In short

A charged capacitor stores energy. For capacitance C held at potential difference V with charge Q = CV on the positive plate,

\boxed{\;U \;=\; \tfrac{1}{2}CV^{2} \;=\; \tfrac{1}{2}QV \;=\; \frac{Q^{2}}{2C}\;}.

The three forms are the same quantity rewritten using Q = CV — reach for whichever form uses the variables you already know. The factor of \tfrac{1}{2} comes from the fact that the first coulomb of charge is pushed onto an empty plate (low voltage), while the last coulomb is pushed onto an almost-full plate (high voltage); the average voltage during charging is V/2, so the total work is \tfrac{1}{2}QV.

The energy does not sit "on the plates" — it sits in the electric field that fills the gap. The energy per unit volume of field (the energy density) is

\boxed{\;u \;=\; \tfrac{1}{2}\varepsilon_{0}E^{2}\;}.

Multiply u by the volume between the plates and you recover \tfrac{1}{2}CV^{2} exactly. This is the first time in your physics course that a field (not a particle, not a charge) carries energy — and that is the key idea that, half a century later, became the theory of electromagnetic waves.

When the photographer in a Delhi wedding studio presses the shutter, a tiny relay inside the camera dumps the charge on a capacitor through the xenon tube of the flash. The entire discharge is over in about one millisecond. For that millisecond, the tube glows with a power of maybe ten thousand watts — more than ten ordinary household bulbs combined. The capacitor had been quietly sipping power from a 3-volt AA cell for ten seconds beforehand, charging up through a step-up transformer to about 300 volts. Ten seconds of slow charging, one millisecond of blinding discharge. That is the job of the capacitor: it is a bucket for electrical energy.

The same pattern — store slowly, release fast — runs a crash cart defibrillator in the cardiology ward of AIIMS Delhi. A bank of capacitors is charged from a mains supply to about 1600 volts. When the doctor shouts "clear", a relay connects the bank across the patient's chest and delivers 200 joules through the torso in roughly 4 milliseconds. The average power during the pulse is about 50 kilowatts — which no battery could deliver, but which a capacitor bank can. It is not that the capacitor makes energy; it collects it over time and lets it out all at once. The formula U = \tfrac{1}{2}CV^{2} is the bookkeeping.

ISRO uses the same trick on an industrial scale. When the lab at Vikram Sarabhai Space Centre fires a pulsed plasma thruster for a satellite-propulsion experiment, a 1-millifarad capacitor bank charged to 2 kV dumps \tfrac{1}{2}(10^{-3})(2000)^{2} = 2000 joules into a gas jet in less than 10 microseconds. Peak power: 200 megawatts. From a 10-watt solar panel. The capacitor is the transformer between slow and fast.

The puzzle: why \tfrac{1}{2}?

Look again at the top formula, U = \tfrac{1}{2}QV. Where does the \tfrac{1}{2} come from?

Here is a plausible-sounding argument that is wrong. "The capacitor has charge Q sitting at potential V, so its potential energy is charge times potential, U = QV." The answer is off by a factor of two.

The error in the plausible argument is treating the charging as if you parachuted a whole coulomb of charge from infinity onto a plate already at potential V. That is not what happens. The charge arrives one electron at a time (or more precisely, one infinitesimal chunk dq at a time), and the voltage across the plates rises as charge accumulates. The first infinitesimal chunk arrives when the voltage is zero — it does zero work against the (nonexistent) field. The last infinitesimal chunk arrives when the voltage is the full V — it does maximum work. On average, each bit of charge does work against a voltage of V/2, and the total work is \tfrac{1}{2}QV.

That picture — the voltage ramping from zero to V as you charge — is the entire physics of this chapter. Once you see it, the formula writes itself.

