Capacitance

Spec mapping: OCR H556 Module 6.1 — Capacitors (capacitance as the ratio of charge to potential difference, $C = Q/V$C=Q/V; the farad as the SI unit of capacitance; capacitance of a parallel-plate capacitor $C = \varepsilon_0 A/d$C=ε0​A/d and the role of relative permittivity $\varepsilon_r$εr​; capacitance of an isolated conducting sphere $C = 4\pi\varepsilon_0 R$C=4πε0​R as an OCR-specific result). Refer to the official OCR H556 specification document for exact wording.

A capacitor is one of the simplest — and yet most important — components in electronics. In its purest form it is two metal plates held a small distance apart, separated by an insulator. Despite this geometric humility, capacitors appear in every radio, every power supply, every camera flash, every computer motherboard, every defibrillator and almost every sensor you are ever likely to meet. The single number that characterises a capacitor — telling you how much charge it can hold for a given voltage — is its capacitance, and that is the subject of this lesson.

This lesson begins Module 6.1 — Capacitors. Module 6 as a whole is about storing and manipulating charge, and it leans heavily on ideas from Module 4 (electricity — current, potential difference, resistance) and Module 5 (fields — gravitational analogues that will be reused for electric and magnetic fields). Capacitance is the bridge between them: it tells us how much charge a conductor can store for a given potential, and the formulae we derive here will unlock every later calculation in the topic, from energy storage to RC time constants to the OCR-specific isolated-sphere result that re-enters in Module 6.2 via the potential of a point charge.

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What Does a Capacitor Do?

Connect a resistor to a battery and current flows through it continuously, dissipating energy as heat. Connect a capacitor to the same battery, and something very different happens. A brief current flows — milliseconds or microseconds — and then stops. The capacitor has stored charge on its plates, and now holds a potential difference equal to the battery e.m.f. It is, in effect, a tiny reservoir of electrical energy.

Key idea: A capacitor does not allow steady current through it. It stores charge on its plates and a corresponding amount of energy in the electric field between them.

Disconnect the battery and connect the charged capacitor to a small lamp. The lamp lights briefly as charge leaves the plates and flows through the lamp, then goes out again. The capacitor has discharged.

This "charge up quickly, release quickly" behaviour is exploited in camera flash units (tens of joules released in a millisecond), defibrillators (hundreds of joules released in a few milliseconds), and the timing circuits that make computers tick. Even the dynamic random-access memory (DRAM) in the laptop you are reading this on is a vast array of microscopic capacitors, each storing a single bit by being either charged or empty.

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Defining Capacitance

Suppose we put a charge $Q$Q on a capacitor and measure the potential difference $V$V that develops across its terminals. Experiment shows that the two quantities are proportional:

$Q \propto V$Q∝V

The constant of proportionality is called the capacitance, symbol $C$C:

$C = \frac{Q}{V}$C=VQ​

The SI unit of capacitance is the farad (F), named after Michael Faraday. From the defining equation,

$1\ \text{F} = 1\ \text{C V}^{-1}$1 F=1 C V−1

One farad is an enormous amount of capacitance. Most real capacitors in circuits are measured in microfarads ($\mu$μF, $10^{-6}$10−6 F), nanofarads (nF, $10^{-9}$10−9 F) or picofarads (pF, $10^{-12}$10−12 F). A 1 F discrete capacitor is roughly the size of a fist; supercapacitors of tens or hundreds of farads exist but use very different chemistry from the parallel-plate ideal.

Symbol	Meaning	Unit
$C$C	Capacitance	farad (F)
$Q$Q	Charge stored on one plate	coulomb (C)
$V$V	Potential difference between plates	volt (V)

Watch the notation. The letter C appears twice in capacitor calculations — as the symbol for capacitance (italicised: C) and as the symbol for the unit coulomb (upright: C). Context always makes it clear, but written exam answers must use the correct unit. A capacitor of capacitance $C = 47\ \mu\text{F}$C=47 μF holding charge $Q = 5.6 \times 10^{-4}\ \text{C}$Q=5.6×10−4 C uses both meanings of the letter in a single sentence.

