# Energy Densities of Electric and Magnetic Fields

That electric (${\displaystyle E}$) and magnetic (${\displaystyle B}$) fields hold energy shouldn’t be surprising; the electromagnetic fields transporting energy from the Sun to Earth warm us up (and give us sunburns) when we stand outside. But the fact that macroscopic kinetic energy could disappear into an invisible field and be recovered later did indeed come as a surprise to 18th and 19th century physicists.

In astrophysical applications ranging from radiative transport to magnetohydrodynamics (the behavior of conducting fluids such as plasmas), our understanding of the energy dynamics would be fundamentally lacking if we didn’t include the energy of the electric and magnetic fields in our tally. The short answer is that, in CGS units, the energy density ${\displaystyle U_{tot}}$ of the ${\displaystyle E}$ and ${\displaystyle B}$ fields is given by

${\displaystyle U_{tot,CGS}={\frac {1}{8\pi }}E^{2}+{\frac {1}{8\pi }}B^{2}\,\!}$

It intuitively makes sense that energy should be proportional to the square of the amplitude of the ${\displaystyle E}$ and ${\displaystyle B}$ fields, just as power is proportional to the square of voltage in electrical circuits. What may be less intuitive is that the scalar amplitudes of the electric and magnetic fields are intrinsically densities, in the sense that they are per-volume quantities. This is because the ${\displaystyle E}$- and ${\displaystyle B}$-field amplitudes are local quantities intrinsic to a point in space, and must be integrated over volume to become energy.

As an aside, the symmetry in expression for ${\displaystyle U_{tot}}$ with respect to ${\displaystyle E}$ and ${\displaystyle B}$ is much more apparent in the CGS expression than in MKS, where the presence of ${\displaystyle \epsilon _{0}}$ and ${\displaystyle \mu _{0}}$ are distracting:

${\displaystyle U_{tot,MKS}={\frac {\epsilon _{0}}{2}}E^{2}+{\frac {1}{2\mu _{0}}}B^{2}\,\!}$

to me, this symmetry is one of the more compelling reasons why I prefer CGS to MKS; ${\displaystyle E}$ and ${\displaystyle B}$ fields are already in the same units!

### Derivation

#### Deriving Energy Density in an Electric Field Using a Capacitor

Recall that for a parallel-plate capacitor, two plates close together create a constant electric field.

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The electric field due to just one plate is

${\displaystyle E={\frac {Q}{2A\epsilon _{0}}},\,\!}$

where ${\displaystyle Q}$ is the charge, ${\displaystyle A}$ is the area of the plate, and ${\displaystyle \epsilon _{0}}$ is the permittivity of free space (${\displaystyle 9\times 10^{-12}}$ ${\displaystyle {\frac {C^{2}}{Nm^{2}}}}$). Note how this equation is independent of the distance between the plates; as long as the charge and the area of a plate is constant, the electric field is the same no matter how much you pull the plates apart.

For a parallel-plate capacitor, what is the force that one plate exerts on another? Well, if you break a capacitor plate into a distribution of a bunch of point charges, each charge will have the same force acting on it because the electric field of the opposite plate is constant. So the force on the opposite plate will be:

${\displaystyle F=Q_{\text{opposite}}E\,\!}$
${\displaystyle F={\frac {Q^{2}}{2A\epsilon _{0}}}\,\!}$

Now, we pull the plates apart by a distance ${\displaystyle \Delta s}$; to do this, we must do work on one of the plates.

${\displaystyle W=\Delta U=F\Delta s,\,\!}$

where

${\displaystyle \Delta U={\frac {Q^{2}}{2A\epsilon _{0}}}\Delta s.\,\!}$

By doing work on one of the capacitor plates, we’ve increased the energy by increasing the distance between the plates (when dealing with a constant electric field).

Since the energy has changed by a change in distance, how does the energy change per volume for the parallel-plate capacitor?

The change in energy by the change in volume is the energy density for that new area (which now incorporates that distance of ${\displaystyle \Delta s}$).

${\displaystyle {\frac {\Delta U}{\Delta V}}={\frac {Q^{2}\Delta s}{2\epsilon _{0}A}}{\frac {1}{A\Delta s}}\,\!}$
${\displaystyle {\frac {\Delta U}{\Delta V}}={\frac {1}{2}}{\frac {Q^{2}}{\epsilon _{0}A^{2}}},\,\!}$

which can be reconfigured to

${\displaystyle {\frac {\Delta U}{\Delta V}}={\frac {1}{2}}\epsilon _{0}{\Bigg (}{\frac {Q}{A\epsilon _{0}}}{\Bigg )}^{2}.\,\!}$

Recall that the electric field for a parallel-plate capacitor is the sum of the electric fields of each plate. This becomes

${\displaystyle E_{//-plate}={\frac {Q}{\epsilon _{0}A}},\,\!}$

and so the energy density equation for an electric field can be written as

${\displaystyle {\frac {\Delta U}{\Delta V}}={\frac {1}{2}}\epsilon _{0}E^{2},\,\!}$

where ${\displaystyle E}$ is the magnitude of the electric field.

