Capacitance and dielectrics

Assembling charge takes work, and the previous lesson left the question of where that work goes while the charges sit still.

Answering it needs a device rather than an argument, because the answer is a claim about energy density and energy density has to be measured somewhere. The device is the capacitor: two conductors, one holding $+ Q$ and the other $- Q$ , with a field between them. It is the simplest arrangement in which a definite amount of electrostatic energy is stored in a definite volume, and it turns the question into arithmetic.

Capacitance

Take any two conductors, isolated from everything else, and move charge $Q$ from one to the other, leaving $+ Q$ on one and $- Q$ on the other. A potential difference $V$ appears between them. Double the charge and, because the field everywhere doubles by superposition and the potential difference is a line integral of the field, $V$ doubles too. The ratio is therefore a constant of the geometry alone:

C = \frac{Q}{V}

This is the capacitance, in coulombs per volt, called farads. Note that $Q$ here means the charge on the positive conductor, not the net charge of the pair, which is zero.

The farad is an absurdly large unit, and seeing why is the fastest way to get a feel for the quantity. Take two parallel plates of area $A$ separated by $d$ , with $d$ small enough that the field between them is uniform and edge effects can be ignored. Gauss's law at the surface of a conductor gives $E = σ / ε_{0} = Q / (ε_{0} A)$ between the plates, and since the field is uniform the potential difference is simply $V = E d = Q d / (ε_{0} A)$ . So

C = \frac{ε_{0} A}{d}

Plates of one square metre a millimetre apart give $C = (8.854 \times 10^{- 12}) (1) / (10^{- 3}) = 8.85$ nF. To reach one farad at that spacing you would need $1.13 \times 10^{8}$ m² of plate, an area of 113 square kilometres. Real capacitors of a farad and more exist, and they get there by making $d$ molecular rather than by making $A$ absurd, which is the trick described at the end of this lesson.

An isolated conductor also has a capacitance, with the second conductor taken to be a sphere at infinity. For a sphere of radius $R$ at potential $k Q / R$ , this gives $C = R / k = 4 π ε_{0} R$ . The Earth, with $R = 6.37 \times 10^{6}$ m, comes out at 709 μF, which is why connecting something to ground is a good way to make its potential stop changing: it is being connected to a large capacitor.

Example. Two parallel plates of area 20 cm² are separated by 0.50 mm of air and connected to a 12 V battery. Find the capacitance, the charge on each plate and the field between them.

$C = ε_{0} A / d = (8.854 \times 10^{- 12}) (20 \times 10^{- 4}) / (5.0 \times 10^{- 4}) = 3.54 \times 10^{- 11}$ F, that is 35.4 pF. The charge is $Q = C V = (3.54 \times 10^{- 11}) (12) = 4.25 \times 10^{- 10}$ C, or 425 pC. The field is uniform, so $E = V / d = 12 / (5.0 \times 10^{- 4}) = 2.4 \times 10^{4}$ V/m, comfortably below the 3 × 10⁶ V/m at which air breaks down.

Now you. Two parallel plates of area 50 cm² are separated by 1.0 mm of air and connected to a 9.0 V battery. Find the capacitance, the charge and the field.

Answer

$C = (8.854 \times 10^{- 12}) (50 \times 10^{- 4}) / (1.0 \times 10^{- 3}) = 4.43 \times 10^{- 11}$ F, or 44.3 pF. Then $Q = C V = 3.98 \times 10^{- 10}$ C, and $E = V / d = 9.0 / 10^{- 3} = 9.0 \times 10^{3}$ V/m.

Combinations

Two capacitors side by side, both connected across the same pair of nodes, are in parallel. Each sees the same $V$ , and the charges add, so $Q = Q_{1} + Q_{2} = (C_{1} + C_{2}) V$ and

C_{parallel} = C_{1} + C_{2}

which is exactly what putting two capacitors side by side does geometrically: it adds their plate areas.

Two capacitors joined end to end, so that the same wire carries charge from one to the next, are in series. Here the argument that matters is about the isolated conductor between them. It started neutral and nothing can reach it, so whatever charge $- Q$ is drawn to one side must leave $+ Q$ on the other: every capacitor in a series chain carries the same charge, regardless of its size. The voltages add, so $V = Q / C_{1} + Q / C_{2}$ and

\frac{1}{C_{series}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

The series combination is always smaller than either member, which again matches the geometry: putting two capacitors in series is like increasing the separation.

