Notes · PhysicsUK · A-Levels

Fields and their consequences

This chapter treats gravitational, electric and magnetic fields together, exploiting the deep analogy between them. It develops gravitational fields, potential and orbits; electric fields and the inverse-square Coulomb law; capacitance, energy storage and the exponential charge and discharge of a capacitor; and magnetic forces and electromagnetic induction, the physics behind generators and transformers.

5 sections~15 min reading time3 competenciesLevel Standard 1 · Advanced 4

T·0777 / 13

Exam profile

AO1 · State the field laws, define field strength and potential, and describe capacitor behaviour and electromagnetic inductionAO2 · Calculate field strengths, potentials, orbital quantities, capacitor energy, time constants and induced emfsAO3 · Analyse exponential capacitor-discharge data on log-linear axes and draw the gravitational-electric field analogy

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basic level

This is A2 (full A-Level) content. It requires the inverse-square field laws, field strength and potential, capacitance and energy, exponential charge/discharge, and the laws of electromagnetic induction.

higher level

The full A-Level demands the gravitational-electric analogy, orbital and Kepler analysis, the derivation of the capacitor discharge exponential, and quantitative Faraday/Lenz problems including transformers and generators.

Depth

Text

Contents · 5 sections

Fields and their consequences

Gravitational fields, potential and orbits#

AdvancedAQA 7408 3.7.2DfE GCE Physics - gravitational fields

Radial gravitational field of a mass

Fig. 1The field of a point (or spherical) mass is radial, its lines pointing inward towards the mass; a dashed circle is an equipotential surface, perpendicular to the field lines.

A gravitational field is a region in which a mass experiences a force. Newton's law of gravitation gives the attractive force between two point masses as

F = \dfrac{G m_1 m_2}{r^2}

, where

G = 6.67\times10^{-11}\,\text{N m}^2\,\text{kg}^{-2}

is the gravitational constant. It is an inverse-square law: doubling the separation quarters the force. The gravitational field strength is the force per unit mass,

g = \dfrac{F}{m} = \dfrac{GM}{r^2}

, in

\text{N kg}^{-1}

, and points towards the mass producing it.

Near a planet's surface the field is effectively uniform, giving the familiar

g = 9.81\,\text{N kg}^{-1}

; on the scale of orbits it is radial, weakening as

1/r^2

. Gravitational potential

V

is the work done per unit mass in bringing a small mass from infinity to a point,

V = -\dfrac{GM}{r}

. It is negative because gravity is attractive - work is released as a mass approaches. The gravitational potential energy of a mass

m

E_p = mV = -\dfrac{GMm}{r}

, and the field strength is the negative gradient of potential,

g = -\dfrac{\Delta V}{\Delta r}

For a satellite in a circular orbit, gravity provides exactly the centripetal force:

\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}

. Cancelling and rearranging gives the orbital speed

v = \sqrt{GM/r}

- closer orbits are faster. Substituting

v = 2\pi r/T

leads to Kepler's third law,

T^2 = \dfrac{4\pi^2}{GM}r^3

, so the square of the period is proportional to the cube of the orbital radius.

A geostationary satellite has an orbital period of exactly one day and orbits above the equator, so it stays over the same point on the Earth - ideal for communications. Setting

T = 24\,\text{hours}

in Kepler's law gives an orbital radius of about

4.2\times10^{7}\,\text{m}

. The escape velocity, the minimum speed to leave a body's field entirely, comes from equating kinetic energy to the depth of the potential well:

v_{\text{esc}} = \sqrt{2GM/r}

F = \dfrac{G m_1 m_2}{r^2}, \qquad g = \dfrac{GM}{r^2}

Newton's law and field strength

The inverse-square force and the field strength (force per unit mass).

V = -\dfrac{GM}{r}

Gravitational potential

Work per unit mass from infinity; negative because gravity is attractive.

T^2 = \dfrac{4\pi^2}{GM}r^3

Kepler's third law

Period squared is proportional to orbital radius cubed.

Field strength against distance

Fig. 2Both gravitational and electric field strength obey an inverse-square law: the field falls to a quarter when the distance doubles.

Worked example

Speed of a low-Earth-orbit satellite

A satellite orbits $400\,\text{km}$ above the Earth's surface. Find its orbital speed. Earth's mass $= 6.0\times10^{24}\,\text{kg}$ , radius $= 6.4\times10^{6}\,\text{m}$ , $G = 6.67\times10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ .

Orbital radius
$r = 6.4\times10^{6} + 4.0\times10^{5} = 6.8\times10^{6}\,\text{m}$ .
Equate gravity to centripetal force
$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ gives $v = \sqrt{GM/r}$ .
Substitute
$v = \sqrt{(6.67\times10^{-11})(6.0\times10^{24})/(6.8\times10^{6})}$ .
$v = \sqrt{5.88\times10^{7}} = 7.7\times10^{3}\,\text{m}\,\text{s}^{-1}$

Result: The satellite orbits at about $7.7\,\text{km s}^{-1}$ .

