linkage: $N Φ = NB A cos θ$ for a coil of $N$ turns, since each turn links the same flux.

Key results

Weber defined: $1 Wb = 1 T m^{2}$ .

Notes

$θ = 0°$ (field along the coil's normal, passing straight through the coil face) gives maximum flux, $Φ = B A$ ; $θ = 90°$ (field in the plane of the coil) gives zero flux — a common point of confusion is measuring $θ$ from the coil's plane rather than from its normal.
Flux linkage can change for three genuinely different reasons: the field strength $B$ changing, the area of the coil actually within the field changing, or the coil's orientation (angle $θ$ ) changing — any one alone is enough to induce an emf.

Induction: Faraday's and Lenz's laws 3.7.5.4

Definitions

Lenz's law: the induced EMF always opposes the change that causes it — captured by the minus sign in Faraday's law.

Key results

Faraday's law: $ε = - \frac{d ( N Φ )}{d t}$ , or $∣ ε_{mean} ∣ = N \frac{ΔΦ}{Δ t}$ .
A conducting rod of length $ℓ$ sliding at speed $v$ across a uniform field $B$ (rod, velocity and field all mutually perpendicular): $∣ ε ∣ = B ℓ v$ .

Notes

No EMF is induced by a steady field with no relative motion or change — induction always requires flux linkage to be actively changing.
Lenz's law is a direct consequence of energy conservation: if the induced current instead reinforced the change producing it, the changing flux would accelerate itself indefinitely, generating energy from nothing.
For the sliding rod: the induced current direction is exactly whatever direction produces a magnetic force on the rod opposing its motion ('magnetic drag') — the mechanical work needed to keep the rod moving against this drag is precisely what supplies the electrical energy transferred to the circuit's resistance.
An EMF can exist across an open circuit (e.g. the rod alone, disconnected) even though no current can then flow — current additionally requires a complete, closed conducting path.

Rotating-coil EMF and alternating quantities 3.7.5.4–3.7.5.5

Key results

A coil of $N$ turns, area $A$ , rotating at constant angular speed $ω$ in a uniform field $B$ : flux linkage $N Φ = NB A cos (ω t)$ , induced EMF $ε = NB A ω sin (ω t) = ε_{0} sin (ω t)$ , with peak EMF $ε_{0} = NB A ω$ .
RMS values for a sinusoidal quantity: $V_{rms} = \frac{V _{0}}{2}$ , $I_{rms} = \frac{I _{0}}{2}$ ; peak-to-peak $V_{pp} = 2 V_{0}$ .
Mean power dissipated in a resistor: $\overline{P} = V_{rms} I_{rms} = I_{rms}^{2} R$ — the same heating effect as the equivalent steady d.c. value.

Notes

Flux and EMF are exactly out of step with each other: EMF is zero when flux is at its maximum (coil momentarily parallel to the field, so flux is momentarily not changing), and EMF is at its maximum when flux is momentarily zero (coil perpendicular to the field, so flux is changing fastest) — since EMF depends on the RATE of change of flux, not flux itself.
Reading an oscilloscope trace: peak-to-peak voltage $=$ (peak-to-peak divisions) $\times$ (volts/division); period $=$ (divisions per cycle) $\times$ (seconds/division); frequency $= 1/ T$ .
UK mains, quoted as $230 V$ , is an RMS value — its actual peak voltage is about $325 V$ ( $230 \times 2$ ), and its peak-to-peak voltage is about $651 V$ .

Transformers and power transmission 3.7.5.6

Key results

Transformer voltage ratio: $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ . Ideal transformer (no losses): $V_{p} I_{p} = V_{s} I_{s}$ , so $\frac{I _{p}}{I _{s}} = \frac{N _{s}}{N _{p}}$ .
Efficiency: $η = \frac{P _{out}}{P _{in}} = \frac{V _{s} I _{s}}{V _{p} I _{p}}$ .

Notes

A rotating coil in a magnetic field (a simple a.c. generator) induces a sinusoidal EMF because $Φ = B A cos θ$ varies sinusoidally as the coil turns; a transformer instead uses a fixed coil pair, with an alternating primary current producing a changing flux that links the secondary — steady d.c. in the primary gives no sustained secondary EMF at all, since flux would then be constant.
Real transformer losses: eddy currents induced in the iron core (reduced by using thin, electrically insulated laminations instead of one solid core), hysteresis losses from repeatedly magnetising and demagnetising the core material (reduced with a magnetically soft core), and ordinary resistive ( $I^{2} R$ ) heating in the windings themselves (reduced with low-resistance windings). A shared core also reduces flux leakage between the two windings.
For power transmission at fixed power $P$ : cable current $I = P / V$ , and cable heating loss $= I^{2} R_{cable}$ — transmitting at a much higher voltage gives a much lower current for the same power, and so dramatically less resistive heating loss in the cables, the entire economic reason for high-voltage transmission lines and step-up/step-down transformers at each end.

Investigating magnetic fields 3.7.5.1, 3.7.5.3

Method

Required practical 10 (force on a wire): support a current-carrying wire (separately from a pair of magnets, with no direct contact) so it passes through the magnets' field, resting the magnets on a top-pan balance; the wire's magnetic force on the magnets (and, by Newton's third law, their force on the wire) shows up as a change in the balance reading, $Δ m$ .
Since $F = ∣Δ m ∣ g = B I l$ , varying current $I$ at fixed $B$ and $l$ and plotting $F$ against $I$ gives a straight line of gradient $Bl$ , letting $B$ be found.
Required practical 11 (search coil and oscilloscope): place a small, $N$ -turn search coil in a region where $B$ varies with time (e.g. near an a.c. electromagnet), and observe the induced EMF on an oscilloscope; for a sinusoidally varying field of amplitude $B_{0}$ and angular frequency $ω$ , peak EMF $∣ ε_{0} ∣ = N A ω B_{0} ∣ cos θ ∣$ ( $θ$ : angle between the coil's normal and the field).
Investigate how peak EMF depends on the number of turns $N$ , coil area $A$ , field amplitude, or coil angle $θ$ — varying one at a time, and keeping the field's own frequency fixed throughout any single comparison.

