Energy stored in capacitors

Every camera flash, defibrillator paddle and photographic strobe relies on the same trick: charge a capacitor slowly, then release its stored energy almost instantly. That energy did not appear from nowhere — it is the work done pushing charge onto the plates against an electric field that grows stronger with every bit of charge already there. This closing topic of ISC Class 12 Electrostatics gives that stored energy a precise formula, $U=\tfrac12CV^2$ , and shows the same energy can be pictured as living inside the electric field itself, spread through the volume between the plates as an energy density. Both pictures — energy "on the plates" and energy "in the field" — describe the same physical quantity and agree exactly for a parallel-plate capacitor.

The core insight

Charging a capacitor to potential difference $V$ stores energy $U=\tfrac12CV^2=\tfrac12QV=Q^2/2C$ because the average potential during charging is only $V/2$ , and this same energy can equivalently be located in the field between the plates as a density $u=\tfrac12\varepsilon_0E^2$ .

The core idea

Why the factor of one-half. As a capacitor charges, the potential difference across it rises steadily from $0$ (uncharged) to $V$ (fully charged), because each new bit of charge deposited makes it harder to push the next bit on. The work done moving a tiny charge $dq$ against the instantaneous potential $q/C$ is $dq\times(q/C)$ , and summing this over the whole charging process gives an average potential of just $V/2$ , not the final value $V$ . That single fact is the origin of every $\tfrac12$ in this topic — it is what separates charging a capacitor from moving a fixed charge $Q$ through a constant potential difference $V$ (which would do work $QV$ , no half).

Three equivalent forms. Once a capacitor of capacitance $C$ reaches potential difference $V$ and final charge $Q=CV$ , the energy stored is

U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}

where: $U$ is energy in joules, $C$ is capacitance in farads, $V$ is potential difference in volts, and $Q$ is charge in coulombs. The three forms are algebraically identical (swap between them using $Q=CV$ ); which one to reach for depends only on which two quantities a problem hands you — $\tfrac12CV^2$ when $C,V$ are known, $\tfrac12QV$ when $Q,V$ are known, and $Q^2/2C$ whenever charge is the quantity held fixed, such as after a charged capacitor is disconnected from its battery and something about it (plate separation, a dielectric) is then changed.

Energy lives in the field. For a parallel-plate capacitor of plate area $A$ , separation $d$ and uniform field $E=V/d$ between the plates, substituting $C=\varepsilon_0A/d$ into $U=\tfrac12CV^2$ gives $U=\tfrac12\varepsilon_0E^2\,(Ad)$ . Dividing by the volume $Ad$ occupied by the field leaves a quantity that depends only on the local field strength, the energy density:

u = \frac{U}{Ad} = \frac{1}{2}\varepsilon_0E^2

where: $u$ is energy per unit volume (J/m³) and $\varepsilon_0=8.85\times10^{-12}\ \text{F/m}$ is the permittivity of free space. This reframes the stored energy as belonging to the electric field occupying the space between the plates rather than sitting "on" the charges — a view that generalises to any electric field, not just the uniform one inside a parallel-plate capacitor.

Figure

Charge against potential difference graph for a charging capacitor with the triangular area under the line equal to the stored energy

Derivation AEnergy stored during charging, from work done moving charge

The goal is to show, starting from nothing more than the definition of potential and a single integral, that charging a capacitor of capacitance $C$ to a final potential difference $V$ stores energy $U=\tfrac12CV^2$ — turning the "average potential is $V/2$ " statement above into an exact result rather than a hand-wave, and confirming why the factor is one-half and not one.

Step 1. At some intermediate instant during charging, suppose charge $q$ has already been placed on the plates ( $0\le q\le Q$ ). The potential difference across the capacitor at that instant is $v=q/C$ , directly from the definition of capacitance $C=q/v$ .

Step 2. Moving the next infinitesimal charge $dq$ onto the plates requires work $dW=v\,dq=\dfrac{q}{C}\,dq$ , since work done moving a charge $dq$ through potential difference $v$ is $v\,dq$ by definition of potential.

Step 3. Total work done charging the capacitor from $q=0$ to $q=Q$ is the sum (integral) of these small contributions:

W=\int_0^Q \frac{q}{C}\,dq=\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q=\frac{Q^2}{2C}

Step 4. This work is stored as electrical potential energy, $U=W=Q^2/2C$ . Substituting $Q=CV$ gives $U=(CV)^2/2C=\tfrac12CV^2$ , and substituting instead $C=Q/V$ gives $U=\tfrac12QV$ — the same three forms met earlier, now derived rather than merely stated. The integral itself is the rigorous version of "average potential is $V/2$ ": integrating the straight line $v=q/C$ from $0$ to $Q$ literally averages it to $V/2$ , so $U=Q\times(V/2)=\tfrac12QV$ .