Definition vs derived formula — a crucial distinction

$C = Q/V$C=Q/V is the defining equation of capacitance. It applies to any capacitor of any shape, made of any material, regardless of geometry. The capacitor itself is whatever device you choose to build. By contrast the formula $C = \varepsilon_0 A/d$C=ε0​A/d that we will derive below applies only to the parallel-plate geometry in vacuum (or air). The isolated-sphere formula $C = 4\pi\varepsilon_0 R$C=4πε0​R applies only to an isolated sphere. The defining equation $C = Q/V$C=Q/V subsumes them all.

The OCR examiner regularly tests whether you understand this distinction: "Explain why $C = Q/V$C=Q/V does not depend on the size of the charge placed on the capacitor." The answer is that $C$C is a geometric property — it is determined by the shape, size and separation of the plates and by the material between them, not by the charge you happen to put on. If you double $Q$Q, then $V$V doubles too, and $C$C stays the same. The defining equation works because of this proportionality.

Worked Example 1 — Charge on a Capacitor

A 47 $\mu$μF capacitor is connected to a 12 V battery. How much charge is stored on each plate?

$$\begin{aligned} Q &= CV \\ &= 47 \times 10^{-6} \times 12 \\ &= 5.64 \times 10^{-4}\ \text{C} \\ &= 564\ \mu\text{C} \end{aligned}$$Q​=CV=47×10−6×12=5.64×10−4 C=564 μC​

The plates carry equal and opposite charges: $+564\ \mu$+564 μC on the plate connected to the positive terminal, $-564\ \mu$−564 μC on the other. The capacitor as a whole is electrically neutral; what it "stores" is the separation of charge, not any net charge. This is a subtle but important point: a charged capacitor weighed on a sufficiently precise balance would weigh exactly the same as an uncharged one.

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Parallel-Plate Capacitors

The simplest capacitor geometry is two parallel metal plates of area $A$A, separated by a small distance $d$d. For a vacuum (or, to an excellent approximation, air) between the plates, the capacitance is

$C = \frac{\varepsilon_0 A}{d}$C=dε0​A​

where $\varepsilon_0 = 8.85 \times 10^{-12}$ε0​=8.85×10−12 F m$^{-1}$−1 is the permittivity of free space.

This formula makes intuitive sense:

• Larger plates (larger $A$A) $\Rightarrow$⇒ more room for charge to spread out at a given surface density $\Rightarrow$⇒ more charge for a given voltage $\Rightarrow$⇒ larger $C$C.
• Smaller separation (smaller $d$d) $\Rightarrow$⇒ the electric field between the plates is larger for a given $V$V (because $E = V/d$E=V/d), but the charge for a given field is also larger (because $\sigma = \varepsilon_0 E$σ=ε0​E), so $Q$Q for a given $V$V is larger $\Rightarrow$⇒ larger $C$C.

A clean schematic of the geometry, with the field shown:

Worked Example 2 — Parallel-Plate Capacitance

Two square metal plates of side 15 cm are held 2.0 mm apart in air. Calculate the capacitance.

$$\begin{aligned} A &= 0.15 \times 0.15 = 0.0225\ \text{m}^2 \\ d &= 2.0 \times 10^{-3}\ \text{m} \\ C &= \frac{\varepsilon_0 A}{d} \\ &= \frac{(8.85 \times 10^{-12})(0.0225)}{2.0 \times 10^{-3}} \\ &= 9.96 \times 10^{-11}\ \text{F} \\ &\approx 100\ \text{pF} \end{aligned}$$AdC​=0.15×0.15=0.0225 m2=2.0×10−3 m=dε0​A​=2.0×10−3(8.85×10−12)(0.0225)​=9.96×10−11 F≈100 pF​

Even a reasonably large pair of plates in air gives only about 100 picofarads — a tiny capacitance. This is why real capacitors are built with very thin insulators and often with rolled or interleaved layers of metal foil to get large areas into small volumes. An electrolytic capacitor used in a power supply rolls a metre-square of aluminium foil with a sub-micrometre oxide dielectric into a tube the size of a thumb to reach hundreds of microfarads — an effective $A/d$A/d ratio about $10^7$107 times larger than the lab demonstration above.

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Dielectrics

If the space between the plates is filled with an insulator — called a dielectric — the capacitance increases. The formula becomes

$C = \frac{\varepsilon_0 \varepsilon_r A}{d}$C=dε0​εr​A​

where $\varepsilon_r$εr​ is the dimensionless relative permittivity (sometimes called the dielectric constant) of the material.

Material	Relative permittivity $\varepsilon_r$εr​
Vacuum	1 (exactly)
Air	$\approx 1.0006$≈1.0006
Paper	2.0 – 4.0
Polythene	2.3
Mica	7
Ceramic (low-K)	10
Ceramic (high-K)	up to 10 000
Water (at 20°C)	$\approx 80$≈80

Why does a dielectric raise the capacitance? When the dielectric is placed in the field between the plates, its molecules become polarised: their positive and negative charge centres shift slightly in opposite directions. The surfaces of the dielectric develop layers of induced charge that partially cancel the field due to the plates. For the same charge $Q$Q on the plates, the field — and hence the voltage $V$V across the plates — is reduced, so $C = Q/V$C=Q/V is larger. The molecular polarisation can be thought of as creating little internal capacitors in series with the original geometry, all of which add to the overall ability of the device to store charge.

Worked Example 3 — Dielectric Capacitor

The plates from the previous example are now separated by a 2.0 mm sheet of mica with $\varepsilon_r = 7.0$εr​=7.0. What is the new capacitance?

$$\begin{aligned} C &= \frac{\varepsilon_0 \varepsilon_r A}{d} \\ &= 7.0 \times (9.96 \times 10^{-11}\ \text{F}) \\ &\approx 7.0 \times 10^{-10}\ \text{F} \\ &\approx 700\ \text{pF} \end{aligned}$$C​=dε0​εr​A​=7.0×(9.96×10−11 F)≈7.0×10−10 F≈700 pF​

Inserting a dielectric has multiplied the capacitance by $\varepsilon_r = 7$εr​=7, exactly as the formula predicts. Real high-K ceramics can multiply by several thousand, which is why a 1 cm$^3$3 ceramic capacitor with a barium-titanate dielectric can rival a much larger air-spaced capacitor.

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Isolated Sphere — An OCR-Specific Formula

OCR H556 is unusual among the major A-Level boards in requiring students to know the capacitance of an isolated conducting sphere. A conducting sphere of radius $R$R held at potential $V$V relative to a notional "zero" at infinity carries charge

$Q = 4\pi\varepsilon_0 R V$Q=4πε0​RV

Dividing both sides by $V$V gives the capacitance:

$C = 4\pi\varepsilon_0 R$C=4πε0​R

This formula tells you how much charge a sphere can hold per volt of potential — and the answer is surprisingly small. Even a sphere the size of the Earth has a capacitance of only a few hundred microfarads. The reason is that the "other plate" of an isolated sphere is infinity, so $d$d is effectively huge.

The derivation re-uses the point-charge potential $V = Q/(4\pi\varepsilon_0 r)$V=Q/(4πε0​r) from Module 6.2: for a uniformly charged sphere, all the charge can be treated (outside the sphere) as if concentrated at the centre, so at the surface $r = R$r=R and $V_\text{surface} = Q/(4\pi\varepsilon_0 R)$Vsurface​=Q/(4πε0​R). Re-arranging gives $C = Q/V = 4\pi\varepsilon_0 R$C=Q/V=4πε0​R.

Worked Example 4 — Isolated Sphere

A metal sphere of radius 5.0 cm is isolated in air. Calculate its capacitance.

$$\begin{aligned} C &= 4\pi\varepsilon_0 R \\ &= 4\pi \times (8.85 \times 10^{-12}) \times 0.050 \\ &= 5.56 \times 10^{-12}\ \text{F} \\ &\approx 5.6\ \text{pF} \end{aligned}$$C​=4πε0​R=4π×(8.85×10−12)×0.050=5.56×10−12 F≈5.6 pF​

A 5 cm sphere is therefore essentially a 5.6 pF capacitor. Contrast this with the 100 pF we calculated for two 15 cm plates: the parallel-plate geometry wins easily, which is why every practical capacitor is built that way.

Exam tip. OCR loves the isolated-sphere formula because it ties capacitance back to the potential of a point charge (Module 6.2). If you can derive $C = 4\pi\varepsilon_0 R$C=4πε0​R from $V = Q/(4\pi\varepsilon_0 R)$V=Q/(4πε0​R) in the exam, you will pick up marks other candidates miss.

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Typical Capacitor Values

Real capacitors span about 12 orders of magnitude:

Device	Typical capacitance
RF tuning capacitor	1 – 100 pF
Ceramic disc (bypass)	1 – 100 nF
Film capacitor	10 nF – 10 $\mu$μF
Electrolytic (power supply)	1 $\mu$μF – 10 000 $\mu$μF
Supercapacitor (memory backup)	0.1 – 100 F
Earth as an isolated sphere	$\approx 700\ \mu$≈700 μF

The fact that the Earth itself has a finite capacitance (about 700 $\mu$μF when treated as an isolated sphere of radius 6371 km — try the calculation) explains why even modest amounts of static charge generate kilovolts. Walk across a nylon carpet on a dry day and you can charge your body, considered as an isolated capacitor of about 100 pF, to several thousand volts — enough to draw a visible spark to the next door handle.

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Synoptic Links

• Charge and current (Module 4.1): A capacitor is the only common component for which the relationship between $Q$Q and $V$V is the primary defining equation, rather than current and voltage as for a resistor (Ohm's law) or a battery (e.m.f.). $C = Q/V$C=Q/V is the analogue of $R = V/I$R=V/I — but with charge instead of current as the responding variable. The fact that capacitors do not permit a steady current is the most synoptic possible link to Module 4.1: the rate of charge flow through a capacitor branch is $i = C\,\mathrm{d}V/\mathrm{d}t$i=CdV/dt, so a constant $V$V gives zero current.
• Electric field strength (Module 6.2): The parallel-plate geometry gives a uniform field $E = V/d$E=V/d — the standard worked example of a uniform electric field. The isolated-sphere capacitance derives directly from the point-charge potential $V = Q/(4\pi\varepsilon_0 R)$V=Q/(4πε0​R), which is in turn an integral of the radial field $E = Q/(4\pi\varepsilon_0 r^2)$E=Q/(4πε0​r2). Capacitors and fields are not two topics; they are two sides of the same Module-6 coin.
• Potential energy and energy stored (Module 6.1, lesson 3): Once you can compute $C$C from geometry, you can also compute the energy stored as $\tfrac{1}{2}CV^2$21​CV2. The capacitor energy formulae of lesson 3 are applications of the defining $C = Q/V$C=Q/V relationship developed here.

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Specimen question modelled on the OCR H556 paper format

Question (8 marks): A student builds a parallel-plate capacitor by mounting two square aluminium foil plates of side 12.0 cm on a perspex frame. The plates are separated by an air gap of 1.5 mm.