Notice that the rules are the reverse of the ones resistors will follow in the next lesson. It is not worth memorising either; both fall out in two lines from what is shared and what adds.

Example. A 4.0 μF and a 12 μF capacitor are connected in series, and that pair is connected in parallel with a 5.0 μF capacitor across a 24 V supply. Find the total capacitance and the voltage across the 4.0 μF capacitor.

The series pair gives $1 / C = 1 / 4 + 1 / 12 = 4 / 12$ , so $C = 3.0$ μF. In parallel with 5.0 μF the total is 8.0 μF. The series branch is across the full 24 V, so it carries $Q = (3.0 \times 10^{- 6}) (24) = 72$ μC, and that same 72 μC sits on the 4.0 μF capacitor. Its voltage is therefore $V = Q / C = 72 / 4.0 = 18$ V, leaving $72 / 12 = 6$ V across the other, which correctly sums to 24 V. The smaller capacitor takes the larger share of the voltage.

Now you. A 6.0 μF and a 3.0 μF capacitor in series are connected in parallel with a 4.0 μF capacitor across 30 V. Find the total capacitance and the voltage across the 3.0 μF capacitor.

Answer

Series: $1 / C = 1 / 6 + 1 / 3 = 1 / 2$ , so $C = 2.0$ μF, and the total is $2.0 + 4.0 = 6.0$ μF. The series branch carries $Q = (2.0 \times 10^{- 6}) (30) = 60$ μC, so the 3.0 μF capacitor has $V = 60 / 3.0 = 20$ V across it, and the 6.0 μF has 10 V.

The energy of a charged capacitor

Charging a capacitor is a matter of moving charge from one plate to the other, and it gets harder as it goes: the first electron crosses a zero potential difference, and the last crosses the full $V$ . If the capacitor holds $q$ at some moment, its voltage is $q / C$ , and moving a further $d q$ costs $(q / C) d q$ . Integrating from empty to $Q$ ,

U = \int_{0}^{Q} \frac{q}{C} d q = \frac{Q^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V

The factor of one half is the average voltage during the charging, and it is where the energy that a battery delivers but a capacitor does not store goes: charge a capacitor through a resistor from a battery of EMF $V$ and the battery supplies $Q V$ while the capacitor keeps $\frac{1}{2} Q V$ , the other half being dissipated in the resistance no matter how small that resistance is.

The numbers get large quickly. A photographic flash capacitor of 100 μF charged to 200 V holds $\frac{1}{2} (10^{- 4}) (200)^{2} = 2.0$ J, which is unremarkable until it is released in a millisecond and becomes two kilowatts. A defibrillator is the same idea scaled up.

Example. A defibrillator stores energy in a 32 μF capacitor charged to 5.0 kV, and delivers it over 4.0 ms. How much energy, and what average power?

$U = \frac{1}{2} C V^{2} = \frac{1}{2} (32 \times 10^{- 6}) (5.0 \times 10^{3})^{2} = \frac{1}{2} (32 \times 10^{- 6}) (2.5 \times 10^{7}) = 400$ J. Average power is energy over time, $400 / 0.0040 = 1.0 \times 10^{5}$ W. A hundred kilowatts from a device that runs off a small battery, because the battery spends seconds filling the capacitor and the capacitor empties in milliseconds. Capacitors are poor stores of energy and superb stores of power.

Now you. A capacitor of 100 μF is charged to 300 V and discharged in 2.0 ms. How much energy, and what average power?

Answer

$U = \frac{1}{2} (100 \times 10^{- 6}) (300)^{2} = \frac{1}{2} (10^{- 4}) (9.0 \times 10^{4}) = 4.5$ J, and the average power is $4.5 / 0.0020 = 2.25 \times 10^{3}$ W.

The energy is in the field

Now the question the previous lesson left open. Take the parallel plate result and rewrite it, substituting $C = ε_{0} A / d$ and $V = E d$ :

U = \frac{1}{2} C V^{2} = \frac{1}{2} \frac{ε_{0} A}{d} (E d)^{2} = \frac{1}{2} ε_{0} E^{2} \times (A d)

and $A d$ is precisely the volume between the plates, the region where the field is. Dividing,

u = \frac{1}{2} ε_{0} E^{2}

joules per cubic metre, wherever there is a field. Every reference to charge, plates or geometry has vanished; what is left refers only to the local field strength.