Worked example

A capacitor of capacitance $20\ \mu\text{F}$ is charged to a potential difference of $200\ \text{V}$ . Find the charge stored and the energy stored.

Step 1. Charge stored: $Q=CV=(20\times10^{-6})\times200=4\times10^{-3}\ \text{C}=4\ \text{mC}$ .

Step 2. Energy stored using $U=\tfrac12CV^2$ : $U=\tfrac12\times(20\times10^{-6})\times(200)^2=\tfrac12\times(20\times10^{-6})\times(4\times10^4)=\tfrac12\times0.8=0.4\ \text{J}$ .

Step 3. Cross-check with the alternate form $U=\tfrac12QV=\tfrac12\times(4\times10^{-3})\times200=\tfrac12\times0.8=0.4\ \text{J}$ — consistent. Energy stored is $\boxed{0.4\ \text{J}}$ , with charge $4\ \text{mC}$ .

Competitive stretch

Now this same $10\ \mu\text{F}$ -scale idea is pushed further: a $10\ \mu\text{F}$ capacitor is charged by a $100\ \text{V}$ battery, disconnected, then connected in parallel with an identical uncharged $10\ \mu\text{F}$ capacitor. Step 1. Charge before connecting is fixed at $Q=CV=10\times10^{-6}\times100=1000\ \mu\text{C}$ , and this total redistributes over $C_{\text{eq}}=20\ \mu\text{F}$ , giving a common potential $V_{\text{common}}=Q/C_{\text{eq}}=1000/20=50\ \text{V}$ . Step 2. Energy before connecting was $U_{\text{before}}=\tfrac12CV^2=\tfrac12(10\times10^{-6})(100)^2=0.05\ \text{J}$ ; energy after is $U_{\text{after}}=\tfrac12C_{\text{eq}}V_{\text{common}}^2=\tfrac12(20\times10^{-6})(50)^2=0.025\ \text{J}$ — exactly half is gone, dissipated as heat and radiation during the sudden, non-quasi-static redistribution, even though the total charge $1000\ \mu\text{C}$ is fully conserved throughout.

Common traps

Writing $U=QV$ instead of $U=\tfrac12QV$ — the average potential during charging is $V/2$ , not the final voltage $V$ ; only the last drop of charge moves through the full potential difference $V$ .
Assuming energy is always conserved when capacitors are connected together — charge is always conserved, but energy is lost to resistance and radiation whenever the redistribution is sudden rather than slow and quasi-static.
Reaching for $\tfrac12CV^2$ in a fixed-charge scenario (capacitor disconnected from the battery, then $C$ changed by a dielectric or by moving the plates) — $Q^2/2C$ is the safer form because $Q$ , not $V$ , is what stays constant.
Confusing the local energy density $u=\tfrac12\varepsilon_0E^2$ (per unit volume, valid at a point in the field) with the total stored energy $U$ , which needs an extra multiplication by the volume $Ad$ .

Ask the tutor: want to work through why pulling a disconnected capacitor's plates apart increases its stored energy, or practise choosing the right energy formula for a "battery disconnected, then something changes" problem?

Why is the energy half Q V and not Q V, when work done equals charge times potential difference?

Work equals charge times potential difference is exactly right for moving a charge through a constant potential difference, like a fixed battery voltage. But while a capacitor charges, the potential difference across it is not constant — it climbs from 0 up to V as more charge piles onto the plates, since each new bit of charge has to work against the field left by all the charge already there. Averaging that rising potential over the whole process gives V over 2, so the total work done, and hence the energy stored, is Q times V over 2, which is half Q V.

If I disconnect a charged capacitor and pull its plates farther apart, does the stored energy change?

Yes, and the safest way to see how is to hold onto whichever quantity is physically fixed. With the battery disconnected, no charge can flow on or off the plates, so Q stays fixed while capacitance C decreases as separation increases (C is inversely proportional to separation for a parallel-plate capacitor). Using U equals Q squared over 2C, a smaller C with the same Q means larger U — the stored energy actually increases, because you do external work pulling the attracting plates apart against their mutual electrical attraction, and that work becomes extra stored energy.

Does energy density u equal half epsilon naught E squared apply only inside capacitors?

No — it is a general statement about any electric field, not a special property of capacitors. The parallel-plate capacitor is simply the easiest place to derive it, because the field between its plates is uniform and the volume it occupies is easy to state as A times d. The same formula for the energy carried by an electric field applies point by point anywhere a field exists, which is one reason physicists find it useful to think of the field itself, not just the charges, as a genuine carrier and store of energy.

Energy stored in capacitors

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads