Derivation: Energy of the Electric Field

Let us consider a conducting sphere (e.g. a metal sphere) with radius $r$. The sphere is charged step by step with small charge portions $\text{d}Q$ coming from infinity. After the total charge $Q$ is brought onto the sphere, the sphere produces the following electric potential (we assume this to be known here):

$$ \begin{align} \phi_{\text e}(r) ~=~ \frac{1}{4\pi\,\varepsilon_0} \, \frac{ Q }{ r } \end{align} $$

The energy $\text{d}W$ of a charge portion $\text{d}Q$ brought to the surface of the sphere with potential $\phi_{\text e}(r) $ is:

$$ \begin{align} \text{d}W ~=~ \phi_{\text e}(r) \, \text{d}q \end{align} $$

Substitute the potential 1 into Eq. 2:

$$ \begin{align} \text{d}W ~=~ \frac{1}{4\pi\,\varepsilon_0} \, \frac{ Q }{ r } \, \text{d}Q \end{align} $$

To get the total energy $W_{\text{e}}$, we need to integrate the left side of Eq. 3 over the energy from 0 to $W_{\text{e}}$ and integrate the right side over the charge from $0$ to $Q$:

$$ \begin{align} \int_0^{W_{\text e}} \, \text{d}W ~&=~ \frac{1}{4\pi\,\varepsilon_0 \, r} \, \int_0^Q Q \, \text{d}Q \\\\
W_{\text e} ~&=~ \frac{1}{4\pi\,\varepsilon_0 \, r} \, \int_0^Q Q \, \text{d}Q \end{align} $$

Let's integrate the right side:

$$ \begin{align} W_{\text e} ~=~ \frac{1}{4\pi\,\varepsilon_0 \, r} \, \left[ \frac{1}{2} \, Q^2 \right]^{Q}_0 \end{align} $$

Inserting the integration limits results:

Electrical energy of a charged sphere

$$ \begin{align} W_{\text e} ~=~ \frac{1}{4\pi\,\varepsilon_0 \, r} \, \frac{1}{2} \, Q^2 \end{align} $$

To eliminate the geometric factor, the radius $r$, in 7, we replace it with the capacitance $C = 4\pi\,\varepsilon_0 \, r$ of a sphere:

Electrical energy via charge

$$ \begin{align} W_{\text e} ~=~ \frac{1}{2C} \, Q^2 \end{align} $$

Since we have eliminated the radius of the sphere and expressed the energy in terms of the capacitance $C$, the equation 8 applies not only to a sphere but also to other charged bodies to which a capacitance can be assigned.

Since the charge $Q$ is not so well accessible experimentally, we can express formula 9 by voltage $U$. The definition of the capacitance $ C = Q/U \Leftrightarrow Q = C\,U$ is used for this purpose. The we insert $Q$ in 8:

Electrical energy via voltage

$$ \begin{align} W_{\text e} ~=~ \frac{1}{2} \, C \, U^2 \end{align} $$

The assumption that the energy 9 is stored in the electric field can be motivated as follows: Consider a plate capacitor with plate area $A$ and plate distance $d$. Let $U$ be the voltage between the capacitor plates. The electric field $E$ inside the plate capacitor is given by $ E = U/d \Leftrightarrow U = E\,d$ and the capacitance by $C = \varepsilon_0 A / d $. (see Derivation). Voltage and capacitance used in 9 results:

$$ \begin{align} W_{\text e} ~=~ \frac{1}{2} \, \varepsilon_0 \, \frac{A}{d} \, E^2 \, d^2 \end{align} $$

Distance $d$ can be canceled once:

$$ \begin{align} W_{\text e} ~=~ \frac{1}{2} \, \varepsilon_0 \, A \, E^2 \, d \end{align} $$

Here $ A \, d$ is the volume $V$ enclosed between the capacitor plates:

Electrical energy via E-field

$$ \begin{align} W_{\text e} ~=~ \frac{1}{2} \, \varepsilon_0 \, V \, E^2 \end{align} $$

The relation 12 between the energy and the E-field motivates the assumption that the energy $W_{\text e}$ is stored in the electric field $E$, which is in the volume $V$. The energy density $w_{\text e} = W_{\text e}/V $ of the E-field is then:

$$ \begin{align} w_{\text e} ~=~ \frac{1}{2} \, \varepsilon_0 \, E^2 \end{align} $$

Note that...

All derived formulas for energy are valid only in vacuum. In order to make them also valid for E-fields in matter, they must be multiplied by the dielectric constant $\varepsilon_{\text r}$ of the material.

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