# Energy density



My notes of the excellent lectures 7 and 12 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

To assemble charges, I have to do work. We called that electrostatic energy, $$U$$.

Here we will evaluate that energy in terms of the electric field.

## Energy in terms of electric field

Continuing the charged plates example. If the plates are close together $$h\ll\sqrt{A}$$, we can approximate the field between these plates as that coming from a pair of infinite planes.

We saw that the electric field in between the plates is $$E=\sigma/\varepsilon_0$$.

We now pull the top plate upwards over distance $$x$$. This creates an electric field, in the shaded area, that wasn’t there before. That field has the same strength $$\sigma/\varepsilon_0$$, because the charges on the plates stayed the same.

These plates are attracted to each other due to the opposite charges. The force on the charge on the top plate is due to the electric field of the bottom plate only ($$\frac{\sigma}{2\varepsilon_0}$$). To move the top plate upwards, we need to overcome this electric field.

The electric field is defined as force per unit charge, an with $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec E \parallelsum \vec F$$, the dot-product becomes a multiplication of scalars \begin{align*} \frac{\sigma}{2\varepsilon_0} &= -\frac{F}{Q} \\ \Rightarrow F &=-\frac{\sigma}{2\varepsilon_0}\,Q & \left(Q=-\sigma\,A\right) \\ &= \frac{1}{2}\frac{\sigma^2}{\varepsilon_0} A & \left(\times\frac{\varepsilon_0}{\varepsilon_0}\right) \\ &= \frac{1}{2}\varepsilon_0\frac{\sigma^2}{\varepsilon_0^2}\,A\ & \left(\frac{\sigma}{\varepsilon_0}=E\right) \\ &= \frac{1}{2}\varepsilon_0\,E^2\,A \end{align*}

The work to move the top plate, follows as the force $$F$$ times the distance $$x$$ \begin{align} W &= F\,x \nonumber \\ &= \frac{1}{2}\varepsilon_0\,E^2\,\underline{A\,x} \label{eq:W} \end{align}

## Field energy density

The term $$Ax$$ in equation $$\eqref{eq:W}$$ is the new volume, in which we created electric field. The work per unit volume $$\shaded{ \frac{W}{\rm{volume}} = \frac{1}{2} \varepsilon_0 E^2 } \quad \left[\frac{\rm{J}}{\rm m^3}\right] \nonumber$$

Since the work created electric field, we call this field energy density.

It can be proven that this applies to the general case, not only for this charge configuration.

## A new expression for energy

This gives us a new way of looking at the energy that it takes to assemble charges. Earlier, we calculated the work to put charges in place. Now, we could calculate that the electrostatic potential energy by integrating over volume $$\shaded{ U = \iiint_\text{all space} \frac{1}{2} \varepsilon_0 E^2\,dV } \nonumber$$

We look at as the energy in the electric field, and we no longer think of it as the work that you have done to assemble the charges.

To calculate the electrostatic energy, we can now choose between

• calculating the work to bring all the charges in place, or
• take the electric field everywhere in space, and integrate over all space.

### Example

The total potential energy of these parallel plates $$U = \iiint \frac{1}{2} \varepsilon_0 E^2\, dV \nonumber$$

Since the outside field is zero everywhere, and the internal field $$\vec E$$ is constant \begin{align*} U &= \int_\text{inside}\ldots+\int_\text{outside}\ldots \\ &= \frac{1}{2} \varepsilon_0 E^2\iiint_\text{inside} dV + 0 \\ &= \frac{1}{2} \varepsilon_0 E^2\,Ah \\ &= \frac{1}{2} \varepsilon_0 EA\,\underline{Eh} \end{align*}

Remember: the potential difference is defined as

\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} V &= \int \vec E\cdot d\vec l & \left(\vec E\parallelsum d\vec l\right) \\ &= \underline{E\,h} \end{align*} \nonumber

Substituting the potential difference between the plates $$V$$, in the expression for $$U$$ \begin{align*} U &= \frac{1}{2} \varepsilon_0 EA\,V & \left(E=\frac{\sigma}{\varepsilon_0}\right) \\ &= \frac{1}{2} \bcancel{\varepsilon_0} \frac{\sigma}{\bcancel{\varepsilon_0}} A\,V & \left(\sigma A=Q\right) \end{align*}

So $$\shaded{ U = \frac{1}{2} Q V } \nonumber$$

## Note

All of these are equivalent:

• the total energy in the field,
• the total work you have to do to assemble these charges,
• the total work you have to do to create the electric fields.

## Complete example (Lec.12)

Two very large conducting plates with a thickness, and separation $$d$$. Charged with density $$+\sigma$$ and $$-\sigma$$.

In the conductor, there is no current, so $$E=0$$. We already derived that the field inbetween the plates using Gauss’ law and is $$E=\frac{\sigma}{\varepsilon_0}$$. The electric field outside the plates is very close to zero. We found that through the superposition principle.

### Potential difference

What is the potential difference $$V_P-V_S$$ over this capacitor?