This can be read two ways. Conservatively, it is an identity: an algebraic rearrangement of $\frac{1}{2} C V^{2}$ that happens to have a suggestive form, proved only for a parallel plate capacitor. Read strongly, it says the energy is genuinely located in the field, spread through space at a density $\frac{1}{2} ε_{0} E^{2}$ , and that charges are irrelevant to the accounting.

Nothing in electrostatics can distinguish the two readings, because in electrostatics the charges and the field always come together. The strong reading is nonetheless correct, and the proof is in the twelfth lesson: an electromagnetic wave carries energy through empty space at exactly the density this formula gives, with no charge anywhere near it, and that energy can be absorbed and weighed. Take this formula as a promissory note.

Meanwhile it says something sobering about capacitors as batteries. Air breaks down at $3 \times 10^{6}$ V/m, so the largest energy density an air capacitor can reach is $\frac{1}{2} (8.854 \times 10^{- 12}) (3 \times 10^{6})^{2} = 39.8$ J/m³. Petrol stores about $3.4 \times 10^{10}$ J/m³. The chemical fuel wins by a factor near $10^{9}$ , and even commercial supercapacitors, which use tricks well beyond a plain dielectric, reach only a few per cent of a lithium ion cell's energy per kilogram. Capacitors are used where power density and cycle life matter, not where energy is wanted.

What a dielectric does

Fill the gap between the plates with an insulating material and the capacitance rises, by a factor $κ$ called the dielectric constant or relative permittivity of the material. Faraday measured this in 1837 and the definition is still his: $κ$ is the ratio of the capacitance with the material to the capacitance in vacuum.

Material	Dielectric constant $κ$	Breakdown field / MV m⁻¹
Vacuum	$1$ exactly	no breakdown
Air	$1.00059$	$3$
Paper	$3.7$	$16$
Mica	$5.4$	$100$
Water at 20 degrees Celsius	$80.4$	conducts, so unusable
Barium titanate	$1200$ and above	$8$

The mechanism is polarisation. Every molecule in the material develops, or already has, a dipole moment that partly lines up with the applied field. The dipoles inside the bulk cancel against their neighbours, but at the two faces of the slab there is an uncancelled layer of charge: negative at the face nearest the positive plate, positive at the other. That bound surface charge is of opposite sign to the free charge on the adjacent plate, so it partly cancels it, and the field inside the material drops to $E = E_{0} / κ$ .

From there everything follows. If the capacitor is held at fixed charge, the field drops by $κ$ , so the voltage drops by $κ$ , so $C = Q / V$ rises by $κ$ . If it is held at fixed voltage by a battery, the field is unchanged and more charge flows in, which again means $C$ rose by $κ$ . Either way

C = \frac{κ ε_{0} A}{d}

and the energy density becomes $u = \frac{1}{2} κ ε_{0} E^{2}$ .

The two columns of the table have to be read together, because they pull in opposite directions. Water's $κ$ of 80 comes from whole molecules rotating into alignment, which is a large effect and a slow one: it collapses above about 10 GHz, where the molecules can no longer keep up, and water is useless as a capacitor dielectric anyway because it conducts. Mica has a modest $κ$ and an excellent breakdown field, so a mica capacitor can be run at high voltage. Since the stored energy goes as $κ E^{2}$ and $E$ is capped by breakdown, a material with a high $κ$ and a poor breakdown field can easily be the worse choice.

Practical capacitors take the dielectric idea to its limit. An electrolytic capacitor grows an aluminium oxide layer, sometimes under 100 nm thick, on an etched foil whose roughness multiplies its effective area many times over. With $d$ that small, $C = κ ε_{0} A / d$ delivers millifarads from something the size of a thumb, which is how the 113 square kilometres of the opening estimate is avoided.

The limits of the electrostatic picture

Five lessons have now built a complete theory of charges at rest: a force law, a field, two ways of computing that field, an energy, and a device that stores it. Everything in it is exact, and everything in it assumes nothing moves.

That assumption is doing more work than it looks. A conductor was defined as a material in which charges are free to move, and then every result about conductors was obtained by waiting until they had stopped. The waiting time was never asked about. The capacitor was charged, and how the charge got there was never modelled. Even the sentence "connect it to a battery" is outside the theory: nothing so far explains what a battery is, or why charge would keep flowing through a wire instead of arranging itself once and stopping.

Charge in steady motion is a genuinely different regime, with its own quantities and its own laws, and it is also the regime in which every application of electricity lives. It is next.