Capacitors & Stored Energy

Set a capacitor, a supply and a resistor, choose how many steps the supply climbs in, and step through the charge to see where the supply's energy goes: into the capacitor, or into heat. Further down, a pair of plates pulled apart with the battery disconnected and with it left on.

Worked example — Capacitors & Stored Energy

Charge on the capacitor	1.200 mC
Energy stored	7.200 mJ
Energy drawn from the supply	14.40 mJ
Heat in the resistor	7.200 mJ
Share of the supply's energy kept	50.00%
Current at the instant of each jump	12.00 mA
Capacitance at the start	354.2 pF
Energy stored at the start	1.771 µJ

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Capacitors & Stored Energy

A capacitor is two conductors separated by an insulator. Put +Q on one and −Q on the other and a voltage V appears between them; the ratio C = Q/V is the capacitance, in farads. Two flat plates of area A a distance d apart have C = ε₀εᵣA/d. A charged capacitor holds the energy U = ½CV² = Q²/2C. A supply at a fixed V pays QV to move the charge Q, which is twice what ends up stored.

What each symbol means

C: capacitance, in farads: the charge per volt. A farad is enormous; the capacitors on this page are microfarads and picofarads
Q: the charge on each plate, in coulombs, equal and opposite on the two
V: the voltage between the plates. For a supply, the voltage it holds whatever current flows
U: the energy stored in the field between the plates, in joules
εᵣ: relative permittivity of what fills the gap: 1 for vacuum, very close to 1 for air, and the factor by which an insulator raises the capacitance
d: the gap between the plates. Capacitance goes as 1/d, so halving the gap doubles it
N: the number of equal steps the supply climbs in. Each jump is V/N

Assumes: Ideal parts: a capacitor with no leakage and no resistance of its own, a supply that holds its voltage exactly, and a resistor that obeys Ohm's law. Each step of the staircase is left to settle for fifteen time constants before the next, which leaves 3 parts in 10 million of it undone. The plates are ideal too: the field between them is taken as uniform and the field that bulges out past the edges is ignored, which is fair while the gap is small compared with the width of the plates. The plate is moved slowly, so the current that flows to or from the battery never heats anything.

Ten steps waste a tenth as much 🖖

Press Ten steps. The capacitor finishes where it did after one jump, at 12 V holding 7.200 mJ, but the resistor has made 720.0 µJ of heat instead of 7.200 mJ. Each jump costs ½C(ΔV)², and the square is what pays: ten jumps of 1.2 V cost ten times ½C(1.2 V)², which is a tenth of ½C(12 V)². The share of the supply's energy that is kept climbs from 1/2 to N/(N+1), so 90.91% at ten steps and 99.01% at a hundred, and a supply that rises smoothly and slowly enough loses almost nothing. Circuit designers call this stepwise charging, one of the adiabatic techniques used to drive capacitive loads such as the gates of large power transistors, clock lines and touch sensors.

The resistor decides how fast, never how much 🖖

Press Through 10 Ω, then Through 1 MΩ. At the instant of the jump the first draws 1.200 A and the second 12.00 µA, a factor of 100,000, and the heating power at that instant differs by the same factor. The heat row reads 7.200 mJ both times. A small resistor spends the energy in a short spike and a large one in a long trickle, and the area under I²R comes out the same. The reason is in the charge: every coulomb leaves the supply at the full 12 V and lands on a capacitor that, averaged over the charge, sat at 6 V. The difference has to be dissipated in whatever stands between them, and the size of the resistor never enters that sum.

Pull the plates apart and the stored energy can fall 🖖

Press Pull to three times the gap. On the left the battery was disconnected first, so the charge is trapped. The field between the plates depends only on that charge and stays at 100 kV/m, so the voltage, field times gap, triples to 300.0 V. The stored energy triples too, and your pull paid for the rise: 3.542 µJ. On the right the battery holds 100 V. The capacitance falls to a third, two-thirds of the charge flows back into the battery, and the stored energy falls to 590.3 nJ. The plates still attract, so you still do work, 1.181 µJ, exactly what the capacitor lost. The battery is recharged by both: 2.361 µJ, twice your work, and that factor of two holds for any area, any voltage and any pair of gaps.

