Capacitors

Beginner

Camera flashes, heart defibrillators and the circuits in your phone all depend on capacitors. A capacitor is one of the simplest electrical parts: two metal plates facing each other across a thin gap of air or plastic. It stores separated electric charge, and with it energy, which it can give back in a split second.

Step 1: moving electrons. Metals contain electrons that are free to move. Connect the plates to a battery and it pulls electrons off one plate and pushes them, through the wires, onto the other. One plate is left positive, the other turns negative. No electron is created or destroyed: each one that lands on one plate was taken from the other, so the two charges are always equal and opposite.

Step 2: nothing crosses the gap. The gap is an insulator, so charge cannot jump across it. Instead the + plate and the − plate attract each other through it, and an electric field fills the space between them.

Step 3: filling up. The more charge the plates already hold, the harder they push back against the battery. Charging is therefore quick at first, then slows, and stops when the capacitor's voltage equals the battery's, just as a tank fed from a water tower stops filling when the two levels match.

Step 4: size matters. Larger plates or a narrower gap let the same battery store more charge.

In the animation the blue electrons race through the wires at first, then slow down and stop, while charge piles up on the plates and the gauge fills ever more slowly.

Check yourself: one plate of a charged capacitor carries +5 units of charge. What does the other carry? (Answer: −5 units; the total is still zero.)

Intermediate

Charge, voltage and capacitance

Before a camera flash fires, a capacitor collects its energy slowly; the flash then releases it in one brief burst. A capacitor is any pair of conductors separated by an insulator, most simply two metal plates across a thin gap. A battery moves charge through the wires from one plate to the other, leaving $+Q$ on one and $-Q$ on the other, so the capacitor as a whole stays neutral and its charge means $Q$. Its voltage $V$ is the potential difference between the plates, the work per coulomb to carry charge from the negative plate to the positive one.

Double the charge and the field between the plates doubles everywhere, and so does the work per coulomb. Charge and voltage are therefore proportional:
$Q=CV$

The constant $C$ is the capacitance, the charge stored per volt, measured in farads: $1\mathrm{F}=1C/V$. A farad is enormous, so most capacitors are rated in picofarads ($10^{-12}\mathrm{F}$), nanofarads or microfarads (µF, $10^{-6}\mathrm{F}$).

What sets the capacitance

For two flat parallel plates of area $A$ a distance $d$ apart, the field between them is uniform, $E=V/d$, and nearly zero outside. The capacitance depends only on this geometry and the insulator:
$C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}$

Here ${\epsilon }_0\approx 8.85\times 10^{-12}F/m$ is the permittivity of free space, and ${\epsilon }_r$ is the relative permittivity, or dielectric constant, of the insulator: 1.0006 for air, about 2.2 for polypropylene, 80 for water.

The charge per square metre fixes the field, $E=Q/({\epsilon }_0{\epsilon }_{r}A)$, so larger plates spread the same charge thinner, and the field and the voltage $V=Ed$ are smaller. A narrower gap keeps the field of the same charge but shortens the path across it. In a dielectric the molecules polarise, their negative ends turning towards the positive plate, and partly cancel the field. Each of these lowers the voltage per coulomb, so the same battery pushes on more charge.

Two plates of 1 cm² with a 0.5 mm air gap have $C=8.85\times 10^{-12}\cdot 10^{-4}/(5\times 10^{-4})\approx 1.77\mathrm{pF}$; pressed to 0.3 mm, as in a capacitive key, they reach 2.95 pF. One farad with a 1 mm air gap would need 113 km² of plate.

Charging through a resistor

In the animation a battery of voltage $V_0$ charges the capacitor through a resistor $R$. The battery's voltage is shared between them, $V_0=RI+V_C$, so only the voltage left over for the resistor drives the current:
$I=\frac{V_0-V_C}{R}$

At the start the capacitor is empty, $V_C=0$, and the current is at its largest, $V_0/R$, as if the capacitor were a plain wire. Each bit of charge that arrives raises $V_C$, leaves less voltage for the resistor and slows the current: the fuller the capacitor, the gentler the push that fills it.