This equation means that there is energy stored in an electric field.

#### Deriving the Energy Density for a Magnetic Field Using a Solenoid

If you want to find out the energy density of a magnetic field, how would you do this?

Recall that the magnetic field of an inductor with ${\displaystyle N}$ coils is

${\displaystyle B=\mu _{0}NI\,\!}$

where ${\displaystyle \mu _{0}}$ the permeability (${\displaystyle 4\pi \times 10^{-7}{\frac {N}{Ampere^{2}}}}$) and ${\displaystyle I}$ is the current in each turn. ${\displaystyle N}$ is further described as the number of turns per unit length.

*Simplified diagram of a solenoid to show how energy density of a magnetic field is derived from self-inductance.

The expression for the coefficient of self-inductance is

${\displaystyle L=\mu _{0}N^{2}\pi r^{2}l,\,\!}$

where ${\displaystyle l}$ is the length of the solenoid, and ${\displaystyle r}$ is the radius of the coil, or turn.

The energy corresponding to the self-inductance is

${\displaystyle U={\frac {1}{2}}LI^{2},\,\!}$

where we can substitute the value for L to get

${\displaystyle U={\frac {1}{2}}\mu _{0}N^{2}\pi r^{2}lI^{2}.\,\!}$

However, we can rearrange this equation again in terms of the magnetic field of an inductor by dividing by ${\displaystyle \mu _{0}}$:

${\displaystyle U={\frac {1}{2\mu _{0}}}\mu _{0}^{2}N^{2}I^{2}\pi r^{2}l.\,\!}$

The area of the solenoid is defined as ${\displaystyle \pi r^{2}}$, and multiplying this area by the length of the solenoid gives us the volume of the solenoid.

${\displaystyle {\frac {U}{V}}={\frac {1}{2\mu _{0}}}(\mu _{0}NI^{2})\,\!}$

The energy stored in a magnetic field per unit volume is

${\displaystyle {\frac {U}{V}}={\frac {1}{2\mu _{0}}}B^{2},\,\!}$

where ${\displaystyle B}$ is the magnitude of the magnetic field.

#### Using the Propagation Speed of Electromagnetic Waves

The fact that the energy density from an electric field is equal to the energy density from a magnetic field is a very powerful statement.

Setting the energy densities equal to each other and canceling out the factors of ${\displaystyle {\frac {1}{2}}}$ on both sides we find

${\displaystyle {\frac {B^{2}}{\mu _{0}}}=\epsilon _{0}E^{2}\,\!}$
${\displaystyle {\frac {1}{\sqrt {\mu _{0}\epsilon _{0}}}}={\frac {E}{B}}.\,\!}$

Recall that ${\displaystyle {\frac {1}{\sqrt {\mu _{0}\epsilon _{0}}}}}$ is the speed of the wave that travels by the interaction of electric and magnetic fields. This is the speed of light in a vacuum, reiterating the fact that light is an electromagnetic wave!

#### Total Energy Density in terms of E or B Only:

Now that we know that

${\displaystyle B=E{\sqrt {\epsilon _{0}\mu _{0}}},\,\!}$

we can describe the total energy density of an EM wave as

${\displaystyle {\frac {U}{V}}=u=u_{B}+u_{E}\,\!}$
${\displaystyle u={\frac {1}{2\mu _{0}}}B^{2}+{\frac {1}{2}}\epsilon _{0}E^{2}\,\!}$

Plugging in B as a function of the electric field, we find that the total energy density in terms of the magnitude of the electric field only is:

${\displaystyle u={\frac {1}{2}}{\frac {(E{\sqrt {\epsilon _{0}\mu _{0}}})^{2}}{\mu _{0}}}+{\frac {1}{2}}\epsilon _{0}E^{2}\,\!}$
${\displaystyle u=\epsilon _{0}E^{2}\,\!}$

Similarly, we can describe:

${\displaystyle E={\frac {B}{\sqrt {\epsilon _{0}\mu _{0}}}}\,\!}$
${\displaystyle u={\frac {1}{2}}\epsilon _{0}({\frac {B}{\sqrt {\epsilon _{0}\mu _{0}}}})^{2}+{\frac {1}{2}}{\frac {B^{2}}{\mu _{0}}}\,\!}$
${\displaystyle u={\frac {B^{2}}{\mu _{0}}}.\,\!}$