Typical mistakes

Confusing the gravitational constant $G$ with the surface field strength $g$ .
Dropping the minus sign in gravitational potential, or forgetting that potential is zero at infinity, not at the surface.

Active revision

Calculate the orbital speed and period of a satellite $500\,\text{km}$ above the Earth's surface. Take the Earth's mass $6.0\times10^{24}\,\text{kg}$ , radius $6.4\times10^{6}\,\text{m}$ and $G = 6.67\times10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ .

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Electric fields and the gravitational-electric analogy#

AdvancedAQA 7408 3.7.3DfE GCE Physics - electric fields

An electric field is a region in which a charge experiences a force. Coulomb's law gives the force between two point charges as

F = \dfrac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}

, where

\varepsilon_0 = 8.85\times10^{-12}\,\text{F m}^{-1}

is the permittivity of free space. Like gravity it is an inverse-square law, but unlike gravity it can be attractive (unlike charges) or repulsive (like charges). The electric field strength is the force per unit positive charge,

E = \dfrac{F}{Q} = \dfrac{Q}{4\pi\varepsilon_0 r^2}

, in

\text{N C}^{-1}

\text{V m}^{-1}

Between two parallel plates the field is uniform,

E = \dfrac{V}{d}

, directed from the positive to the negative plate. A charge

Q

in this field feels a constant force

F = EQ

, which is how charged particles are accelerated and deflected in oscilloscopes and mass spectrometers. Electric potential is the work done per unit positive charge from infinity,

V = \dfrac{Q}{4\pi\varepsilon_0 r}

, positive near a positive charge; as with gravity, field strength is the negative potential gradient.

The analogy between gravitational and electric fields is one of the most elegant ideas in the specification. Both have an inverse-square force law, both define field strength as force per unit 'charge' (mass or electric charge), both have a potential that is the negative gradient of which gives the field, and both have radial and uniform configurations. The crucial difference is that mass produces only attraction, whereas electric charge produces both attraction and repulsion - so gravitational potential is always negative while electric potential can be either sign.

The two fields also differ enormously in strength. For two protons the electric repulsion exceeds the gravitational attraction by a factor of about

10^{36}

, which is why gravity is utterly negligible inside atoms yet dominant on astronomical scales, where matter is electrically neutral. Recognising which field matters at which scale is a recurring synoptic theme.

F = \dfrac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}

Coulomb's law

The inverse-square force between two point charges.

E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{V}{d}

Electric field strength

Radial field of a point charge; uniform field between parallel plates.

Worked example

Force between two charges

Two point charges of $+4.0\,\text{nC}$ and $+6.0\,\text{nC}$ are $30\,\text{mm}$ apart in a vacuum. Find the force between them. Take $\dfrac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ .

Substitute into Coulomb's law
$F = \dfrac{Q_1 Q_2}{4\pi\varepsilon_0 r^2} = 8.99\times10^{9} \times \dfrac{(4.0\times10^{-9})(6.0\times10^{-9})}{(0.030)^2}$ .
Evaluate
Numerator $= 8.99\times10^{9} \times 2.4\times10^{-17} = 2.16\times10^{-7}$ ; divide by $9.0\times10^{-4}$ .
$F = \dfrac{2.16\times10^{-7}}{9.0\times10^{-4}} = 2.4\times10^{-4}\,\text{N}$

Result: The charges repel with a force of $2.4\times10^{-4}\,\text{N}$ (both positive, so repulsion).

Typical mistakes

Forgetting the $4\pi\varepsilon_0$ factor, or confusing $\varepsilon_0$ with a relative permittivity.
Treating electric potential as always negative like gravitational potential - it takes the sign of the charge.

Active revision

Two point charges of $+3.0\,\text{nC}$ and $-5.0\,\text{nC}$ are $20\,\text{mm}$ apart. Calculate the force between them, and the electric field strength midway between them due to the positive charge alone.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Capacitance and energy storage#

StandardAQA 7408 3.7.4.1AQA 7408 3.7.4.2DfE GCE Physics - capacitance

A capacitor stores charge and energy in the electric field between two conductors separated by an insulator (a dielectric). Its capacitance is the charge stored per unit potential difference,

C = \dfrac{Q}{V}

, measured in farads (one farad is one coulomb per volt, a very large unit, so microfarads and picofarads are usual). A parallel-plate capacitor has

C = \dfrac{A\varepsilon_0\varepsilon_r}{d}

, where

A

is the plate area,

d

their separation and

\varepsilon_r

the relative permittivity of the dielectric.