The pace is set by the time constant $\tau =RC$; ohms times farads give seconds. After one time constant the capacitor has 63 % of its final voltage, and after $5\tau$ more than 99 %. The exact laws, derived at the advanced level, are exponential:

• $\tau =RC$
• $V_C=V_0(1-e^{-t/RC})$
• $I=\left(\frac{V_0}{R}\right)e^{-t/RC}$

Stored energy and the voltage–charge graph

Charging takes work: each bit of charge must be carried across the voltage already built up, almost zero for the first bit and the full $V$ for the last. On a graph of voltage against charge, $V=Q/C$ is a straight line from the origin, and the work is the area under it, a triangle of base $Q$ and height $V$:
$W=\frac{1}{2}C{V}^2=\frac{1}{2}QV$

The ½ appears because the average voltage during charging is half the final one. Plotted as charge against voltage, the same line has slope $C$, so a larger capacitance gives a steeper line. Because $W$ grows with the square of the voltage, doubling the voltage stores four times the energy, held in the field between the plates.

Worked example: the circuit in the animation

With the animation's $V_0=12\mathrm{V}$, $R=10\mathrm{k\Omega }$ and $C=100\mu \mathrm{F}$, the initial current, time constant, final charge and energy are
$I_0=\frac{V_0}{R}=\frac{12\mathrm{V}}{10^4\Omega }=1.2\mathrm{mA}$
$\tau =RC=10^4\Omega \cdot 10^{-4}\mathrm{F}=1\mathrm{s}$
$Q=C{V}_0=10^{-4}\mathrm{F}\cdot 12\mathrm{V}=1.2\mathrm{mC}$
$W=\frac{1}{2}C{V}_0^2=\frac{1}{2}\cdot 10^{-4}\cdot 12^2=7.2\mathrm{mJ}$

These are the readouts' end points: $I$ starts at 1.20 mA, and once charging is over $V_C$ reads 12.0 V, $Q$ 1.20 mC and $I$ 0.00 mA. At $t=\tau =1\mathrm{s}$ they show $V_C=7.6\mathrm{V}$ (63 %), $Q=0.76\mathrm{mC}$ and $I=0.44\mathrm{mA}$, as the loop rule confirms: $(12-7.6)\mathrm{V}/10\mathrm{k\Omega }=0.44\mathrm{mA}$.

The blue dots are electrons. They leave the top plate for the battery's positive terminal, and as many are pushed from the negative terminal onto the bottom plate, so the top plate becomes $+Q$ (red-orange dots) and the bottom one $-Q$ (blue dots). The gold arrow $I$, the conventional current, points against the electrons' drift and shrinks as the current dies. The purple arrows $E$ between the plates grow in proportion to $Q$, and no dot ever crosses the gap.

Worked example: a camera flash

A camera flash stores its energy in a capacitor of, say, 150 µF charged to 300 V. It holds
$Q=CV=150\times 10^{-6}\cdot 300=0.045\mathrm{C}$
$W=\frac{1}{2}C{V}^2=\frac{1}{2}\cdot 150\times 10^{-6}\cdot 300^2=6.75\mathrm{J}$

Suppose the charger acts like a 300 V source behind 8 kΩ. Then $\tau =8000\cdot 150\times 10^{-6}=1.2\mathrm{s}$ and the current starts at $300/8000=37.5\mathrm{mA}$. After $2\tau =2.4\mathrm{s}$ the voltage is $300(1-e^{-2})\approx 259\mathrm{V}$, 86 %, but the energy, which goes with $V^2$, is only $0.865^2\approx 0.75$ of the maximum, about 5.0 J.

The tube then empties the fully charged capacitor in about a millisecond, so the average power of the flash is $6.75\mathrm{J}/10^{-3}\mathrm{s}\approx 7\mathrm{kW}$. A discharge time constant of 1 ms needs a resistance of only $10^{-3}/(150\times 10^{-6})\approx 7\Omega$. The same capacitor fills slowly and empties fast: the resistance in its path sets the pace.

