Back to the on-screen lesson ·
The work of charging, $U = \tfrac{1}{2}CV^2 = Q^2/2C$, energy density $\tfrac{1}{2}\varepsilon_0E^2$, energy lost in charge sharing, and forces from energy.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the energy stored in capacitors and fields, and follow it when capacitors change or share charge.
From the last lesson you can find capacitances and charges. From the potential lesson you know that moving a charge $q$ through a potential difference $V$ takes work $qV$. This lesson adds up that work as a capacitor charges, and then asks where the energy is stored: the answer, in the field itself, is one of the deepest ideas in electromagnetism.
| Term | What it means |
|---|---|
| Stored energy | $U = \tfrac{1}{2}CV^2 = Q^2/2C = \tfrac{1}{2}QV$, the work done charging a capacitor. |
| Energy density | $u = \tfrac{1}{2}\varepsilon_0E^2$, energy per unit volume of an electric field, in J/m³. |
| Self-energy | The energy needed to assemble a charge distribution from far-apart pieces. |
| Charge sharing | Connecting a charged capacitor to another, which conserves charge but not energy. |
| Electrostatic pressure | The inward pull on a charged surface, $\tfrac{1}{2}\varepsilon_0E^2$ per unit area. |
| Defibrillator | A device that stores energy in a capacitor and releases it as a shock to the heart. |
| Joule | The SI unit of energy; one farad at one volt stores half a joule. |
Charging a capacitor means moving charge from one plate to the other against the voltage already there. When the charge is $q$, the voltage is $q/C$, and moving a little more charge $dq$ takes work $(q/C)\,dq$. Adding up from empty to full,
$$U = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C} = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV.$$
The factor of one half appears because the average voltage during charging is half the final one. For parallel plates, $\tfrac{1}{2}CV^2 = \tfrac{1}{2}(\varepsilon_0A/d)(Ed)^2 = \tfrac{1}{2}\varepsilon_0E^2 \cdot Ad$, and $Ad$ is the volume between the plates. So the energy is stored in the field, with energy density
$$u = \tfrac{1}{2}\varepsilon_0E^2.$$
This result holds for any field, not just a capacitor's.
Another way: picture
Picture filling a water tower with a pump. The first bucket goes up easily; each later bucket must be lifted higher, to the level of the water already there. The total work is the full height times the water, halved, since on average the water was lifted half way. A capacitor fills the same way, with voltage playing the role of height.
Another way: steps
Since $Q = CV$, the energy can be written three ways: $\tfrac{1}{2}CV^2$, $Q^2/2C$, and $\tfrac{1}{2}QV$. They are always equal, but the right one to use depends on what is held fixed. Connected to a battery, $V$ is fixed, and $\tfrac{1}{2}CV^2$ shows that raising $C$ raises the energy. Isolated, $Q$ is fixed, and $Q^2/2C$ shows that raising $C$ lowers it.
Both are right. With the battery connected, raising $C$ draws more charge, and the battery does work $V\Delta Q$, twice the increase in stored energy; the other half goes into whatever moved the plates or the dielectric. Isolated, lowering the energy means the capacitor does work on its surroundings, pulling the dielectric in, for example.
The capacitor result, $u = \tfrac{1}{2}\varepsilon_0E^2$, turns out to hold everywhere. For a charged sphere, integrating $\tfrac{1}{2}\varepsilon_0(kQ/r^2)^2$ over all space outside it, in shells of volume $4\pi r^2dr$, gives $kQ^2/2R$, exactly $Q^2/2C$ with $C = 4\pi\varepsilon_0R$. The two views, energy of charges and energy of fields, always agree for static charges.
For changing fields, only the field view survives. Light carries energy across space with no charges in it, and that energy is the $\tfrac{1}{2}\varepsilon_0E^2$ of its electric field, plus a matching magnetic term. The energy density is where electromagnetism stores energy, and the waves lesson will use it directly.
For a field that varies, divide space into small volumes where $E$ is nearly constant, and add $u\,dV$. With spherical symmetry, use shells $dV = 4\pi r^2dr$; with cylindrical symmetry, cylindrical shells $dV = 2\pi r\ell\,dr$.