Voltage ramps linearly from zero to V as charge accumulates from zero to QTriangular area under a straight line from origin to point Q,V represents the work done. The triangle has base Q and height V; its area is one-half Q V, which is the energy stored.charge on plate q → Qvoltage across platesV0Qarea = ½ Q V = Uv(q) = q / C(Q, V)
The voltage rises linearly from zero to $V$ as charge accumulates from zero to $Q$. The work to push the next $dq$ onto the plate is $v\,dq$ — a thin vertical strip. Summing the strips gives the triangular area $\tfrac{1}{2}QV$, which is the stored energy.

The triangle under the line is the visual proof of the \tfrac{1}{2}. We will now make this rigorous.

Deriving U = \tfrac{1}{2}CV^{2}

Start the capacitor uncharged — both plates at zero potential, no charge anywhere. Connect a battery of EMF V through a wire and a resistor (the resistor is there so the current is finite — an ideal short would dump all the charge instantly, which is a separate calculation). Wait until charging stops.

At some moment during the charging, let the charge on the positive plate be q. The potential difference across the plates at that moment is

v \;=\; \frac{q}{C}.

Why: this is the defining equation of capacitance, C = Q/V, rearranged. It holds at every moment during charging, not only at the end — because capacitance is a geometric property that does not depend on how much charge is currently on the plates.

Now imagine pushing an additional infinitesimal charge dq from the negative plate, through the external wire, up to the positive plate. (In reality it is electrons moving the other way, but the energy bookkeeping is identical.) To move this dq against the existing potential difference v, you must do work

dW \;=\; v \, dq \;=\; \frac{q}{C}\,dq.

Why: the work to move a charge dq through a potential difference v is dW = v\,dq, straight from the definition of potential (work per unit charge). The source of this work is the battery — the battery pushes the charge uphill against the growing field.

Integrate from q = 0 (uncharged) to q = Q (fully charged):

U \;=\; \int_{0}^{Q} \frac{q}{C}\,dq \;=\; \frac{1}{C}\,\int_{0}^{Q} q\,dq \;=\; \frac{1}{C}\cdot\frac{Q^{2}}{2}.

Why: C is a constant (set by geometry), so it comes outside the integral. The integrand q\,dq integrates to q^{2}/2.

So

\boxed{\;U \;=\; \frac{Q^{2}}{2C}\;}. \tag{1}

Two algebraic rewrites give the other standard forms. Using Q = CV:

U \;=\; \frac{(CV)^{2}}{2C} \;=\; \tfrac{1}{2}CV^{2}, \tag{2}
U \;=\; \frac{Q^{2}}{2C} \;=\; \frac{Q \cdot Q}{2C} \;=\; \frac{Q \cdot (CV)}{2C} \;=\; \tfrac{1}{2}QV. \tag{3}

Why: these are three faces of one quantity. If you know Q and C, use (1). If you know C and V, use (2). If you know Q and V, use (3). Pick the form that uses the variables you already have and you avoid one substitution.

Sanity check — the units

Each of the three forms should reduce to joules.

All three are joules, as they must be.

Where did the factor of \tfrac{1}{2} come from, physically?

The battery did work W_\text{battery} = QV — it pushed total charge Q through the full potential difference V (from its own terminals' perspective, the voltage across it was always V). But only \tfrac{1}{2}QV is stored in the capacitor. Where did the other \tfrac{1}{2}QV go?

It was dissipated as heat in the charging resistor. This is a real result — no matter how small the resistance, exactly half the battery's work is dissipated, with the other half stored in the capacitor. Make the resistor smaller and the current gets larger and the discharge finishes faster, but the total heat dissipated stays at \tfrac{1}{2}QV. This is one of the deepest results in circuit theory, and it is a preview of the radiation-resistance physics that shows up in antenna theory — an ideal circuit cannot charge a capacitor with 100% efficiency from a DC voltage source. You need to charge through an inductor (a "ringing" LC circuit) or use a resonant converter if you want better efficiency.

For this article, the takeaway is simpler: the \tfrac{1}{2} in U = \tfrac{1}{2}CV^{2} is telling you that only half of what the battery supplies is actually stored — the other half always leaks away during charging.

Where the energy lives — energy density of the field

Up to this point the formulas tell you how much energy is stored, but not where it sits. The electrons are on the plates, but the plates themselves are just metal; nothing interesting about a piece of metal makes it "contain" joules. So where is the energy?