Common mistakes

• Current does not flow through the gap: charge flows onto one plate as an equal charge flows off the other, so both wires carry the same current, 1.2 mA at the start in the animation, and none once the capacitor is full. A capacitor blocks a steady current.
• A charged capacitor carries no net charge: the animation's plates end with $+1.2\mathrm{mC}$ and $-1.2\mathrm{mC}$.
• The stored energy is $\frac{1}{2}QV$, not $QV$: 7.2 mJ in the animation, not 14.4 mJ, because the voltage rose from zero while the charge was moved.
• One time constant does not fill the capacitor: it gives 63 % of the final voltage and only 40 % of the final energy. Allow about $5\tau$.
• Kilo-ohms times microfarads give milliseconds: 2.2 kΩ with 47 µF gives $\tau \approx 0.10\mathrm{s}$, not 103 s.

Where you meet it

Camera flashes and defibrillators collect energy slowly and release it in milliseconds. Smoothing capacitors in power supplies fill up at the peaks of the alternating voltage and supply the current in between. Touchscreens sense the change of capacitance when a finger comes near, and each bit of a computer's main memory is a tiny capacitor, charged or empty.

The resistor and the loop rule are developed in Electric Current and Resistance and in DC Circuits and Ohm's Law. Joined to a coil (Electromagnetic Induction), a capacitor makes a circuit whose charge oscillates as in Simple Harmonic Motion, and resistance makes it a case of Damping and Resonance. The changing field in the gap of a charging capacitor acts as the displacement current, which completes the field equations and predicts Electromagnetic Waves.

Advanced

Why charge and voltage are proportional

Every result below rests on one fact: for any pair of conductors, the charge is proportional to the voltage. It is worth seeing why before using it.

Put $+Q$ on one conductor and $-Q$ on the other, with no other charges nearby. The field they make obeys linear equations: the field of several charges is the sum of their separate fields. Multiply every charge by the same factor and the field is multiplied by that factor at every point. The scaled arrangement still has zero field inside the metal and each conductor still at one potential, so it is the true arrangement for the larger charge. The voltage is the line integral of the field from the positive conductor to the negative one, $V=\int \vec{E}\cdot d\vec{l}$, so it grows by the same factor as well.

Charge and voltage therefore rise together in a fixed ratio, $Q=CV$. The capacitance $C$ is set by the shape and spacing of the conductors and by the insulator between them, never by $Q$ or $V$. The recipe for finding it follows directly: assume a charge $Q$, find the field, integrate it to get $V$, and divide.

Parallel plates from Gauss's law

The simplest case is two flat plates of area $A$ a small distance $d$ apart. Gauss's law gives the field between them in one step, and Figure 1 shows how.

Legend

$Q$	Charge on each plate: $+Q$ on the left plate, $-Q$ on the right one.	$\mathrm{C}$
$\sigma$	Charge per square metre on the inner surface of the positive plate, $\sigma =Q/A$ for plates of area $A$.	$C/{\mathrm{m}}^2$
$dA$	Area of the right face of the Gauss box, the only face the field crosses.	${\mathrm{m}}^2$
$E$	Electric field in the gap: uniform, from the positive plate to the negative one; zero inside the metal.	$N/C$
$d$	Width of the gap between the plates.	$\mathrm{m}$
$V$	Voltage between the plates: the line integral of the field from the positive plate to the negative one, $V=Ed$.	$\mathrm{V}$

Figure 1 shows the plates side on, with $+Q$ on the left plate, $-Q$ on the right one and the gap $d$ between them. The positive charge sits on the inner surface of its plate, the side facing the negative plate, with a charge $\sigma$ on each square metre; for plates of area $A$, $\sigma =Q/A$. The small box drawn across that surface is a Gauss box: an imaginary closed surface to which Gauss's law is applied.

Look at the two large faces of the box. The left face lies inside the metal, marked $E=0$. In a conductor at rest the field must vanish, or its free charges would still be moving. The right face, a patch of area $dA$, lies in the gap, and the field $E$ crosses it at right angles. The other faces either run along the field or lie in the metal, so nothing flows through them. The flux out of the box is therefore $EdA$, and the charge inside it is $\sigma dA$.

