University Physics II · Capacitance and Dielectrics · 5.3
Energy Stored in Capacitors
Every charge moved onto a plate raises the voltage the next one has to climb. Integrate that rising cost and the factor of one half appears — then find the same energy again, sitting in the field between the plates.
Build the model
Connect the measurement to the mechanism.
Moving charge onto a capacitor is not free, and the price rises as you go. With q already on the plates the potential difference is q/C, so the next dq costs (q/C) dq of work; integrating from empty to Q gives U = Q²/(2C). Because the voltage climbed linearly from zero, the average coulomb crossed half the final ΔV, and that is the entire origin of the one half in ½C(ΔV)².
Two consequences matter more than the formula. A source held at fixed ΔV pays the full QΔV, so exactly half of what a battery spends never reaches the capacitor, whatever the circuit resistance. And the stored energy can be relocated: rewrite ½C(ΔV)² for parallel plates and it becomes ½ε₀E² times the volume between them, which says the energy is in the field rather than on the plates.
Once you accept that, forces and energy changes follow from the field alone.
- Simple definition
- The energy stored in a capacitor is the work done against its own rising potential difference while the charge is put on: U = Q²/(2C) = ½QΔV = ½C(ΔV)², the three forms agreeing because Q = CΔV.
- Example
- Charge a 10 μF capacitor to 12 V and it holds Q = 120 μC and U = ½(1.0×10⁻⁵ F)(12 V)² = 0.72 mJ. The battery spent QΔV = 1.44 mJ putting it there; the other half never arrived.
The cost per coulomb rises linearly from zero, so the total is half the final ΔV times Q.
q/C is the ΔV already across the plates; U in J, Q in C, C in F.
Use Q²/(2C) when the charge is trapped and ½C(ΔV)² when a source holds the voltage.
Equal only because Q = CΔV; all three in joules.
Taking a 10 μF capacitor to 12 V costs the battery 1.44 mJ and stores 0.72 mJ.
Ideal source held at ΔV; the loss is independent of the series resistance.
E = 1.0×10⁶ V m⁻¹ gives only u = 4.43 J m⁻³, which is why capacitors are bulky.
J m⁻³, which is also N m⁻²; ε₀ = 8.854×10⁻¹² F m⁻¹.
One number, two pictures: energy on the plates, or energy spread through the gap.
Ad is the volume between the plates, in m³; fringing ignored.
An isolated sphere of radius R carrying Q holds U = Q²/(8πε₀R) in the field outside it.
Integrate wherever the field reaches, not only inside the gap.
The first coulomb is free, the last pays full price
Start with the capacitor empty. To move another dq from one plate to the other when charge q is already in place, you must carry it across the potential difference the plates already have, q/C, so the work is dW = (q/C) dq. Charging creates no charge; it moves charge from one plate to the other, and the device stays neutral overall. Integrate from empty to the final charge: U = ∫₀Q (q/C) dq = Q²/(2C). The factor of one half is the only thing here worth committing to memory, and it is not mysterious. Plot ΔV = q/C against q and you get a straight line through the origin, so the work — the area beneath it — is the area of a triangle, ½ × Q × (Q/C). The first coulomb crossed nothing; the last crossed the full final voltage. For a 10 μF capacitor taken to 12 V the final charge is Q = CΔV = 1.2×10⁻⁴ C, and U = (1.2×10⁻⁴ C)²/(2 × 1.0×10⁻⁵ F) = 7.2×10⁻⁴ J, or 0.72 mJ.
Three forms, and how to pick one
Substituting Q = CΔV turns one result into three: U = Q²/(2C) = ½QΔV = ½C(ΔV)². They are algebraically identical, so any of them returns the right number for a capacitor sitting still. They stop being interchangeable the moment something changes, because a real circuit holds one variable fixed and lets the other move. A capacitor still wired to a battery has ΔV pinned, so ½C(ΔV)² is the form whose supposedly fixed quantity really is fixed. A capacitor that has been disconnected has Q pinned instead, and Q²/(2C) is the honest form there. The middle form, ½QΔV, is the one that keeps the accounting visible, since QΔV is exactly what a constant-voltage source pays out. Note the powers while you are here: at fixed capacitance U goes as (ΔV)², so running a capacitor at twice the voltage stores four times the energy. That is why the voltage rating, rather than the capacitance, usually decides how much energy a bank can hold.