The answer is surprising, beautiful, and correct: the energy sits in the electric field between the plates.

Here is the argument. Take a parallel-plate capacitor with plate area A, gap d, and potential difference V. From the parallel-plate chapter, you already know

C \;=\; \frac{\varepsilon_{0}A}{d}, \qquad V \;=\; E\,d.

Why: the first is the parallel-plate capacitance you derived from Gauss's law. The second is the field-potential relation in a uniform field: the voltage difference across a region of uniform field E is E times the distance d across the region.

Substitute into (2):

U \;=\; \tfrac{1}{2}CV^{2} \;=\; \tfrac{1}{2}\cdot\frac{\varepsilon_{0}A}{d}\cdot (Ed)^{2} \;=\; \tfrac{1}{2}\,\varepsilon_{0}\,A\,d\,E^{2}.

Now notice that A \cdot d is the volume between the plates — the volume of the region where the field exists. Call this volume \mathcal{V}. Then

U \;=\; \tfrac{1}{2}\varepsilon_{0}E^{2}\cdot \mathcal{V}.

Dividing both sides by \mathcal{V} gives the energy per unit volume of the field:

\boxed{\;u \;=\; \frac{U}{\mathcal{V}} \;=\; \tfrac{1}{2}\varepsilon_{0}E^{2}\;}. \tag{4}

Why: the total energy is proportional to the volume of field, which means the energy must be thought of as spread through the field region at a density \tfrac{1}{2}\varepsilon_{0}E^{2} at every point. The plates do not "contain" the energy; the field does.

This is one of the most important formulas in all of electromagnetism. It says that an electric field — not a charged particle, not a capacitor, but the field itself, as a quantity defined at every point in empty space — carries energy, at a density of \tfrac{1}{2}\varepsilon_{0}E^{2} joules per cubic metre at every point.

Why does this matter for the wiki reader today?

Formula (4) is derived for a parallel-plate geometry, but it is general: the same energy density applies in any electrostatic field, because you can always break a non-uniform field into infinitesimal parallel-plate-like pieces and add up their contributions.

Explore the formula

The interactive figure below fixes the capacitance at C = 1000\ \muF (a typical electrolytic capacitor) and lets you drag the voltage from 0 to 500 V. The stored energy U = \tfrac{1}{2}CV^{2} grows as a parabola — doubling the voltage quadruples the energy, which is why high-voltage capacitor banks are so dangerous.

Interactive: energy stored grows as V squaredParabolic curve U = 0.5 times 1000 microfarad times V squared showing energy in joules plotted against voltage in volts. A draggable red dot on the voltage axis lets you choose a voltage; a second dot tracks the matching energy on the parabola. A readout panel shows the numerical values.voltage V (volts)energy U (joules)04080120100200300400500U = ½ C V²C = 1000 μF (fixed)drag the red dot along the V axis
Fix $C = 1000\ \mu$F. Drag the red point to change $V$ between 0 and 500 volts. Watch $U$ grow as $V^{2}$ — at 200 V you hold 20 J, at 400 V you hold 80 J (four times as much, not two times).

Three things to notice as you drag:

Worked examples

Example 1: A camera flash capacitor

A compact camera's electronic flash uses a C = 100\ \muF capacitor charged to V = 300 V. When the shutter fires, the capacitor discharges through the xenon tube in about 1 millisecond. Find the energy stored, the charge on the plates, and the average power delivered during the flash.

Schematic of a camera flash capacitor circuitA battery charges a 100 microfarad capacitor through a step-up transformer. A switch labelled shutter connects the capacitor across a xenon flash tube when triggered. Labels show 300 volts across the capacitor and 1 millisecond discharge time.3 Vbatterystep-upC = 100 μFV = 300 VshutterxenonΔt = 1 ms
A 3-volt AA cell charges a 100 μF capacitor to 300 volts through a step-up transformer (over about 10 seconds). When the shutter closes, the capacitor dumps its charge through the xenon tube in 1 millisecond.