(a) Calculate the capacitance $C_0$C0​ of the capacitor when the gap contains air. Use $\varepsilon_0 = 8.85 \times 10^{-12}$ε0​=8.85×10−12 F m$^{-1}$−1. [3]

(b) The student now slides a 1.5 mm-thick sheet of polythene ($\varepsilon_r = 2.3$εr​=2.3) completely between the plates without changing $d$d. State, with a reason, what happens to (i) the capacitance and (ii) the charge stored if the plate voltage is held constant at 9.0 V by a battery. Compute the new charge. [3]

(c) The student then disconnects the battery and withdraws the polythene with the capacitor still electrically isolated. Without further calculation, explain what happens to the charge on the plates, the potential difference across the plates, and the capacitance. [2]

AO breakdown

Mark	AO	Awarded for
1	AO2	$A = 0.144 \times 10^{-2}\ \text{m}^2 = 1.44 \times 10^{-2}\ \text{m}^2$A=0.144×10−2 m2=1.44×10−2 m2 used (or computed correctly to one s.f.)
2	AO2	Substitution $C_0 = \varepsilon_0 A/d$C0​=ε0​A/d shown
3	AO2	$C_0 \approx 8.5 \times 10^{-11}\ \text{F} = 85$C0​≈8.5×10−11 F=85 pF, to 2 s.f.
4	AO1	(i) Capacitance increases by factor of $\varepsilon_r = 2.3$εr​=2.3
5	AO2	(ii) New $C = 2.3 \times 85 \approx 196$C=2.3×85≈196 pF; new $Q = CV = 196 \times 10^{-12} \times 9.0 \approx 1.76 \times 10^{-9}$Q=CV=196×10−12×9.0≈1.76×10−9 C
6	AO3	Reason: dielectric polarises and reduces internal field for given charge, so $V$V falls for given $Q$Q and $C = Q/V$C=Q/V rises
7	AO3	Isolated capacitor: charge $Q$Q unchanged (no path for it to flow), $C$C falls back to $C_0$C0​, hence $V = Q/C$V=Q/C rises by factor of $\varepsilon_r$εr​
8	AO3	Comment that work has been done against the attractive forces between induced dielectric surface charges and the plate charges (energy of system rises)

AO split: AO1 = 1, AO2 = 4, AO3 = 3.

Mid-band response (4/8):

(a) $A = 0.144$A=0.144 m$^2$2. $C = \varepsilon_0 A/d = 8.85 \times 10^{-12} \times 0.144 / 0.0015 = 8.5 \times 10^{-10}$C=ε0​A/d=8.85×10−12×0.144/0.0015=8.5×10−10 F.

(b) Capacitance increases. New charge $= CV = 8.5 \times 10^{-10} \times 9 \times 2.3 = 1.76 \times 10^{-8}$=CV=8.5×10−10×9×2.3=1.76×10−8 C.

(c) Capacitance falls. Voltage stays the same.

Examiner commentary: The next-band move is to fix the area: $12\ \text{cm} = 0.12$12 cm=0.12 m, so $A = 0.12 \times 0.12 = 0.0144\ \text{m}^2$A=0.12×0.12=0.0144 m2, not $0.144\ \text{m}^2$0.144 m2. The factor-of-10 area error has propagated through the rest of the calculation. Marks 1 lost; mark 2 awarded for correct substitution form. Marks 4 and 5 awarded for correct reasoning that $C$C scales with $\varepsilon_r$εr​. Mark 6 lost — no mention of polarisation as the mechanism. Mark 7 lost — the candidate incorrectly says $V$V stays the same when the dielectric is withdrawn from an isolated capacitor; in fact $Q$Q is unchanged and $V$V must rise as $C$C falls.

Stronger response (7/8):

(a) $A = 0.12 \times 0.12 = 1.44 \times 10^{-2}\ \text{m}^2$A=0.12×0.12=1.44×10−2 m2, $d = 1.5 \times 10^{-3}$d=1.5×10−3 m. $C_0 = \varepsilon_0 A/d = (8.85 \times 10^{-12})(1.44 \times 10^{-2})/(1.5 \times 10^{-3}) = 8.5 \times 10^{-11}$C0​=ε0​A/d=(8.85×10−12)(1.44×10−2)/(1.5×10−3)=8.5×10−11 F $\approx 85$≈85 pF.