Gauss's law sets the flux out of a closed surface equal to the charge inside divided by ${\epsilon }_0$:
$EdA=\frac{\sigma dA}{{\epsilon }_0}$
The patch area $dA$ cancels, so the size of the box does not matter, and the field in the gap is
$E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{{\epsilon }_{0}A}$

Because $\sigma$ is the same all over a large plate, this field has the same value everywhere in the gap, as the evenly spaced field lines of the figure show. Outside the pair, the fields of the two plates cancel.

Now the voltage. The arrow $V$ in the figure runs across the gap from the positive plate to the negative one. Along it the field has the same value $E$ at every step, so the line integral is simply $E$ times the length $d$:
$V=Ed=\frac{Qd}{{\epsilon }_{0}A},C=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d}$
As promised, $Q$ has dropped out. A larger area holds more charge per volt; a wider gap holds less. An insulator of relative permittivity ${\epsilon }_r$ filling the gap makes $C$ larger by the factor ${\epsilon }_r$, as the section on dielectrics explains.

The same recipe works for other shapes. Outside an isolated sphere of radius $a$ the field is that of a point charge, and integrating it from the surface out to infinity gives $V=Q/(4\pi {\epsilon }_{0}a)$. In a coaxial cable with inner radius $a$ and outer radius $b$, the field between the conductors falls as $1/r$, so integrating it from $a$ to $b$ gives a logarithm. For the sphere, and for a cable of length $\ell$ with insulation of relative permittivity ${\epsilon }_r$:
$C=4\pi {\epsilon }_{0}a,\frac{C}{\ell }=\frac{2\pi {\epsilon }_0{\epsilon }_r}{\ln (b/a)}$

The numbers are telling. The whole Earth, as a sphere, has only about 0.71 mF, less than an ordinary 1000 µF electrolytic capacitor. A cable with polyethylene insulation (${\epsilon }_r=2.3$) and $b/a=3.5$ has about 100 pF per metre.

The charging equation and its solution

A capacitor is never charged instantly: some resistance in the circuit always limits the current. This section works out how the charge builds up over time, the law behind every timing circuit built from a resistor and a capacitor.

Legend

$V_0$	Battery voltage: the rise in potential across the battery.	$\mathrm{V}$
$R$	Resistance of the resistor.	$\Omega$
$C$	Capacitance of the capacitor.	$\mathrm{F}$
$I$	Current, the same all round the loop: the rate at which charge reaches the upper plate, $I=dq/dt$.	$\mathrm{A}$
$q$	Charge on the capacitor at the instant drawn: $+q$ on the upper plate, $-q$ on the lower one.	$\mathrm{C}$
$V_R$	Drop in potential across the resistor, $V_R=RI$.	$\mathrm{V}$
$V_C$	Drop in potential across the capacitor, $V_C=q/C$.	$\mathrm{V}$

Figure 2 shows the loop: a battery of voltage $V_0$, a resistor $R$ and a capacitor $C$ in series. There is only one path, so the same current $I$ flows through every part. At the instant drawn the capacitor carries $+q$ on its upper plate and $-q$ on its lower one; $q$ grows from zero as the current brings charge.

Follow the potential once round the loop in the direction of $I$. The battery raises it by $V_0$. The resistor lowers it by $V_R=RI$ (Ohm's law), and the capacitor lowers it by $V_C=q/C$. Back at the starting point the potential must have its original value, so the rise equals the two drops together:
$V_0=V_R+V_C$

The current is the rate at which charge arrives on the upper plate, $I=dq/dt$. Writing $V_R=Rdq/dt$ and $V_C=q/C$ turns the loop rule into an equation for the charge alone:
$R\frac{dq}{dt}+\frac{q}{C}=V_0$
It is first order, because only the first derivative appears, and linear, because $q$ appears only to the first power.