Problem solved in full

Where to put the tap 6 steps
A 100 µF capacitor has to be charged to 12 V through 1 kΩ, and the power supply has one spare output you can switch to first. How much heat does a single jump to 12 V make? How much if the spare output is 4 V, and how much at 6 V? Where should a single intermediate output sit, and how many equal steps would keep the heat to 1% of the stored energy?

Open this problem in the tool
1. One jump first. The supply moves 1.2 mC through 12 V and pays 14.40 mJ; the capacitor keeps ½CV² = 7.200 mJ, so the resistor makes 7.200 mJ of heat. This is the tool with N = 1.
  
  $Q = C V = 1.2 mC, Q V = 14.40 mJ, \frac{1}{2} C V^{2} = 7.200 mJ$
2. A jump of ΔV onto a capacitor costs ½C(ΔV)² in heat, whatever the resistor and whatever voltage the capacitor started from. So a tap at x volts costs two jumps: ½C[x² + (12 − x)²].
  
  $E_{heat} = \frac{1}{2} C [x^{2} + (12 - x)^{2}]$
3. At 4 V: ½ × 100 µF × (16 + 64) V² = 4.000 mJ. A tap at 8 V gives the same, since the two jumps just swap order.
  
  $x = 4 V : \frac{1}{2} \cdot 100 μ F \cdot (16 + 64) V^{2} = 4.000 mJ$
4. At 6 V: ½ × 100 µF × (36 + 36) V² = 3.600 mJ, half the single jump. Set N = 2 on the tool and the heat row agrees.
  
  $x = 6 V : \frac{1}{2} \cdot 100 μ F \cdot (36 + 36) V^{2} = 3.600 mJ$
5. Where is the best tap? x² + (12 − x)² is smallest where its slope 2x − 2(12 − x) is zero, at x = 6 V, the midpoint. The same argument for any number of jumps: with their sum fixed at V, a sum of squares is smallest when all the jumps are equal, and then it is N(V/N)² = V²/N.
  
  $\frac{d}{d x} [x^{2} + (12 - x)^{2}] = 2 x - 2 (12 - x) = 0 \Rightarrow x = 6 V$
6. So equal steps are the best that steps can do, which is why the tool only offers equal ones, and the heat is the stored energy divided by N. For 1% of 7.200 mJ, N has to reach 100.
  
  $N (\frac{V}{N})^{2} = \frac{V ^{2}}{N}, \frac{7 . 200 mJ}{N} \leq 0.072 mJ \Rightarrow N \geq 100$
Answer

7.200 mJ in one jump, 4.000 mJ via 4 V, 3.600 mJ via 6 V. The middle is the best place for a single tap, and for several taps the best spacing is equal, which leaves heat = ½CV²/N. Keeping it to 1% of the stored energy takes 100 steps, and the step table's last row shows the 72.00 µJ that leaves. The tool stops at equal steps; an unequal set, like the 4 V tap, is always worse, and the sum of squares in step 2 says by how much.

$4.000 mJ (4 V) > 3.600 mJ (6 V), N = 100 : 72.00 μ J$

Learning path

What is not a resistor

References (2)

The parallel-plate capacitance and the three forms of the stored energy in the definition and the results table: S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2. OpenStax. §8.1 "Capacitors and Capacitance" (Equation 8.3) and §8.3 "Energy Stored in a Capacitor" (Equation 8.10).
Stepwise charging as a real circuit technique, for the last sentence of the first insight: E. J. Carlson and J. R. Smith, "An Analytical Model for Stepwise Adiabatic Driver Energy Consumption." arXiv:2210.16680, 29 October 2022. §1.

By the 42Math team · Reviewed against published sources. Corrections welcome.