Step 1. Identify the knowns and pick the formula.

You know C = 100\ \mu\text{F} = 10^{-4} F and V = 300 V. You do not yet know Q. The form U = \tfrac{1}{2}CV^{2} uses only the knowns — use it.

Why: pick the form of the formula whose variables you already have. Reaching for U = \tfrac{1}{2}QV here would force you to compute Q first, a detour.

Step 2. Compute the stored energy.

U \;=\; \tfrac{1}{2}CV^{2} \;=\; \tfrac{1}{2}\,(10^{-4})(300)^{2} \;=\; \tfrac{1}{2}\,(10^{-4})(9\times 10^{4})
U \;=\; \tfrac{1}{2}\times 9 \;=\; 4.5\ \text{J}.

Why: convert μF to F first (one microfarad is 10^{-6} F, so 100 μF is 10^{-4} F) so you are working in pure SI units. Mixing in μF without converting is the single most common arithmetic error in capacitor problems.

Step 3. Compute the charge using Q = CV.

Q \;=\; CV \;=\; (10^{-4})(300) \;=\; 3\times 10^{-2}\ \text{C} \;=\; 30\ \text{mC}.

Why: now that the energy is found, compute Q for cross-checking. Thirty millicoulombs is a lot — about 2\times 10^{17} electrons sitting on one plate.

Step 4. Cross-check U using the other form.

U \;=\; \tfrac{1}{2}QV \;=\; \tfrac{1}{2}(3\times 10^{-2})(300) \;=\; \tfrac{1}{2}(9) \;=\; 4.5\ \text{J}. \checkmark

Step 5. Compute the average power during the 1-ms flash.

P_\text{avg} \;=\; \frac{U}{\Delta t} \;=\; \frac{4.5\ \text{J}}{10^{-3}\ \text{s}} \;=\; 4.5\times 10^{3}\ \text{W} \;=\; 4.5\ \text{kW}.

Why: the defining equation of power is energy delivered divided by time. A 4.5 kW burst is about what 15 household ceiling fans running full blast would draw — for one thousandth of a second. That is how the flash gets bright enough to light up a Diwali sparkler mid-throw.

Result: U = 4.5 J, Q = 30 mC, P_\text{avg} \approx 4.5 kW.

What this shows: A 3-volt battery cannot deliver 4.5 kW — the internal resistance is too high. But 10 seconds of slow charging into a capacitor, followed by a fast discharge, gives you a brief burst of kilowatts. The capacitor is the impedance-matcher between slow sources and fast loads.

Example 2: Comparing the three forms on a defibrillator bank

A hospital defibrillator stores 200 J of energy at 1600 V for a cardiac shock. (a) What is its capacitance? (b) What charge does it hold at full voltage? (c) If the 200 J are delivered through the patient in 4 ms, what is the average power, and how does it compare with the continuous power from the mains plug?

Step 1. Solve (a) using U = \tfrac{1}{2}CV^{2}.

C \;=\; \frac{2U}{V^{2}} \;=\; \frac{2(200)}{(1600)^{2}} \;=\; \frac{400}{2.56\times 10^{6}} \;=\; 1.56\times 10^{-4}\ \text{F} \;\approx\; 156\ \mu\text{F}.

Why: rearrange U = \tfrac{1}{2}CV^{2} to isolate C. A 156 μF capacitor at 1600 V is a genuinely large and dangerous device — hospital staff are trained not to touch the leads until the internal relay has discharged them through a bleeder resistor.

Step 2. Solve (b) using Q = CV.

Q \;=\; CV \;=\; (1.56\times 10^{-4})(1600) \;=\; 0.25\ \text{C} \;=\; 250\ \text{mC}.

Why: a quarter of a coulomb is an enormous amount of charge to sit on a pair of plates. For comparison, a typical car ignition coil puts about 10^{-5} C through a spark plug — this defibrillator holds 25,000 times more.

Step 3. Cross-check with U = Q^{2}/(2C).