To solve it, multiply by $C$ and rearrange: $RCdq/dt=C{V}_0-q$. The right-hand side is the charge still missing from the final value $C{V}_0$. Dividing both sides by it puts everything with $q$ on one side and everything with $t$ on the other:
$\frac{dq}{C{V}_0-q}=\frac{dt}{RC}$
Integrate from an empty capacitor, $q=0$ when the switch closes at $t=0$, up to the charge $q$ at time $t$:
$-\ln \left(1-\frac{q}{C{V}_0}\right)=\frac{t}{RC}$

Undoing the logarithm gives $q=C{V}_0(1-e^{-t/RC})$. Dividing by $C$ gives the capacitor's voltage, and differentiating $q$ gives the current:
$V_C=V_0(1-e^{-t/RC})$
$I=\frac{V_0}{R}{e}^{-t/RC}$
The product $\tau =RC$ has the units of time, since $\Omega \cdot \mathrm{F}=\mathrm{s}$. It is called the time constant, and it sets the pace of everything that follows.

A first-order equation needs one initial condition, here $q(0)=0$. Notice what can and cannot jump when the switch closes. The current jumps at once from zero to $V_0/R$. The charge cannot jump, because changing it in no time at all would need an infinite current; so a capacitor's voltage is always continuous. Discharging through $R$ from a voltage $V_1$ is the same equation with $V_0=0$, and gives $V_C=V_1{e}^{-t/RC}$.

Every quantity in such a circuit, charge, voltages and current alike, follows one pattern. A quantity $x$ starts at $x_0$ and heads for its final value $x_{\infty }$:
$x(t)=x_{\infty }+(x_0-x_{\infty })e^{-t/\tau }$
The reason is that its rate of change is proportional to the distance still to go, $dx/dt=(x_{\infty }-x)/\tau$. As a result, equal times shrink that distance by equal factors: by $e^{-1}\approx 0.37$ in each time constant, and by one half in each $\tau \ln 2\approx 0.69\tau$.

Reading the exponential graphs

The solutions are easiest to grasp as graphs. They are drawn here for the animation's circuit: $V_0=12\mathrm{V}$, $R=10\mathrm{k\Omega }$ and $C=100\mu \mathrm{F}$, so that $\tau =RC=1\mathrm{s}$ and the starting current is $I_0=V_0/R=1.2\mathrm{mA}$.

Legend

$t$	Time since the switch closed.	$\mathrm{s}$
$\tau$	Time constant, $\tau =RC$: the time to cover 63 % of the change.	$\mathrm{s}$
$V_C$	Voltage of the capacitor, rising from zero towards $V_0$.	$\mathrm{V}$
$V_0$	Battery voltage, the final value of $V_C$.	$\mathrm{V}$
$I$	Current in the loop, falling from $I_0$ towards zero; $I=Cd{V}_C/dt$.	$\mathrm{A}$
$I_0$	Starting current, $I_0=V_0/R$.	$\mathrm{A}$
$Q$	Total charge delivered: the shaded area under $I(t)$, equal to $I_0\tau =C{V}_0$.	$\mathrm{C}$

Figure 3 draws the two solutions on a common time axis $t$. The upper graph is the capacitor's voltage $V_C$, climbing from zero towards the battery voltage $V_0$. The lower graph is the current $I$, falling from $I_0$ towards zero. The dashed vertical line in each marks one time constant, $t=\tau$. The animation's expert layer plots the same curves, each as a fraction of its maximum, $V_C$ in blue and $I$ in gold.

Look first at the dots at $t=\tau$. On the upper curve the voltage has reached 63 % of $V_0$; on the lower one the current has fallen to 37 % of $I_0$. These add up to 100 %, and they must do so at every instant, because $V_C+RI=V_0$: whatever voltage the capacitor has not yet taken up, the resistor still carries. As fractions, the two curves are mirror images, and they cross at 50 % when $t=\tau \ln 2\approx 0.69\mathrm{s}$.

Next, the sloping dashed line in the upper graph. It is the tangent to $V_C$ at $t=0$, with slope $V_0/\tau =12V/s$, and it shows what would happen if the current kept its starting value $I_0$: the capacitor would be full after exactly one time constant. The tangent indeed reaches the level $V_0$ at $t=\tau$. This holds from any point on the curve, not only the start: the tangent always reaches $V_0$ one $\tau$ later.