In practiceThe balance reading changes by 2.4 g when 3.0 A flows in a wire with 5.0 cm in the field. Find $B$ .

$F = ∣Δ m ∣ g = 2.4 \times 1 0^{- 3} \times 9.81 = 0.0235 N$
$B = \frac{F}{I l} = \frac{0.0235}{3.0 \times 0.050} = 0.16 T$
For several currents, plot $F$ against $I$ : the gradient is $Bl$ .

Notes

In the force-on-a-wire practical, the wire and the magnets must be mechanically independent (the wire supported separately, not touching the magnets) — only the magnetic force between them should affect the balance reading, not any direct mechanical contact.
The search coil method works because the coil's own induced EMF is a direct, real-time measurement of $\frac{d ( N Φ )}{d t}$ — it turns Faraday's law from an abstract relationship into something read straight off an oscilloscope trace.

Worked examples

Worked example 3.7.5 · 6 marks

$I$ / A	Balance reading $Δ m$ / g
1.0	1.63
2.0	3.27
3.0	4.90
4.0	6.53

In a Required Practical 10 investigation, a straight horizontal wire of length $8.0 cm$ passes at right angles through a uniform magnetic field between two magnadur magnets resting on a top-pan balance (initially zeroed with no current flowing).

The table shows the balance reading for several values of current through the wire.

Use the data to determine the magnetic flux density $B$ between the magnets.

Show worked solution

By Newton's third law, the reading $Δ m$ (converted to a force $F = Δ m g$ ) equals the force the wire exerts on the magnets, which equals $F = B I l$ .

So a graph of $F$ against $I$ has gradient $Bl$ .

Converting the balance readings to force: at $I = 1.0 A$ ,

F = 1.63 \times 1 0^{- 3} \times 9.81 \approx 1.599 \times 1 0^{- 2} N

at $I = 4.0 A$ ,

F = 6.53 \times 1 0^{- 3} \times 9.81 \approx 6.406 \times 1 0^{- 2} N

Gradient:

= \frac{6.406 \times 1 0 ^{- 2} - 1.599 \times 1 0 ^{- 2}}{4.0 - 1.0} \approx 1.602 \times 1 0^{- 2} N A^{- 1}

Since gradient $= Bl$ ,

B = \frac{gradient}{l} = \frac{1.602 \times 1 0 ^{- 2}}{0.080} \approx 0.20 T

Mark scheme · 6 marks

States that, by Newton's third law, the balance reading (as a force) equals the force on the wire, $F = B I l$ 1 mark
Converts at least two balance readings from mass to force using $F = Δ m g$ 1 mark
Calculates the gradient of $F$ against $I$ , $\approx 1.60 \times 1 0^{- 2} N A^{- 1}$ 1 mark
States that this gradient equals $Bl$ 1 mark
Divides by $l = 0.080 m$ to find $B$ 1 mark
Calculates $B \approx 0.20 T$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.5 · 5 marks

A rectangular coil of 250 turns and area $0.015 m^{2}$ rotates at a constant angular frequency of $50 rad s^{- 1}$ in a uniform magnetic field of flux density $0.080 T$ , generating an alternating EMF.

Calculate

(a) the peak EMF generated, and

(b) the r.m.s.

EMF.

Show worked solution

(a) Peak EMF:

ε_{0} = NB A ω = 250 \times 0.080 \times 0.015 \times 50 = 15 V

(b):

ε_{rms} = \frac{ε _{0}}{2} = \frac{15}{2} \approx 10.6 V

Mark scheme · 5 marks

Selects $ε_{0} = NB A ω$ for the peak EMF 1 mark
Substitutes $N$ , $B$ , $A$ and $ω$ correctly 1 mark
Calculates $ε_{0} = 15 V$ 1 mark
Uses $ε_{rms} = ε_{0} / 2$ 1 mark
Calculates $ε_{rms} \approx 10.6 V$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.5 · 4 marks

A step-down transformer, assumed 100% efficient, has a primary coil of 2000 turns connected to a 230 V a.c. supply, and is designed to deliver 12 V to a secondary circuit drawing a current of 3.0 A.

Calculate

(a) the number of turns on the secondary coil, and

(b) the current drawn from the primary supply.

Show worked solution

(a):

\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} \Rightarrow N_{s} = N_{p} \times \frac{V _{s}}{V _{p}} = 2000 \times \frac{12}{230} \approx 104

turns.

(b) For 100% efficiency, input power equals output power:

V_{p} I_{p} = V_{s} I_{s} \Rightarrow I_{p} = \frac{V _{s} I _{s}}{V _{p}} = \frac{12 \times 3.0}{230} \approx 0.157 A

Mark scheme · 4 marks

Selects and rearranges the transformer equation to find $N_{s}$ 1 mark
Calculates $N_{s} \approx 104$ turns 1 mark
States that for 100% efficiency, $V_{p} I_{p} = V_{s} I_{s}$ 1 mark
Calculates $I_{p} \approx 0.157 A$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.