For a coaxial cable with $E = \lambda/2\pi\varepsilon_0r$ between radii $a$ and $b$, the energy per length is $\int_a^b \tfrac{1}{2}\varepsilon_0(\lambda/2\pi\varepsilon_0r)^2 2\pi r\,dr = (\lambda^2/4\pi\varepsilon_0)\ln(b/a)$. Comparing with $Q^2/2C$ recovers the capacitance $2\pi\varepsilon_0/\ln(b/a)$, a neat check that the two methods agree. Most of the energy sits near the inner wire, where the field is strongest.
Connect a charged capacitor to an identical empty one. Charge is conserved, so each ends with half, at half the voltage. The energy was $\tfrac{1}{2}CV^2$; afterward it is $2 \times \tfrac{1}{2}C(V/2)^2 = \tfrac{1}{4}CV^2$. Half the energy is gone.
It went into heating the wires and, as the charge sloshed back and forth, into radiation. Remarkably, the loss does not depend on the wires' resistance: a small resistance gives a large current for a short time, a large one a small current for a long time, and the heat is the same. Whenever charge flows between capacitors at different voltages, energy is lost.
The attraction between capacitor plates can be found from energy. For an isolated capacitor, pulling the plates apart by $dx$ raises the energy by $dU = (Q^2/2\varepsilon_0A)\,dx$, so the force is $F = Q^2/2\varepsilon_0A = \tfrac{1}{2}QE$. The one-half appears because each plate feels only the field of the other, $E/2$.
Per unit area, this force is $\tfrac{1}{2}\varepsilon_0E^2$, the energy density again, now read as a pressure. Fields pull on the surfaces they end on. The same method finds the force pulling a dielectric slab into a capacitor, which is how some microphones and actuators work.
Checking an answer. Units: F times V² is J. Energy must be positive. Sharing charge must lose energy, never gain. And the energy found by integrating the field must equal $Q^2/2C$ for the same arrangement.
The charging integral assumes the capacitor is charged slowly, so that at each moment the charges are in equilibrium and the voltage is $q/C$. The stored energy does not depend on how quickly it was charged, since it is a property of the final arrangement; only the heat lost in the process does.
The energy-density argument for parallel plates relies on the field being uniform between the plates and zero outside. That it gives the right answer for spheres and cables too is not automatic; it follows from the general theorem that the work to assemble any static charge distribution equals the integral of $\tfrac{1}{2}\varepsilon_0E^2$ over all space.
Compared with batteries, capacitors store little energy. A $1000$ μF capacitor at $25$ V holds about $0.3$ J; an AA battery about $10{,}000$ J. The limit is the energy density: air breaks down near $3$ MV/m, where $u$ is only about $40$ J/m³.
What capacitors offer is speed. They can release their energy in microseconds, delivering enormous power: camera flashes, defibrillators, pulsed lasers and the capacitor banks at the National Ignition Facility in California, which store $400$ MJ to fire the world's most energetic laser. Supercapacitors, with huge internal area, close part of the gap with batteries, powering regenerative braking in buses and trains. They charge in seconds and survive a million cycles, where a battery would wear out in a few thousand, so they suit jobs with many quick charges and discharges.
A dielectric changes the energy stored at a given voltage or charge, and the direction of the change depends on what is fixed. With a battery connected, filling the gap with a dielectric of constant $\kappa$ multiplies $C$ by $\kappa$, and the energy $\tfrac{1}{2}CV^2$ rises by the same factor. The battery supplies twice that increase as work; the extra half goes into pulling the slab in, which the fringing field at the edge does willingly.
With the capacitor isolated, the charge stays put, and the energy $Q^2/2C$ falls by the factor $\kappa$ as the slab slides in. The capacitor does work on the slab, drawing it into the gap, and a slab released at the edge will be pulled inside and, without friction, oscillate back and forth through the plates.
Inside a dielectric the energy density is $\tfrac{1}{2}\kappa\varepsilon_0E^2$, larger than in vacuum at the same field, because energy is also stored in stretching the molecules into dipoles. This is why capacitor makers value materials with both a high dielectric constant and a high breakdown field: the energy a capacitor can hold per unit volume grows with $\kappa$ times the square of the largest field the material can survive. Modern film capacitors for electric cars and wind turbines are judged by exactly this figure.
