Energy Stored in a Capacitor

Why a capacitor stores energy

• A capacitor is a device used to store energy. Charging it means transferring electrons from one plate to the other, and the battery must do work to move them against the opposing forces.
• Those forces arise because the electrons are pushed onto the negative plate, which repels them, and removed from the positive plate, which attracts them. In both cases the Coulomb forces oppose the transfer, and as the charge on the plates grows the opposing force grows with it.
• The work done is stored as electrostatic energy in the electric field between the plates, and can later be recovered by discharging the capacitor.

Derivation

Set-up and diagram

• Consider a capacitor of capacitance $C$ being charged by a DC source of $V$ volts. Draw the capacitor connected across the cell through a key.
• During charging, let $q'$ be the instantaneous charge on the capacitor and $V$ the potential difference between the plates, so that

$$C = \dfrac{q'}{V}$$

Work done

• A small amount of work is done if a further small charge $dq$ is transferred between the plates:

$$dW = V\,dq = \dfrac{q'}{C}\,dq$$

• Total work done in transferring the charge from 0 to $Q$:

$$W = \int dW = \int_{0}^{Q}\dfrac{q'}{C}\,dq = \dfrac{1}{C}\int_{0}^{Q}q'\,dq = \dfrac{1}{C}\left[\dfrac{(q')^2}{2}\right]_{0}^{Q} = \dfrac{1}{2}\dfrac{Q^2}{C}$$

The three forms

• This work is stored as the electrical potential energy $U$ of the capacitor, and using $Q = CV$ it can be written in three equivalent ways:

$$\boxed{\,U = \dfrac{1}{2}\dfrac{Q^2}{C} = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV\,}$$

• Use the first form when $Q$ and $C$ are given, the second when $C$ and $V$ are given, and invert the first as $Q = \sqrt{2UC}$ when the charge is wanted.

Energy supplied versus energy stored

• The energy supplied by the battery is $QV$, but the energy stored in the electric field is only $\dfrac{1}{2}QV$.
• The remaining half, $\dfrac{1}{2}QV$, is wasted as heat in the connecting wires and in the battery itself.
• Hence the ratio of energy stored in a capacitor to energy dissipated during charging is $1:1$.

Energy density

• The energy is stored in the field itself, so it is useful to speak of the energy per unit volume of the field:

$$\boxed{\,u = \dfrac{1}{2}\varepsilon_0 kE^2\,}$$

• measured in $\text{J m}^{-3}$. The total energy contained in a volume $V$ of dielectric sitting in a field $E$ is $U = u\times V$ — the working route for the "cube of marble in a field of $300\ \text{V m}^{-1}$" type of numerical.

Energy lost when two capacitors are joined

• If a capacitor charged to $V_1$ is connected to a second capacitor at $V_2$, charge redistributes until both reach a common potential

$$V_{\text{common}} = \dfrac{C_1V_1+C_2V_2}{C_1+C_2}$$

• The total stored energy always falls, the difference being lost as heat and electromagnetic radiation:

$$\Delta U = \dfrac{1}{2}\dfrac{C_1C_2}{C_1+C_2}\left(V_1-V_2\right)^2$$

• The same answer follows by computing $\dfrac{1}{2}C_1V_1^2$ before and $\dfrac{1}{2}(C_1+C_2)V_{\text{common}}^2$ after, and subtracting.

Change of energy when a dielectric is introduced

• With the supply kept connected so that $V$ stays constant, $C$ becomes $kC$ and $U = \dfrac{1}{2}CV^2$ rises to $k$ times its value — the energy of the capacitor increases when the slab is introduced, and falls back when it is removed.
• With the capacitor charged and then isolated so that $Q$ stays constant, $U = \dfrac{1}{2}\dfrac{Q^2}{C}$ falls to $\dfrac{1}{k}$ of its value when the slab goes in, and rises $k$ times when the slab is taken out. Always check first which of $Q$ or $V$ is being held fixed.