The real curve keeps slowing, because the gap still to fill shrinks by a factor $e$ in each time constant, here every second: 12, 4.4, 1.6, 0.60, 0.22 and 0.08 V at $t$ = 0, 1, 2, 3, 4 and 5 s. After about $5\tau$ the capacitor is full for practical purposes.

The two graphs are tied together by $I=Cd{V}_C/dt$: the lower curve is the slope of the upper one, multiplied by $C$. Where the voltage climbs steeply, at the start, the current is large; where it levels off, the current dies away.

Finally, the shaded area under $I(t)$. Current is charge per second, so the area under the current curve is the total charge delivered:
$Q={\int }_0^{\infty }{I}_0{e}^{-t/\tau }dt=I_0\tau =1.2\mathrm{mA}\cdot 1\mathrm{s}=1.2\mathrm{mC}$
This is the animation's final $Q$ readout, and it equals $C{V}_0$, as it must. It is also the area of the dashed rectangle of height $I_0$ and width $\tau$: the long exponential tail holds exactly the charge that the rectangle misses.

The stored energy, $W=\frac{1}{2}C{V}_C^2$, lags behind the voltage because it goes with the square of $V_C$. At $t=\tau$ the voltage is at 63 %, but the energy readout shows 2.88 mJ, only 40 % (that is, $0.63^2$) of the final 7.2 mJ.

Where the energy goes

Charging a capacitor from a battery always wastes energy, and always the same fraction of it. To see why, compare what the capacitor stores with what the battery supplies.

First, the stored energy. Moving a small charge $dq$ from the negative plate to the positive one, against the voltage $q/C$ already there, takes the work $dW=(q/C)dq$. Adding these up from empty to the final charge $Q$ gives
$W=\frac{Q^2}{2C}=\frac{1}{2}C{V}^2$
The ½ is there because the early charge crosses a small voltage, and only the last charge crosses the full $V$.

The battery, in contrast, pushes all of $Q=C{V}_0$ through its full voltage $V_0$ from the very start, so it supplies $Q{V}_0=C{V}_0^2$. In the animation that is 14.4 mJ, twice the 7.2 mJ stored. The difference becomes heat in the resistor, as integrating its power $I^{2}R$ confirms:
${\int }_0^{\infty }{I}^{2}Rdt=\frac{V_0^2}{R}{\int }_0^{\infty }{e}^{-2t/RC}dt=\frac{1}{2}C{V}_0^2$

Exactly half is lost, 7.2 mJ, whatever the value of $R$. A smaller resistance gives a larger current for a shorter time, and the two effects cancel. Most of the heat is made early, while the current is large: by $t=\tau$ the resistor has already turned 6.2 mJ into heat. The worked problem below shows how to lose less.

Energy density of the field

Where in the capacitor is the stored energy kept? The parallel-plate formulas suggest an answer. Put $C={\epsilon }_0{\epsilon }_{r}A/d$ and $V=Ed$ into $\frac{1}{2}C{V}^2$:
$\frac{1}{2}C{V}^2=\frac{1}{2}{\epsilon }_0{\epsilon }_r{E}^2\cdot Ad$
The product $Ad$ is the volume of the gap, exactly the region the field fills. So the energy can be read as spread through the field, with an energy density, or energy per cubic metre, of
$u=\frac{1}{2}{\epsilon }_0{\epsilon }_r{E}^2$
This holds for any electric field, not only between plates, and taking it seriously leads to a surprise.

How the energy enters the gap

If the energy sits in the field between the plates, it must travel there while the capacitor charges. The obvious guess, that it flows along the wires, turns out to be wrong.

