KAHIBARO
Table of Contents
Electric energy in a capacitor
A capacitor stores energy in the electric field created between its charged parts. When charge is moved from one plate to the other, work must be done against the electric forces already present. That work does not disappear. It becomes stored electrical energy.
If a capacitor has capacitance $C$, charge $Q$, and potential difference $V$, the stored energy can be written in several equivalent forms:
$$
U = \frac{1}{2}QV
$$
Using the relation $Q = CV$, this can also be written as
$$
U = \frac{1}{2}CV^2
$$
and
$$
U = \frac{Q^2}{2C}
$$
Important energy formulas for a capacitor:
$$
U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}
$$
These formulas describe the same stored energy, written in different variables.
Why the factor one half appears
When a capacitor is uncharged, the potential difference across it is zero. As more charge is placed on it, the voltage rises. This means the later charge added must be pushed through a larger potential difference than the earlier charge.
Suppose at some moment the capacitor has charge $q$ on it. Its voltage at that moment is
$$
V = \frac{q}{C}
$$
To add a small extra charge $dq$, the small amount of work needed is
$$
dU = V \, dq = \frac{q}{C}dq
$$
Now integrate from $q = 0$ to $q = Q$:
$$
U = \int_0^Q \frac{q}{C}dq = \frac{1}{C}\int_0^Q q \, dq = \frac{Q^2}{2C}
$$
Then, since $Q = CV$, we get
$$
U = \frac{1}{2}QV = \frac{1}{2}CV^2
$$
The factor $\frac{1}{2}$ appears because the voltage is not constant while charging. It increases from $0$ to its final value $V$, so the average voltage during charging is $\frac{V}{2}$.
Interpreting the formulas
Each form of the energy equation is useful in a different situation.
| Known quantities | Best formula |
|---|---|
| $Q$ and $V$ | $U = \frac{1}{2}QV$ |
| $C$ and $V$ | $U = \frac{1}{2}CV^2$ |
| $Q$ and $C$ | $U = \frac{Q^2}{2C}$ |
These forms also show how energy changes when one quantity is held fixed.
If capacitance is fixed, then energy increases with the square of the voltage:
$$
U \propto V^2
$$
So doubling the voltage makes the energy four times larger.
If charge is fixed, then
$$
U = \frac{Q^2}{2C}
$$
In that case, increasing the capacitance decreases the stored energy.
Do not confuse charge with energy.
A larger charge does not always mean larger energy unless the other quantities are considered as well.
Energy and the electric field
A capacitor stores energy in the electric field in the region around and between its plates. For a parallel plate capacitor, most of the field is between the plates, so most of the energy is stored there.
This is an important physical idea. The capacitor is not just a container for charge. The stored energy belongs to the electric field itself.
For a uniform electric field, the energy density, meaning energy per unit volume, is
$$
u = \frac{1}{2}\varepsilon E^2
$$
where $\varepsilon$ is the permittivity of the material and $E$ is the electric field magnitude.
Energy density of an electric field:
$$
u = \frac{1}{2}\varepsilon E^2
$$
This tells us how much energy is stored in each unit volume of the field.
For a parallel plate capacitor with plate area $A$ and separation $d$, the field occupies approximately a volume
$$
Ad
$$
So the total stored energy is
$$
U = u(Ad) = \frac{1}{2}\varepsilon E^2 Ad
$$
This agrees with the capacitor energy formulas.
Charging and external work
To charge a capacitor, an external source such as a battery must move charge. That source does work. The stored energy equals the work done in building up the charge on the capacitor, if we ignore energy losses such as heating.
If charging occurs in a real circuit, some energy may also be lost in resistors as thermal energy. So the energy provided by a battery is not always equal to the final stored energy in the capacitor. The capacitor stores only part of the supplied energy.
This is especially important in simple charging circuits. Even if the battery provides energy $CV^2$, the capacitor ends with only
$$
U = \frac{1}{2}CV^2
$$
The other half is dissipated in the circuit resistance.
Example
Consider a capacitor with
$$
C = 4.0 \,\mu\text{F}, \quad V = 12 \,\text{V}
$$
The stored energy is
$$
U = \frac{1}{2}CV^2
$$
Substitute the values:
$$
U = \frac{1}{2}(4.0 \times 10^{-6})(12)^2
$$
$$
U = 2.0 \times 10^{-6} \times 144
$$
$$
U = 2.88 \times 10^{-4}\,\text{J}
$$
So the capacitor stores
$$
U = 2.88 \times 10^{-4}\,\text{J}
$$
or
$$
0.288 \,\text{mJ}
$$
Practical meaning
Stored capacitor energy matters in many devices. Camera flashes, defibrillators, power supplies, and electronic circuits all rely on capacitors that can store and release energy quickly.
A small capacitor at low voltage stores only a tiny amount of energy. A large capacitor at high voltage can store enough energy to be dangerous.
Safety rule:
A charged capacitor can release energy suddenly. Even after a device is turned off, a capacitor may still hold charge and stored energy.
Summary
A capacitor stores energy because work is required to separate charge and create an electric field. The stored energy is
$$
U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}
$$
This energy is physically stored in the electric field. For a uniform field, the field energy density is
$$
u = \frac{1}{2}\varepsilon E^2
$$
These ideas connect the circuit description of a capacitor to the field description of electricity.