(b) (i) The capacitance increases by factor $\varepsilon_r = 2.3$εr​=2.3, because the dielectric polarises in the field and partially cancels the field due to the plates, reducing $V$V for given $Q$Q so that $C = Q/V$C=Q/V rises. (ii) Voltage is held at 9.0 V, so the battery pushes more charge onto the plates until $Q = CV = 196\ \text{pF} \times 9.0\ \text{V} \approx 1.76 \times 10^{-9}$Q=CV=196 pF×9.0 V≈1.76×10−9 C.

(c) The capacitor is now isolated, so $Q$Q cannot change. Withdrawing the polythene returns $C$C to its original 85 pF, so $V = Q/C$V=Q/C rises back by the factor 2.3 to $V = 9.0 \times 2.3 \approx 21$V=9.0×2.3≈21 V.

Examiner commentary: To lift to top-band the answer would address the energy angle in (c): the system's stored energy is $Q^2/(2C)$Q2/(2C), which rises as $C$C falls at fixed $Q$Q. The extra energy comes from the mechanical work done against the attractive force between the induced surface charges on the dielectric and the plate charges as the dielectric is withdrawn. Marks 1-7 awarded. Mark 8 lost — no energy argument.

Top-band response (8/8):

(a) Side $\ell = 12.0$ℓ=12.0 cm $= 0.120$=0.120 m, so $A = \ell^2 = 1.44 \times 10^{-2}\ \text{m}^2$A=ℓ2=1.44×10−2 m2. With $d = 1.5$d=1.5 mm $= 1.5 \times 10^{-3}$=1.5×10−3 m and $\varepsilon_0 = 8.85 \times 10^{-12}$ε0​=8.85×10−12 F m$^{-1}$−1,

$C_0 = \frac{\varepsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(1.44 \times 10^{-2})}{1.5 \times 10^{-3}} \approx 8.5 \times 10^{-11}\ \text{F} = 85\ \text{pF}.$C0​=dε0​A​=1.5×10−3(8.85×10−12)(1.44×10−2)​≈8.5×10−11 F=85 pF.

(b) (i) Polythene has $\varepsilon_r = 2.3$εr​=2.3, so $C$C rises by factor 2.3 to $\approx 196$≈196 pF. Mechanism: the polythene molecules polarise in the field; the induced surface charges partially cancel the field due to the plates, so the internal field — and hence $V$V — falls for a given $Q$Q, raising $C = Q/V$C=Q/V. (ii) With $V$V held at 9.0 V by the battery,

$Q = CV = (1.96 \times 10^{-10})(9.0) \approx 1.76 \times 10^{-9}\ \text{C}.$Q=CV=(1.96×10−10)(9.0)≈1.76×10−9 C.

(c) The capacitor is isolated, so $Q$Q is conserved at $1.76 \times 10^{-9}$1.76×10−9 C. Removing the polythene returns $C$C to $C_0 = 85$C0​=85 pF, so $V = Q/C$V=Q/C rises by a factor of $\varepsilon_r = 2.3$εr​=2.3 to $\approx 21$≈21 V. Energetically, the stored energy $U = Q^2/(2C)$U=Q2/(2C) rises by the same factor 2.3 — and this extra energy is supplied by the mechanical work the student does to pull the dielectric out, against the attractive force between the induced dielectric surface charges and the plate charges.

Examiner commentary: Full marks. The discriminator moves are (1) explicit treatment of $A$A from $\ell = 0.120$ℓ=0.120 m rather than $0.144$0.144 m, (2) the polarisation mechanism in (b)(i) — not just the formula — and (3) the energy bookkeeping in (c) that explains where the increase in stored energy comes from. Top-band answers always close the energy accounts.

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Common errors and mark-loss patterns

Teacher-voice observations from years of marking parallel-plate questions.