Legend

$I$	Current in the leads, flowing down onto the upper plate.	$\mathrm{A}$
$\vec{E}$	Electric field in the gap, pointing down and growing as the plates charge.	$N/C$
$\vec{B}$	Magnetic field made by the growing $\vec{E}$, circling the axis clockwise seen from above.	$\mathrm{T}$
$\vec{S}$	Energy flux (Poynting vector), $\vec{E}\times \vec{B}/{\mu }_0$: energy per second per square metre, inward all round the rim.	$W/{\mathrm{m}}^2$

Figure 4 shows an air capacitor with circular plates while it charges. The current $I$ in the leads flows down onto the upper plate, so the plates gain charge and the field $\vec{E}$ in the gap, pointing down, grows.

A changing electric field makes a magnetic field, just as a current does. Maxwell called the term ${\epsilon }_{0}d{\Phi }_E/dt$ the displacement current, where ${\Phi }_E$ is the electric flux through the gap. Between the plates it is exactly equal to the current $I$ in the wires, since ${\epsilon }_{0}AdE/dt=dQ/dt$. So a magnetic field $\vec{B}$ circles the axis inside the gap, in the same sense as it circles the leads: clockwise seen from above, by the right-hand rule for a downward current.

Now look at the rim of the gap. There $\vec{E}$ points down and $\vec{B}$ runs along the circle. The energy flux, or Poynting vector, $\vec{S}=\vec{E}\times \vec{B}/{\mu }_0$, is perpendicular to both, and the right-hand rule makes it point radially inward at every point of the rim. The field energy flows in from the surrounding space, through the open sides of the gap, not along the wires. Adding up $\vec{S}$ over the whole rim gives exactly the rate at which the energy $\frac{1}{2}{\epsilon }_0{E}^2$ per cubic metre grows inside.

Forces from energy

The energy also gives the force between the plates, without needing to know how the charge is spread over them. The method, called virtual work, imagines a tiny movement and asks how the energy changes.

Isolate the charged capacitor, so that $Q$ cannot change, and pull one plate away by a small distance $dz$. The work you do, $Fdz$, must equal the rise in stored energy $Q^2/(2C)$. For a gap $z$, $1/C=z/({\epsilon }_{0}A)$, so the stored energy is $Q^{2}z/(2{\epsilon }_{0}A)$, which grows in proportion to $z$. Its rate of change with $z$ is the force:
$F=\frac{Q^2}{2{\epsilon }_{0}A}=\frac{1}{2}QE$
Why ½ and not simply $QE$? Each plate makes half the field in the gap, and a plate is pushed only by the other plate's half. The method works for any shape once $C$ is known as a function of position.

A number shows how weak this force is. Built as an air capacitor with a 1 mm gap, the animation's 100 µF would need a plate area $A=Cd/{\epsilon }_0\approx 1.1\times 10^4{\mathrm{m}}^2$, about one and a half football pitches. At 12 V the plates would attract with $F=W/d=7.2\mathrm{N}$, a pressure of only $6.4\times 10^{-4}\mathrm{Pa}$. That pressure equals the energy density $u$ of the field: force per square metre and energy per cubic metre have the same units, and here the same value.

Dielectrics

An insulator in the gap raises the capacitance by its relative permittivity ${\epsilon }_r$. The cause is polarisation. In the field, the insulator's molecules stretch or turn so that their positive and negative charges separate slightly. Inside the material these shifts cancel out, but they leave bound charges on its two faces: negative next to the positive plate, positive next to the negative one. The bound charges make a field opposite to that of the free charge on the plates, and cancel part of it.

At fixed charge $Q$, the field in the gap, and with it the voltage, falls by the factor ${\epsilon }_r$, so $C=Q/V$ rises by ${\epsilon }_r$. At fixed voltage, ${\epsilon }_r$ times as much charge flows onto the plates. Either way, $C={\epsilon }_0{\epsilon }_{r}A/d$.

Worked problem: charging in two steps

Losing half the energy is the price of charging from a fixed voltage; it is not a law of nature. Take the animation's capacitor and resistor, $C=100\mu \mathrm{F}$ and $R=10\mathrm{k\Omega }$, so $\tau =1\mathrm{s}$, and charge to 12 V in two stages: first from a 6 V supply until full, then from the 12 V battery. How much heat is produced in all?