The potential difference is \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} V_P-V_S &= \int_P^S \vec E\cdot d\vec l & \left(\vec E\parallelsum d\vec l\right) \\ &= E\, d & \left(E=\frac{\sigma}{\varepsilon_0}\right) \\ &= \frac{\sigma}{\varepsilon_0}d \end{align*}

If we had chosen another path, we would have found the same answers, because we’re dealing with conservative fields that are path independent.

What is the potential difference between point $$P$$ and $$T$$? \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} V_P-V_S &= E\, d & \left(E=0\right) \\ &= 0 \end{align*}

### Capacitance

What is the capacitance? It is defined as the charge on one plate divided by the potential difference \begin{align*} C &= \frac{Q}{\Delta V} = \frac{Q}{V_P-V_S}, & \left(Q=\sigma A\right) \\ &= \frac{\bcancel{\sigma} A}{\frac{\bcancel{\sigma}}{\varepsilon_0}d} = \frac{A\,\varepsilon_0}{d} \end{align*}

### Potential energy

What is the electrostatic potential energy? That is the work you have to do to assemble the charges. You can also look at the energy that it takes to create that electric field. For that we derived \begin{align*} U &= \frac{1}{2} Q\,\Delta V, & \left(Q=\sigma A\right) \\ &= \frac{1}{2} \sigma A\,\Delta V, & \left(\Delta V=\frac{\sigma}{\varepsilon_0}d \right) \\ &= \frac{1}{2} \sigma A\,\frac{\sigma}{\varepsilon_0}d \\ &= \frac{\sigma^2\,A\,d}{2\varepsilon_0} \end{align*}

### Charge location

Where is the charge located? On the upper plate: it can’t be inside the plate, because if you make a Gaussian surface around that charge, it will tell met that $$\oint\vec E\cdot d\vec A\neq 0$$ because there is charge inside. But we know the electric field must be zero in the conductor, so the closed surface integral must be zero.

It can’t be on the top surface either, because if I make myself a small pillbox Gaussian surface through the top of the upper plate, the electric field is zero on the top and bottom, and on the sides $$d\vec A\perp \vec E$$. The closed surface integral must be zero, so according to Gauss’ law the contained charge is zero.

That only leaves that nature puts all positive the charge right at the bottom of the top plate, and the negative charge at the top of the bottom plate.

## Example with dielectric (Lec.12)

Aw we saw before, two large parallel plates, that we charge with a power supply. Then we disconnect the power supply. That means the charge is trapped and can never change.

Remember, the formula’s [ 1, 2, 3, 4, 5, 6]

\begin{align} Q_f &= \sigma_f\,A \label{eq:Qf} \\ E &= \frac{\sigma_f}{\varepsilon_0\,\kappa} \label{eq:E} \\ \Delta V &= E\,d \label{eq:dV} \\ C &= \frac{Q_f}{\Delta V} = \frac{A\varepsilon_0}{d}\kappa \label{eq:C} \\ U &= \tfrac{1}{2}Q_F\,\Delta V =\tfrac{1}{2}C\,(\Delta V)^2 \label{eq:U} \end{align}

### Double the distance

With the power supply disconnected, double the distance between the plates, while the dielectric is just air $$\kappa=1$$.

#### Electric field?

What happens with the electric field? $$\vec E$$ can’t change because $$\sigma_f$$ cannot change, since the power supply is disconnected. Based on equation $$\eqref{eq:E}$$, the electric field stays the same.

#### Potential difference?

The potential difference must increase by a factor of 2, based on equation $$\eqref{eq:dV}$$, because we double the distance

#### Capacitance?

The capacitance must decrease by a factor of 2, based on equation $$\eqref{eq:C}$$, because the potential difference doubled.

#### Electrostatic potential energy?

The electrostatic potential energy must increase by a factor of 2, based on equation $$\eqref{eq:U}$$, because the potential difference doubled. When we separate the plates, we do work that increased $$U$$.

### Change the dielectric

Back to the original distance $$d$$, and change $$\kappa=3$$ by moving in a dielectric.

#### Electric field?

What happens with the electric field? $$\vec E$$, based on equation $$\eqref{eq:E}$$, the electric field becomes a third. This is because on the dielectric you introduce opposite charges, that create an opposite induced electric field.

#### Potential difference?

The potential difference must decrease by a factor of 3, based on equation $$\eqref{eq:dV}$$, because the electric field decreased by a factor of 3.

#### Capacitance?

The capacitance must increase by a factor of 3, based on equation $$\eqref{eq:C}$$, because the potential difference decreased by a factor 3.

#### Electrostatic potential energy?

The electrostatic potential energy must decrease by a factor of 3, based on equation $$\eqref{eq:U}$$, because the potential difference decreased by a factor 3. That means that by moving in the dielectric, I did negative work. So in a way, there is a force that pulls it in.

### With power supply connected

What could go through these same exercises, but with the power supply connected. That implies the potential difference can’t change. As a result, you’ll find very different results when increasing $$d$$. E.g., $$E$$ will go down by a factor of two. We leave this as an exercise to the reader.