The AEDs mounted in American airports, schools and offices store energy in a capacitor of about $100$ to $200$ μF charged to $1500$ to $2000$ V, holding $150$ to $360$ J. When the device detects ventricular fibrillation, it discharges that energy through the chest in a few milliseconds, stopping the chaotic rhythm so the heart can restart normally.
A battery could not deliver that much power so quickly: the capacitor takes several seconds to charge from the battery, then releases the energy a thousand times faster. Survival from cardiac arrest falls about ten percent with each minute before a shock, which is why AEDs are placed where anyone can reach them.
At Lawrence Livermore National Laboratory in California, the National Ignition Facility focuses $192$ laser beams onto a pea-sized capsule of hydrogen fuel to make it fuse. The lasers are powered by a bank of capacitors that stores about $400$ MJ, charged over a minute and released in a few hundred microseconds.
In December 2022, NIF produced more fusion energy from the capsule than the laser energy delivered to it, a first. The capacitor bank, filling a building, makes the burst possible: energy trickles in slowly and leaves in an instant. Every pulsed-power machine, from camera flashes to particle accelerators, uses the same principle.
Because charge is conserved when a charged capacitor is connected to an empty one, it is tempting to think energy is conserved too. It is not: connecting identical capacitors loses exactly half the energy, whatever the resistance of the wires, as heat and radiation.
A second error is to drop the one-half and write $U = QV$. That is the work a battery does moving charge $Q$ across a fixed voltage $V$; a capacitor's voltage rises from zero as it charges, so its stored energy is half of that. The other half is lost in the charging circuit.
A $50$ μF capacitor is charged to $200$ V. Find its charge.
$Q = 50 \times 10^{-6} \times 200 = 0.010\ \text{C}$
$Q = CV$.
Find its stored energy.
$U = \tfrac{1}{2} \times 50 \times 10^{-6} \times 200^2 = 1.0\ \text{J}$
$\tfrac{1}{2}CV^2$.
Check with the charge form.
$U = \dfrac{0.010^2}{2 \times 50 \times 10^{-6}} = 1.0\ \text{J}$
$Q^2/2C$ agrees.
Find the work the battery did.
$W = QV = 0.010 \times 200 = 2.0\ \text{J}$
Twice the stored energy.
Account for the difference.
$2.0 - 1.0 = 1.0\ \text{J} \text{ as heat}$
Lost in the charging resistance.
An isolated capacitor with plates of $0.010$ m², $1.0$ mm apart, holds $20$ nC. Find its capacitance.
$C = \dfrac{8.85 \times 10^{-12} \times 0.010}{0.0010} = 88.5\ \text{pF}$
$\varepsilon_0A/d$.
Find its energy.
$U = \dfrac{(20 \times 10^{-9})^2}{2 \times 88.5 \times 10^{-12}} = 2.26\ \mu\text{J}$
$Q$ is fixed, so use $Q^2/2C$.
Pull the plates to $3.0$ mm. Find the new energy.
$U' = 3 \times 2.26 = 6.78\ \mu\text{J}$
$C$ falls to a third.
Find the work done pulling.
$W = 6.78 - 2.26 = 4.52\ \mu\text{J}$
Against the plates' attraction.
Find the force from the work.
$F = \dfrac{4.52 \times 10^{-6}}{0.0020} = 2.26\ \text{mN}$
Constant, since the field does not change.
Check with the formula.
$F = \dfrac{Q^2}{2\varepsilon_0A} = \dfrac{4.0 \times 10^{-16}}{1.77 \times 10^{-13}} = 2.26\ \text{mN}$
They agree.
A sphere of radius $0.10$ m carries $10$ nC. Write the energy density outside it.
$u = \tfrac{1}{2}\varepsilon_0\left(\dfrac{kQ}{r^2}\right)^2$
The field of a point charge outside.
Write the energy in a thin shell.
$dU = u \cdot 4\pi r^2\,dr = \dfrac{kQ^2}{2r^2}\,dr$
Using $\varepsilon_0 = 1/4\pi k$.