Stage one is the animation at half the voltage. The supply moves $C\cdot 6\mathrm{V}=0.6\mathrm{mC}$ through 6 V and so supplies 3.6 mJ. Of this, $\frac{1}{2}\cdot 10^{-4}\cdot 6^2=1.8\mathrm{mJ}$ is stored and the other 1.8 mJ becomes heat: half, as before.

Stage two starts at $V_C=6\mathrm{V}$, so only the difference, 6 V, drives current through the resistor. With $x_0=6$ and $x_{\infty }=12$ in the relaxation law, in volts and milliamperes,
$V_C=12-6e^{-t/\tau },I=0.6e^{-t/\tau }$
The current starts at only 0.6 mA, and the heat is
${\int }_0^{\infty }{I}^{2}Rdt=\frac{(6\mathrm{V}{)}^2}{R}\cdot \frac{\tau }{2}=\frac{36}{10^4}\cdot 0.5\mathrm{J}=1.8\mathrm{mJ}$

The energy bookkeeping agrees. In this stage the battery moves another 0.6 mC through 12 V and supplies 7.2 mJ, while the stored energy rises from 1.8 to 7.2 mJ, by 5.4 mJ. The difference is again 1.8 mJ of heat.

In total 3.6 mJ is lost instead of 7.2 mJ. The general rule is that each step of size $\Delta V$ turns $\frac{1}{2}C(\Delta V{)}^2$ into heat, whatever voltage it starts from, so $N$ equal steps lose $\frac{1}{2}C{V}_0^2/N$: only 0.6 mJ for twelve steps of 1 V. A supply ramped up slowly compared with $\tau$ loses almost nothing, which is the idea behind adiabatic charging in low-power electronics. The price is time, since each step needs about $5\tau$ to finish.

Limits of the simple model

The formula $C={\epsilon }_0{\epsilon }_{r}A/d$ assumes a uniform field that stops sharply at the edges of the plates. In reality the field bulges outward at the edges, and this fringing field adds capacitance: about 4 % for discs 10 cm across and 1 mm apart, and 0.6 % at 0.1 mm.

Real insulators also leak a little current, and every one breaks down, suddenly conducting with a spark, above a field called its dielectric strength. For dry air that is about 3 MV/m, which caps the energy density of an air gap at $\frac{1}{2}{\epsilon }_0{E}^2\approx 40J/{\mathrm{m}}^3$. Polymer films withstand hundreds of megavolts per metre and store a few MJ/m³, still hundreds of times less than a lithium-ion battery. Capacitors win on speed, not on storage.

The circuit model also assumes quasi-static change: slow enough that the field everywhere keeps pace with the charge. Two effects break it at high frequency. First, real parts have inductance: a small ceramic capacitor has about a nanohenry in its leads and plates, so a 100 nF part resonates near 16 MHz and behaves like a coil above that.

Second, even ideal plates fail, because the changing magnetic field of Figure 4 induces an electric field of its own, which alters $\vec{E}$ across the gap. At 1 GHz the field at the rim of plates of 1 cm radius is about 1 % weaker than at the centre, and near 11 GHz such plates become a resonant cavity.

Last, charge is not infinitely divisible. One extra electron raises the voltage of a 1 fF capacitor by only 0.16 mV. But the energy needed to add one electron grows as the capacitance shrinks, and below a few attofarads it exceeds the thermal energy at room temperature, about 0.026 eV. Electrons can then enter only one at a time, an effect called the Coulomb blockade, which single-electron transistors exploit.

Where this leads

Electric Current and Resistance explains the resistor that sets $\tau$, and DC Circuits and Ohm's Law the loop rule behind the charging law. In Electric Power and Energy at Home, capacitors on an alternating voltage store and return energy every cycle. Electromagnetic Induction adds the inductor: together with a capacitor it obeys $L\ddot{q}+q/C=0$, the equation of Simple Harmonic Motion with $\omega =1/\sqrt{LC}$, and resistance turns it into a case of Damping and Resonance. The displacement current in the gap is the term that completes the field equations and predicts Electromagnetic Waves, and the exponential with its time constant returns in Radioactivity and Half-Life.