The source always spends twice what is stored
An ideal source held at ΔV does not care how much charge has already moved: every coulomb it pushes round the circuit is lifted through the full ΔV. Deliver Q in total and it has done W = QΔV = C(ΔV)², while the capacitor keeps ½C(ΔV)². Half of what the source spent is missing, and better wiring does not recover it. Put a resistance R in series and check. The current is I = (ΔV/R)e(−t/RC), so the resistor dissipates ∫₀^∞ I²R dt = (ΔV²/R) × (RC/2) = ½C(ΔV)². R has cancelled: halve it and the peak current doubles while the transient lasts half as long, leaving the total dissipation untouched. Drive R towards zero and the energy leaves as radiation or heats the source's own internal resistance instead. In the 10 μF example the 12 V battery spends QΔV = 1.44 mJ, the capacitor keeps 0.72 mJ, and 0.72 mJ is gone. The only escape is to stop dropping the charge through the whole ΔV at once — charge in voltage steps, or through an inductor.
Put the energy in the field, not on the plates
Where does the stored energy sit? Write ½C(ΔV)² for a parallel-plate capacitor with a vacuum gap, using C = ε₀A/d and ΔV = Ed for the uniform field between the plates: U = ½(ε₀A/d)(Ed)² = ½ε₀E² × (Ad). Both plate quantities have cancelled from the coefficient, leaving the field strength squared multiplied by Ad — precisely the volume the field occupies. So define an energy density u = ½ε₀E², in J m⁻³. Take A = 0.20 m², d = 0.50 mm and E = 1.0×10⁶ V m⁻¹, giving ΔV = Ed = 500 V and C = ε₀A/d = 3.54 nF. Then ½C(ΔV)² = 4.43×10⁻⁴ J, and u = ½(8.854×10⁻¹²)(1.0×10⁶)² = 4.43 J m⁻³ spread over 1.0×10⁻⁴ m³ gives the same 4.43×10⁻⁴ J. The local statement is the more general one, because it still applies where E varies from point to point: U = ∫ ½ε₀E² dV. Electrostatics alone cannot tell the two pictures apart by measurement, but the field version is the one that survives into radiation, where energy travels with no charge to sit on.
Fixed charge or fixed voltage — decide before you compute
Before choosing a formula, ask which quantity the circuit is holding. Take the 10 μF capacitor at 12 V — Q = 120 μC, U = 0.72 mJ — and double the plate separation, halving C to 5.0 μF. Disconnect it first and Q stays at 120 μC: ΔV climbs to Q/C = 24 V and U = Q²/(2C) = 1.44 mJ. The energy doubled, and it came from your hand, since the plates attract and pulling them apart takes work. Leave the battery connected and ΔV is pinned at 12 V instead: the charge falls to 60 μC and U = ½C(ΔV)² = 0.36 mJ, so the capacitor's energy halves. It has not vanished — 60 μC flowed back through the battery, returning ΔQ ΔV = 0.72 mJ to the source, and the balance Wₑₓₜ = ΔU − Wsource = (−0.36 mJ) − (−0.72 mJ) = +0.36 mJ says you still did positive work pulling. Same mechanical action, opposite sign for ΔU, because in the second case the source is inside the energy account.
The plates pull on each other, and the field says how hard
The field picture earns its keep when you want a force. At fixed charge the field between the plates is E = σ/ε₀ = Q/(ε₀A), which does not depend on the separation, so with a gap x the stored energy U = ½ε₀E²(Ax) is linear in x. The plates therefore attract with a force of magnitude dU/dx = ½ε₀E²A = Q²/(2ε₀A) = ½QE. That one half is physics, not carelessness: each plate sits in the field of the other plate alone, which is E/2, not the full gap field. Watch the units of the energy density too — J m⁻³ is N m⁻², so ½ε₀E² is at once an energy per unit volume and the electrostatic pressure pulling the plates together. With A = 0.20 m² and E = 1.0×10⁶ V m⁻¹ that pressure is 4.43 Pa and the attraction is 4.43 × 0.20 = 0.89 N: negligible on a bench, dominant across a micrometre gap, which is how condenser microphones and MEMS actuators work. It is also the force you worked against when the fixed-charge energy went up.