Placing a dielectric between the plates increases the capacitance by the factor

\varepsilon_r

: the dielectric molecules become polarised, partly cancelling the field, so more charge can be stored at the same voltage. A larger plate area and a smaller separation both increase capacitance, which is why practical capacitors use large, thin, rolled or layered plates.

Charging a capacitor to voltage

V

requires work, because each extra bit of charge is pushed onto plates that are already charged. Since

Q = CV

, the work done - stored as electrical potential energy - is the area under the charge-voltage graph, a triangle:

E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}

. The factor of one half distinguishes this from the naive

QV

and is a favourite exam point.

Capacitors deliver their stored energy very quickly, making them ideal where a short, large burst of energy is needed - a camera flash, a defibrillator, or smoothing in power supplies. Unlike a battery, a capacitor stores energy in a field rather than chemically, so it can be charged and discharged almost indefinitely.

C = \dfrac{Q}{V}

Capacitance

Charge stored per unit potential difference.

E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}

Energy stored

The area under the charge-voltage graph.

Worked example

Charge and energy in a capacitor

A $470\,\mu\text{F}$ capacitor is charged to $9.0\,\text{V}$ . Find the charge and the energy stored.

Charge
$Q = CV = 470\times10^{-6} \times 9.0 = 4.23\times10^{-3}\,\text{C}$ .
Energy
$E = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(470\times10^{-6})(9.0)^2$ .
$E = 0.5 \times 470\times10^{-6} \times 81 = 1.9\times10^{-2}\,\text{J}$

Result: The capacitor stores $4.2\,\text{mC}$ of charge and $19\,\text{mJ}$ of energy.

Typical mistakes

Omitting the factor of $\tfrac{1}{2}$ in the energy stored (using $QV$ instead of $\tfrac{1}{2}QV$ ).
Confusing capacitance (a property of the capacitor) with the charge stored (which depends on the applied voltage).

Active revision

A $220\,\mu\text{F}$ capacitor is charged to $12\,\text{V}$ . Calculate the charge stored and the energy stored.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Capacitor charging and discharging#

AdvancedAQA 7408 3.7.4.3AQA 7408 3.7.4.4DfE GCE Physics - RC circuits

Exponential discharge of a capacitor

Fig. 3The charge falls exponentially. After one time constant (

t = RC

) the charge has fallen to about

37\%

of its initial value.

When a charged capacitor discharges through a resistor

R

, the current at any instant is set by the capacitor's voltage,

I = V/R

, but that current removes charge and lowers the voltage, so the discharge slows continuously. This self-limiting behaviour gives an exponential decay. Writing

I = -\dfrac{\text{d}Q}{\text{d}t}

and

V = Q/C

leads to

\dfrac{\text{d}Q}{\text{d}t} = -\dfrac{Q}{RC}

, whose solution is

Q = Q_0\,e^{-t/RC}

- the charge falls exponentially, and so do the voltage and current.

The product

RC

has units of time and is called the time constant

\tau

. It is the time for the charge (or voltage, or current) to fall to

\dfrac{1}{e}

- about

37\%

- of its initial value. After one time constant

37\%

remains, after two about

14\%

, after five under

1\%

- effectively fully discharged. A larger resistance or capacitance stores or restricts the flow more, giving a slower decay.

Charging through a resistor is the mirror image: the charge and voltage rise towards their final values as

Q = Q_0(1 - e^{-t/RC})

, quickly at first and then ever more slowly, while the charging current falls exponentially from its initial value

\varepsilon/R

. The same time constant governs how fast the capacitor approaches full charge.

Because the decay is exponential, taking natural logarithms linearises it:

\ln Q = \ln Q_0 - \dfrac{t}{RC}

, so a graph of

\ln Q

(or

\ln V

) against time is a straight line of gradient

-\dfrac{1}{RC}

. This is how the time constant is measured in the RC required practical, and it is the standard way to extract a decay constant from exponential data - a technique reused for radioactive decay.

Q = Q_0\,e^{-t/RC}

Capacitor discharge

Charge (and voltage and current) decay exponentially.

\tau = RC

Time constant

The time to fall to $1/e$ (about 37%) of the initial value.

\ln Q = \ln Q_0 - \dfrac{t}{RC}

Linearised decay

A graph of $\ln Q$ against $t$ has gradient $-1/RC$ .

Capacitor discharging through a resistor

Fig. 4Closing the switch lets the charged capacitor drive a current through the resistor; the charge and voltage decay exponentially with time constant

RC

Worked example

Discharge voltage after a given time

A $220\,\mu\text{F}$ capacitor charged to $9.0\,\text{V}$ discharges through a $22\,\text{k}\Omega$ resistor. Find the time constant and the voltage after $10\,\text{s}$ .