U \;=\; \frac{(0.25)^{2}}{2(1.56\times 10^{-4})} \;=\; \frac{0.0625}{3.12\times 10^{-4}} \;\approx\; 200\ \text{J}. \checkmark

Why: this is the third face of the formula — using only Q and C. All three forms must give the same number, and they do (up to rounding in the capacitance).

Step 4. Average power during the shock.

P_\text{shock} \;=\; \frac{200\ \text{J}}{4\times 10^{-3}\ \text{s}} \;=\; 5.0\times 10^{4}\ \text{W} \;=\; 50\ \text{kW}.

Step 5. Compare with the mains supply.

An Indian hospital wall socket is 230 V × 10 A = 2.3 kW maximum continuous draw. The defibrillator's instantaneous shock power is 50/2.3 \approx 22 times the continuous mains capability.

Why: the capacitor is not breaking conservation of energy — it is charging slowly from the mains (taking perhaps 10 seconds to fill up), then releasing in 4 ms. Average out over the whole charge-and-fire cycle and the energy balance matches. The capacitor's job is to compress time, turning 10 seconds of 20-watt charging into 4 milliseconds of 50-kilowatt discharge.

Result: C \approx 156\ \muF, Q = 250 mC, P_\text{shock} = 50 kW (about 22 × continuous mains capacity).

What this shows: All three forms of the energy formula give the same result and cross-check each other. The capacitor's utility is as a time-compressor: slow accumulation, fast release. That pattern — reservoir charges slowly, empties quickly — is exactly what a camera flash, a defibrillator, an ISRO plasma thruster, and a car's starter-motor capacitor all have in common.

Example 3: Verifying the field-energy formula

A parallel-plate capacitor has plate area A = 200\ \text{cm}^{2} = 0.02\ \text{m}^{2} and gap d = 1.0 mm. It is charged to 200 V. Compute U two different ways — from \tfrac{1}{2}CV^{2} and from the energy-density formula u = \tfrac{1}{2}\varepsilon_{0}E^{2} — and check that they agree.

Step 1. Capacitance from geometry.

C \;=\; \frac{\varepsilon_{0}A}{d} \;=\; \frac{(8.85\times 10^{-12})(0.02)}{10^{-3}} \;=\; 1.77\times 10^{-10}\ \text{F} \;=\; 177\ \text{pF}.

Why: plug into the parallel-plate result from the previous chapter. Remember to use SI units — plate area in m² (not cm²) and gap in metres (not mm).

Step 2. Energy via \tfrac{1}{2}CV^{2}.

U_{1} \;=\; \tfrac{1}{2}CV^{2} \;=\; \tfrac{1}{2}(1.77\times 10^{-10})(200)^{2} \;=\; \tfrac{1}{2}(1.77\times 10^{-10})(4\times 10^{4})
U_{1} \;=\; 3.54\times 10^{-6}\ \text{J} \;=\; 3.54\ \mu\text{J}.

Step 3. Field between the plates.

E \;=\; \frac{V}{d} \;=\; \frac{200}{10^{-3}} \;=\; 2\times 10^{5}\ \text{V/m}.

Why: in a parallel-plate capacitor, the field is uniform between the plates and relates to the voltage by V = Ed, derived in the field-potential chapter.

Step 4. Energy density.

u \;=\; \tfrac{1}{2}\varepsilon_{0}E^{2} \;=\; \tfrac{1}{2}(8.85\times 10^{-12})(2\times 10^{5})^{2}
u \;=\; \tfrac{1}{2}(8.85\times 10^{-12})(4\times 10^{10}) \;=\; 0.177\ \text{J/m}^{3}.

Step 5. Multiply by the volume of field \mathcal{V} = A\cdot d to get total energy.

\mathcal{V} \;=\; (0.02)(10^{-3}) \;=\; 2\times 10^{-5}\ \text{m}^{3}.
U_{2} \;=\; u \cdot \mathcal{V} \;=\; (0.177)(2\times 10^{-5}) \;=\; 3.54\times 10^{-6}\ \text{J} \;=\; 3.54\ \mu\text{J}.