Change one variable at a time
Make the relationship visible.
Raise the capacitance and the line tilts flatter, so the same ΔV is reached with more charge and a bigger box — but the diagonal always cuts that box exactly in half, and that halving is the whole of the ½ in ½C(ΔV)².
CHARGE Q = CΔV120 μC
STORED U = ½C(ΔV)²720 μJ
SOURCE PAYS QΔV1440 μJ
LOST QΔV − U720 μJ
Live interpretationCHARGE Q = CΔV: 120 μC. STORED U = ½C(ΔV)²: 720 μJ. SOURCE PAYS QΔV: 1440 μJ. LOST QΔV − U: 720 μJ
Catch the common trap
Explain before calculating.
A 10 μF parallel-plate capacitor is charged to 12 V by a battery and then disconnected. Its plate separation is doubled, with vacuum between the plates throughout. What is the stored energy afterwards?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 47 μF capacitor is charged to 9.0 V by a battery. Find the charge it holds, the energy it stores, and the energy the battery spent putting it there.
- Charge first: Q = CΔV = 47 × 10⁻⁶ F × 9.0 V = 4.23 × 10⁻⁴ C.
- Energy stored: U = ½C(ΔV)² = ½ × 47 × 10⁻⁶ F × (9.0 V)² = 1.90 × 10⁻³ J.
- The battery sat at 9.0 V while every one of those coulombs crossed it, so it spent W = QΔV = 4.23 × 10⁻⁴ C × 9.0 V = 3.81 × 10⁻³ J.
- The difference, 3.81 − 1.90 = 1.90 mJ, went to the wiring and the battery's own resistance — and thicker wire would not have saved it.
AnswerQ = 4.2 × 10⁻⁴ C and U = 1.9 mJ; the battery spent 3.8 mJ, so exactly half never reached the capacitor.
MediumA camera flash dumps a 2.2 μF capacitor charged to 250 V through its lamp in 1.5 ms. Find the stored energy and the average power delivered — then say what both become if the capacitor is rated to 500 V and run there instead.
- U = ½C(ΔV)² = ½ × 2.2 × 10⁻⁶ F × (250 V)² = 6.88 × 10⁻² J.
- Average power is the store divided by the time it takes to leave: P = U/t = 6.875 × 10⁻² J ÷ 1.5 × 10⁻³ s = 45.8 W.
- At fixed C the store goes as (ΔV)², so doubling to 500 V quadruples it: U = 4 × 6.875 × 10⁻² J = 0.275 J.
- The same 1.5 ms delivery then averages P = 0.275 J ÷ 1.5 × 10⁻³ s = 183 W — also four times larger.
AnswerU = 69 mJ at an average 46 W; at 500 V it is 0.28 J at 1.8 × 10² W. The voltage rating, not the capacitance, decides how much energy the flash can hold.
HardA parallel-plate capacitor has plates of area 0.015 m² separated by 0.10 mm of vacuum, charged to 300 V. Find the energy density in the gap and the total field energy, then check that answer against ½C(ΔV)².
- The gap field is uniform, so E = ΔV/d = 300 V ÷ 1.0 × 10⁻⁴ m = 3.0 × 10⁶ V m⁻¹.
- u = ½ε₀E² = ½ × 8.854 × 10⁻¹² F m⁻¹ × (3.0 × 10⁶ V m⁻¹)² = 39.8 J m⁻³.
- The field occupies only the gap, whose volume is Ad = 0.015 m² × 1.0 × 10⁻⁴ m = 1.5 × 10⁻⁶ m³.
- U = u × Ad = 39.8 J m⁻³ × 1.5 × 10⁻⁶ m³ = 5.98 × 10⁻⁵ J.
- Now the plate picture: C = ε₀A/d = (8.854 × 10⁻¹² F m⁻¹ × 0.015 m²) ÷ (1.0 × 10⁻⁴ m) = 1.33 × 10⁻⁹ F, so ½C(ΔV)² = ½ × 1.33 × 10⁻⁹ F × (300 V)² = 5.98 × 10⁻⁵ J.
Answeru = 39.8 J m⁻³ and U = 5.98 × 10⁻⁵ J — about 60 μJ, and the two pictures agree because they are the same algebra rearranged.