• Side in centimetres squared in formula. $A = (12\ \text{cm})^2 = 144\ \text{cm}^2$A=(12 cm)2=144 cm2 is correct; substituted into the SI formula as $144$144 (or even $0.144$0.144) gives wrong-order-of-magnitude $C$C. Always convert to metres before squaring: $\ell = 0.12$ℓ=0.12 m gives $A = 0.0144$A=0.0144 m$^2$2.
• Missing $\varepsilon_r$εr​ when a dielectric is present. The formula $C = \varepsilon_0 A/d$C=ε0​A/d is vacuum only. With a dielectric it is $C = \varepsilon_0 \varepsilon_r A/d$C=ε0​εr​A/d. A common slip is to write the vacuum formula with a footnote "with dielectric" — examiners reject this.
• Diameter for radius in the isolated-sphere formula. $C = 4\pi\varepsilon_0 R$C=4πε0​R, where $R$R is the radius. Using $D$D doubles your $C$C.
• Confusing "$Q$Q on the capacitor" with "$Q$Q on each plate" with "total $Q$Q". The OCR convention is that $Q$Q in $C = Q/V$C=Q/V is the magnitude of charge on one plate — the positive plate, say. The other plate carries $-Q$−Q. The capacitor as a whole is neutral; there is no "total $Q$Q" to speak of.
• Confusing C (capacitance) with C (coulomb). Always write the unit explicitly. "$C = 47$C=47 C" is meaningless on its own; "$C = 47\ \mu$C=47 μF" or "$Q = 47\ \mu$Q=47 μC" is what you mean.
• Reading the wrong $\mu/n/p$μ/n/p prefix. A 470 $\mu$μF capacitor is $4.70 \times 10^{-4}$4.70×10−4 F, not $4.70 \times 10^{-7}$4.70×10−7 F (which would be 470 nF). Practise converting until it is automatic; in an exam this kind of slip costs a whole question.
• Forgetting that the proportionality $Q \propto V$Q∝V breaks down at breakdown. Real capacitors have a maximum voltage rating beyond which the dielectric breaks down (an air gap fails at $\approx 3$≈3 MV m$^{-1}$−1). OCR rarely asks for this number but a top-band answer mentions the limit when discussing high-voltage capacitors.

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Going further

The microscopic theory of dielectrics is one of the loveliest crossover topics in undergraduate electromagnetism. At university you will meet the polarisation density $\vec{P}$P — the dipole moment per unit volume of a dielectric — and the electric displacement $\vec{D} = \varepsilon_0\vec{E} + \vec{P}$D=ε0​E+P. For a linear isotropic dielectric, $\vec{P} = \chi_e \varepsilon_0 \vec{E}$P=χe​ε0​E with $\chi_e$χe​ the electric susceptibility, and the relative permittivity is $\varepsilon_r = 1 + \chi_e$εr​=1+χe​. The A-Level formula $C = \varepsilon_0\varepsilon_r A/d$C=ε0​εr​A/d is the special case for a uniform linear dielectric filling a parallel-plate gap.

Capacitance at radio frequencies behaves like an impedance $Z_C = 1/(\mathrm{i}\omega C)$ZC​=1/(iωC) — the capacitor lets high-frequency signals through with low impedance but blocks DC entirely. This is the principle behind every audio coupling capacitor on every guitar amplifier and the high-pass filters in radio receivers. Two Oxbridge interview-style prompts that probe this material:

1. "A capacitor is charged from a battery through a small resistance. Half of the battery's work ends up stored in the capacitor and half is dissipated in the resistor, regardless of the resistance. Where does the energy go if the resistance is made arbitrarily small?" (Answer: it is radiated as a brief pulse of electromagnetic energy — the limit is set by the transmission-line impedance of the wires, not by ohmic dissipation.)
2. "Why is the capacitance of an isolated conducting sphere finite and equal to $4\pi\varepsilon_0 R$4πε0​R? What is the 'other plate'?" (Answer: a sphere at infinity, held at zero potential. The integration runs from $R$R to $\infty$∞, and the inverse-square law makes the integral converge to give the finite result.)
3. "A parallel-plate capacitor is connected to a battery and a dielectric is then inserted. The energy stored rises. Where does the extra energy come from?" (Answer: the battery does extra work pushing more charge onto the plates; the dielectric is pulled into the gap by an attractive force which does positive work, but the battery does still more work than is stored — half of the battery's extra work is dissipated as transient resistive heat. A full energy account is a beautiful undergraduate problem.)