Step 6. Compare.

U_{1} \;=\; U_{2} \;=\; 3.54\ \mu\text{J}. \checkmark

Why: the two formulas are logically equivalent — one derived from the other — but it is still reassuring to see the numbers match, especially with the very different-looking arithmetic along the way. The agreement is the concrete confirmation that the energy really is "in the field" throughout the volume between the plates.

Result: Both methods give U = 3.54\ \muJ. The energy density in the 1-mm gap is a modest 0.18 J/m³ — far below the 44 J/m³ where dry air would break down at E = 3 MV/m.

What this shows: The energy in a capacitor is physically located in the electric field, not on the metal plates. Computing U as "capacitor energy" or as "integrated field energy density" gives identical answers — because they are the same thing, viewed from opposite ends.

Common confusions

If you came here to understand the three energy formulas, use them in problems, and grasp the field-energy idea, you have what you need. What follows is for readers who want the calculus-free proof of the resistance-independent heat loss, the force between the plates from energy considerations, and the general energy-integral that extends the formula to arbitrary capacitor shapes.

Charging a capacitor through a resistor — where the other half goes

Suppose you charge a capacitor C from zero to voltage V_{0} through a resistor R and an ideal battery of EMF V_{0}. The current at any moment is

i(t) \;=\; \frac{V_{0}}{R}\,e^{-t/RC}.

(This comes from Kirchhoff's voltage law around the loop, V_{0} = iR + q/C, and integrating — a full derivation lives in the RC-circuits article.) The instantaneous power dissipated in the resistor is

P_{R}(t) \;=\; i^{2}R \;=\; \frac{V_{0}^{2}}{R}\,e^{-2t/RC}.

Integrate this from t = 0 to t = \infty to get the total heat dissipated:

W_{R} \;=\; \int_{0}^{\infty}\frac{V_{0}^{2}}{R}\,e^{-2t/RC}\,dt \;=\; \frac{V_{0}^{2}}{R}\cdot\frac{RC}{2} \;=\; \tfrac{1}{2}CV_{0}^{2}.

Why: the integral \int_{0}^{\infty}e^{-2t/RC}\,dt = RC/2 is a standard exponential integral. The R's cancel, leaving \tfrac{1}{2}CV_{0}^{2} — independent of the resistance.

So exactly half the battery's work W_\text{bat} = CV_{0}^{2} (total charge Q = CV_{0} moved through voltage V_{0}) is dissipated as heat, regardless of how large or small R is. The other half ends up as stored field energy in the capacitor, \tfrac{1}{2}CV_{0}^{2}. This is the rigorous version of the "50% efficiency limit" mentioned in the main text.

Two consequences:

  • Dumping charge into a capacitor via an ideal wire (R → 0) still loses half. As R \to 0, the current spike becomes larger and briefer, but the integrated i^{2}R heat stays at \tfrac{1}{2}CV_{0}^{2}. What happens in the zero-resistance limit is that the energy goes into radiation — a high-frequency electromagnetic pulse leaves the wire. Still lost.
  • To beat 50%, you need an inductor. Charging through an inductor (an LC circuit) produces a resonant oscillation. You can arrange to open the switch at the peak — capturing the full CV_{0}^{2} in the capacitor without dissipation. That is the basis of switched-mode power supplies, resonant-converter efficiency, and why ISRO plasma-thruster capacitor banks use inductive charging circuits.

Force between the plates from energy considerations

A surprising use of the energy formula: you can compute the mechanical force pulling the plates together without writing down any force law for individual charges.

Hold the charge Q fixed (battery disconnected) and imagine pulling the plates apart by a tiny distance dx. The capacitance changes: C = \varepsilon_{0}A/d decreases as d grows. Since U = Q^{2}/(2C) and C decreases, U increases — you have done work against an attractive force between the plates.

The force is

F \;=\; -\frac{dU}{dx}\bigg|_{Q} \;=\; -\frac{d}{dx}\left[\frac{Q^{2}}{2C}\right].