Integrate from the surface outward.
$U = \displaystyle\int_{0.10}^{\infty} \dfrac{kQ^2}{2r^2}\,dr = \dfrac{kQ^2}{2 \times 0.10}$
The integral of $1/r^2$.
Evaluate the energy.
$U = \dfrac{9.0 \times 10^9 \times (10^{-8})^2}{0.20} = 4.5\ \mu\text{J}$
All of it in the field outside.
Check with the capacitance.
$C = 4\pi\varepsilon_0(0.10) = 11.1\ \text{pF}, \quad \dfrac{Q^2}{2C} = 4.5\ \mu\text{J}$
The two views agree.
Find how much lies within $0.20$ m of the center.
$\dfrac{kQ^2}{2}\left(\dfrac{1}{0.10} - \dfrac{1}{0.20}\right) = 2.25\ \mu\text{J}$
Half the energy is within one radius of the surface.
Find the pressure on the surface.
$\tfrac{1}{2}\varepsilon_0E^2 = \tfrac{1}{2}(8.85 \times 10^{-12})(9000)^2 = 3.6 \times 10^{-4}\ \text{Pa}$
A tiny outward push on the charged surface.
Find the charge.
$Q = 10 \times 10^{-6} \times 100 = 1.0\ \text{mC}$
$Q = CV$.
Write the energy.
$U = \tfrac{1}{2} \times 10 \times 10^{-6} \times 100^2$
$\tfrac{1}{2}CV^2$.
Evaluate the energy.
A charged parallel-plate capacitor stores $5$ mJ. It is disconnected from its battery, and then its plates are pulled apart to twice their separation. How much energy does it store now?
Complete the worked solution: a $3$ μF capacitor charged to $30$ V is connected across an uncharged $1$ μF capacitor. Find the common voltage afterward in V, and the total stored energy before and after, in μJ.
Conserve the charge to find the common voltage.
$V' = \dfrac{C_1V}{C_1 + C_2} =$ w
The charge spreads over both capacitors.
Evaluate the energy before.
$U = \tfrac{1}{2}C_1V^2 =$ b
Only the first capacitor is charged.
Evaluate the energy after.
$U' = \tfrac{1}{2}(C_1 + C_2)V'^2 =$ e
Both at the common voltage.
Account for the difference.
$U' = \dfrac{C_1}{C_1 + C_2}U$
The rest is lost as heat and radiation while the charge flows.
Match each description to its energy expression.
| $\tfrac{1}{2}CV^2$ | $Q^2/2C$ | $\tfrac{1}{2}\varepsilon_0E^2$ | $kQ^2/2R$ | |
|---|---|---|---|---|
| known voltage | ||||
| known charge | ||||
| energy density | ||||
| charged sphere |
A $5$ μF capacitor is charged to $20$ V and disconnected. It is then connected across an identical, uncharged capacitor. Fill in its charge before in μC, its energy before in μJ, and the total energy of the pair afterward in μJ.
| value | |
|---|---|
| charge before (μC) | |
| energy before (μJ) | |
| energy after (μJ) |
A $1$ μF capacitor starts uncharged and is charged slowly. Write the work needed to bring its charge up to $q$ μC, in μJ, as a formula in $q$.
Answer:
The field between a capacitor's plates is $1.5$ MV/m, in air. With $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m, what is the energy stored per cubic meter of that field, in J/m³?
Answer: J/m³ in the field
An automated external defibrillator on the wall of a Chicago airport terminal charges a $150$ μF capacitor to $1500$ V before delivering a shock. How much energy does the capacitor store, in joules?
Answer: J stored for the shock
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $5$ μF capacitor starts uncharged and is charged slowly. Write the work needed to bring its charge up to $q$ μC, in μJ, as a formula in $q$.
Answer:
You can account for energy in electric fields. Explain to someone why connecting a charged capacitor to an empty one loses energy.
20. Your turn: a $10$ μF capacitor is charged to $100$ V. How much energy does it hold, and what charge?, step 3
$U = 0.050\ \text{J}$
Fifty millijoules.