Recommended reading: Griffiths, Introduction to Electrodynamics, Chapter 4 — the cleanest pedagogical treatment of dielectrics and polarisation at the undergraduate level. Chapter 4.4 gives the energy argument for the dielectric-pulled-in problem in full.

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A-Level misconceptions

• "$C = Q/V$C=Q/V is a derived formula like $C = \varepsilon_0 A/d$C=ε0​A/d." No. $C = Q/V$C=Q/V is the definition of capacitance and applies to any capacitor of any shape. $C = \varepsilon_0 A/d$C=ε0​A/d is the parallel-plate vacuum derived formula. Mixing the two up is the single most common conceptual slip on this topic.
• "Adding charge to a capacitor changes its capacitance." Wrong. Capacitance is a geometric property — it depends on shape, size, separation and dielectric, not on how much charge is currently stored. If you double $Q$Q, then $V$V also doubles and $C$C stays the same. This is why $C = Q/V$C=Q/V works.
• "The plates of a charged capacitor carry charge $+Q$+Q and $+Q$+Q." Wrong. They carry $+Q$+Q and $-Q$−Q — equal and opposite. The capacitor as a whole is electrically neutral.
• "Capacitance is measured in volts." Wrong. The volt is the unit of potential; the farad is the unit of capacitance. 1 F = 1 C V$^{-1}$−1.
• "A 1 farad capacitor is tiny because the farad is a tiny unit." Wrong. The farad is an enormous unit; 1 F discrete capacitors are physically large (or use exotic electrochemistry, as in supercapacitors). Most everyday capacitors are micro-, nano- or picofarads.
• "The factor $\varepsilon_r$εr​ is the same as $\varepsilon_0$ε0​." No — $\varepsilon_0$ε0​ is a fundamental constant (with units F m$^{-1}$−1), while $\varepsilon_r$εr​ is a dimensionless material property that multiplies $\varepsilon_0$ε0​ when a dielectric is present. The actual permittivity of a material is $\varepsilon = \varepsilon_r \varepsilon_0$ε=εr​ε0​.

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Summary

• Capacitance is defined as $C = Q/V$C=Q/V, with $Q$Q the magnitude of charge on one plate and $V$V the potential difference between the plates. Unit: the farad (F), with 1 F $=$= 1 C V$^{-1}$−1.
• For a parallel-plate geometry in vacuum, $C = \varepsilon_0 A/d$C=ε0​A/d, where $A$A is the plate area, $d$d the separation and $\varepsilon_0 = 8.85 \times 10^{-12}$ε0​=8.85×10−12 F m$^{-1}$−1.
• Inserting a dielectric of relative permittivity $\varepsilon_r$εr​ multiplies the capacitance: $C = \varepsilon_0 \varepsilon_r A/d$C=ε0​εr​A/d. Mechanism: polarisation reduces the field for given charge.
• An isolated conducting sphere has $C = 4\pi\varepsilon_0 R$C=4πε0​R — an OCR-specific result that ties capacitance to the point-charge potential of Module 6.2.
• Practical capacitors span 12 orders of magnitude, from a few picofarads (RF tuning) to hundreds of farads (supercapacitors).
• $C = Q/V$C=Q/V is the definition; the geometric formulae are consequences of the definition applied to particular shapes. The defining equation is what you reach for first in any unfamiliar problem.

Master these five relationships and you have the foundation of every later capacitor calculation in the topic — including series and parallel combinations (lesson 2), energy storage (lesson 3) and RC charge/discharge dynamics (lessons 4 and 5).

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Reference: OCR A-Level Physics A (H556) specification 6.1 — Capacitors (refer to the official OCR H556 specification document for exact wording).