Time constant
$\tau = RC = 22\times10^{3} \times 220\times10^{-6} = 4.84\,\text{s}$ .
Voltage decays exponentially
$V = V_0 e^{-t/RC} = 9.0 \times e^{-10/4.84}$ .
Evaluate
$10/4.84 = 2.07$ , so $V = 9.0 \times e^{-2.07}$ .
$V = 9.0 \times 0.126 = 1.1\,\text{V}$

Result: The time constant is $4.8\,\text{s}$ and the voltage has fallen to about $1.1\,\text{V}$ after $10\,\text{s}$ .

Typical mistakes

Thinking the capacitor loses a fixed amount of charge per second - the decay is exponential, not linear.
Muddling the charging equation $(1 - e^{-t/RC})$ with the discharging equation $(e^{-t/RC})$ .

Active revision

A $100\,\mu\text{F}$ capacitor charged to $12\,\text{V}$ discharges through a $50\,\text{k}\Omega$ resistor. Find the time constant and the voltage after $8.0\,\text{s}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Magnetic fields, forces and electromagnetic induction#

AdvancedAQA 7408 3.7.5AQA 7408 3.7.6DfE GCE Physics - magnetic fields and induction

A transformer

Fig. 5An alternating current in the primary coil sets up a changing flux in the iron core, which links the secondary coil and induces an alternating emf; the turns ratio sets the voltage ratio.

A current in a magnetic field experiences a force. When a wire of length

L

carrying current

I

lies at right angles to a field of flux density

B

, the force is

F = BIL

, its direction given by Fleming's left-hand rule. Flux density

B

, measured in tesla, is defined by this equation. A charged particle of charge

Q

moving at speed

v

perpendicular to the field feels

F = BQv

; because this force is always perpendicular to the velocity, it makes the particle move in a circle, the basis of the cyclotron and the mass spectrometer.

Magnetic flux through an area

A

perpendicular to a field is

\Phi = BA

, in webers; for a coil of

N

turns the flux linkage is

N\Phi = BAN

. Electromagnetic induction is the production of an emf whenever the flux linkage through a circuit changes - by moving a magnet, moving a conductor, or changing a current in a nearby coil. Faraday's law states that the induced emf equals the rate of change of flux linkage,

\varepsilon = -\dfrac{\text{d}(N\Phi)}{\text{d}t}

The minus sign is Lenz's law: the induced emf (and any current it drives) opposes the change producing it - a consequence of conservation of energy, since an induced current that aided the change would create energy from nothing. This is why moving a magnet into a coil is resisted, and why eddy-current braking works: the faster you try to change the flux, the harder the induced effects push back.

A transformer applies these ideas to alternating current. An alternating current in the primary coil (

N_p

turns) produces a changing flux in an iron core, which links the secondary coil (

N_s

turns) and induces an alternating emf there. For an ideal transformer

\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}

, so turns set the voltage ratio, and with

100\%

efficiency

V_pI_p = V_sI_s

. Transformers step voltage up for efficient transmission (high voltage, low current, low

I^2R

loss) and down for safe use - the backbone of the national grid.

F = BIL, \qquad F = BQv

Magnetic forces

Force on a current-carrying wire and on a moving charge.

\varepsilon = -\dfrac{\text{d}(N\Phi)}{\text{d}t}

Faraday's and Lenz's laws

Induced emf equals the rate of change of flux linkage; the sign opposes the change.

\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}, \qquad V_pI_p = V_sI_s

Ideal transformer

Turns set the voltage ratio; power is conserved when efficient.

Worked example

Transformer turns and current

A transformer steps $240\,\text{V}$ down to $8.0\,\text{V}$ to run a lamp drawing $1.5\,\text{A}$ . The primary has $1500$ turns. Assuming $100\%$ efficiency, find the secondary turns and the primary current.

Secondary turns
$\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}$ , so $N_s = 1500 \times \dfrac{8.0}{240} = 50$ turns.
Primary current from power conservation
$V_pI_p = V_sI_s$ , so $I_p = \dfrac{V_sI_s}{V_p} = \dfrac{8.0 \times 1.5}{240}$ .
$I_p = \dfrac{12}{240} = 0.050\,\text{A}$

Result: The secondary has $50$ turns and the primary draws $0.050\,\text{A}$ - a small current at high voltage.

Typical mistakes

Forgetting that $F = BIL$ needs the field perpendicular to the current (otherwise a $\sin\theta$ factor applies).
Omitting the minus sign / Lenz's law, or thinking a steady flux induces an emf - only a changing flux does.

Active revision

A transformer steps $230\,\text{V}$ down to $12\,\text{V}$ . If the primary has $920$ turns, find the number of secondary turns, and the primary current when the secondary delivers $2.0\,\text{A}$ (assume $100\%$ efficiency).

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Fields and their consequences

Key points

Speed of a low-Earth-orbit satellite

Key points

Force between two charges

Key points

Charge and energy in a capacitor

Key points

Discharge voltage after a given time

Key points

Transformer turns and current

Sources