Compute:

\frac{dC}{dx} \;=\; \frac{d}{dx}\left[\frac{\varepsilon_{0}A}{x}\right] \;=\; -\frac{\varepsilon_{0}A}{x^{2}}.
\frac{dU}{dx} \;=\; \frac{Q^{2}}{2}\cdot\frac{d}{dx}\left[\frac{1}{C}\right] \;=\; \frac{Q^{2}}{2}\cdot\left(-\frac{1}{C^{2}}\right)\cdot\frac{dC}{dx}.
\frac{dU}{dx} \;=\; \frac{Q^{2}}{2C^{2}}\cdot\frac{\varepsilon_{0}A}{x^{2}}.

Substitute Q = \sigma A and C = \varepsilon_{0}A/x:

\frac{dU}{dx} \;=\; \frac{\sigma^{2}A^{2}}{2}\cdot\frac{x^{2}}{\varepsilon_{0}^{2}A^{2}}\cdot\frac{\varepsilon_{0}A}{x^{2}} \;=\; \frac{\sigma^{2}A}{2\varepsilon_{0}}.

So the attractive force between the plates is

F \;=\; \frac{\sigma^{2}A}{2\varepsilon_{0}} \;=\; \tfrac{1}{2}\sigma E \cdot A \;=\; \tfrac{1}{2}QE.

Why: each plate feels a force \tfrac{1}{2}QE (not QE — there is a factor of \tfrac{1}{2}) because the field E in the gap is produced by both plates together, while the force on plate 1 comes only from plate 2's field, which is E/2. The energy method recovers this subtle factor of \tfrac{1}{2} automatically, without having to parse the self-field-vs.-external-field distinction by hand.

This is the foundation of electrostatic micro-actuators — the tiny MEMS deflection mirrors in projection displays use exactly this force, derived from the energy formula.

General energy integral for an arbitrary capacitor shape

For an arbitrary conductor geometry, the total electrostatic energy is

U \;=\; \frac{1}{2}\int_\text{all space} \varepsilon_{0}E^{2}\, dV,

integrated over the entire region where the field is nonzero. This is the general version of equation (4). For the parallel-plate case the field is uniform and nonzero only in the gap, so the integral reduces to \tfrac{1}{2}\varepsilon_{0}E^{2}\cdot \mathcal{V} as before.

Two worked-out examples using this integral:

  • Isolated charged sphere of radius R, charge Q. Outside the sphere, E(r) = Q/(4\pi\varepsilon_{0}r^{2}); inside, E = 0. Integrating:
U \;=\; \frac{\varepsilon_{0}}{2}\int_{R}^{\infty}\left(\frac{Q}{4\pi\varepsilon_{0}r^{2}}\right)^{2} 4\pi r^{2}\,dr \;=\; \frac{Q^{2}}{8\pi\varepsilon_{0}}\int_{R}^{\infty}\frac{dr}{r^{2}} \;=\; \frac{Q^{2}}{8\pi\varepsilon_{0}R}.

Compare with U = Q^{2}/(2C) using C_\text{sphere} = 4\pi\varepsilon_{0}R:

U \;=\; \frac{Q^{2}}{2\cdot 4\pi\varepsilon_{0}R} \;=\; \frac{Q^{2}}{8\pi\varepsilon_{0}R}. \checkmark

Perfect agreement.

  • Self-energy of a point charge. Take R \to 0 in the above. U \to \infty. This is the famous self-energy divergence of a point charge — one of the deep puzzles classical electromagnetism leaves unsolved and that quantum field theory only partially resolves (through renormalisation). For any ordinary physics problem, you ignore self-energy; it is a constant you subtract.

The field-integral formulation also makes it obvious that the capacitance C is a property of geometry and the medium — the same integral applied to the same geometry always gives the same C. The \tfrac{1}{2}\varepsilon_{0}E^{2} energy density is the physical quantity; the formulas U = \tfrac{1}{2}CV^{2} are just the integrated form for specific shapes.

Where this leads next