# The Unapologetic Mathematician

## Energy and the Electric Field

Okay, now let’s consider the electric field from the perspective of energy. We have an idea that this might be interesting because we know that the field produces a force, and forces and energies interact in interesting ways.

So recall that if we have a “test charge” $q$ at a point $p$ in an electric field $E$ it experiences a force $F=qE(p)$. As we saw when discussing Faraday’s law, for a static electric field we can write $E=-\nabla\phi$ for some “electric potential” function $\phi$. Thus we can also write $F=-\nabla U$ for the potential energy function $U=q\phi$.

Now, say the field is generated by a charge distribution $\rho$; how much potential energy is contained in the force the field exerts on the little bit of charge at $p$? We count $U=\rho(p)\phi(p)$, but this is too much — half of it is due to the rest of the distribution acting on the bit of charge at $r$ and half of it comes from $r$ acting back. We can thus find the total potential energy by integrating

$\displaystyle U=\frac{1}{2}\int\limits_{\mathbb{R}^3}\rho(p)\phi(p)\,d^3p$

Now, Gauss’ law tells us that $\rho=\epsilon_0\nabla\cdot E$, so we substitute:

$\displaystyle U=\frac{1}{2}\int\limits_{\mathbb{R}^3}\epsilon_0(\nabla\cdot E)\phi\,dV$

Next we use a form of the product rule — $\nabla\cdot(fV)=(\nabla f)\cdot V+f(\nabla\cdot V)$ — and run it backwards to write:

\displaystyle\begin{aligned}U&=\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}\nabla\cdot(\phi E)-(\nabla\phi)\cdot E\,dV\\&=\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}\nabla\cdot(\phi E)\,dV+\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}(-\nabla\phi)\cdot E\,dV\\&=\frac{\epsilon_0}{2}\lim\limits_{R\to\infty}\int\limits_{B_R}\nabla\cdot(\phi E)+\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}E\cdot E\,dV\end{aligned}

where we evaluate the first integral over space by evaluating it over the solid ball of radius $R$ and taking the limit as $R$ goes off to infinity. The divergence theorem says we can write:

\displaystyle\begin{aligned}U&=\frac{\epsilon_0}{2}\lim\limits_{R\to\infty}\int\limits_{S_R}\phi E\cdot dA+\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}E\cdot E\,dV\\&=\frac{\epsilon_0}{2}\int\limits_{\mathbb{R}^3}E\cdot E\,dV\end{aligned}

where, as usual, we have taken the charge distribution to be compactly supported, so as our sphere gets large enough, the potential energy $\phi$ goes to zero. Yes, this is very hand-wavy, but this is how the physicists do it.

Anyway, what does this tell us? It means that a static electric field contains energy with a density

$\displaystyle u_E=\frac{1}{2}\epsilon_0\lvert E\rvert^2$

which we can integrate over any region of space to find the electrostatic potential energy contained in the field.

We can also check the units here; the electric field has units of force per unit charge:

$\displaystyle\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{C}\cdot\mathrm{s}^2}$

while the electric constant has units of farads per meter:

$\displaystyle\frac{\mathrm{s}^2\cdot\mathrm{C}^2}{\mathrm{m}^3\cdot\mathrm{kg}}$

Putting these together — two factors of $E$ and one of $\epsilon_0$ we find the units:

$\displaystyle\frac{\mathrm{s}^2\cdot\mathrm{C}^2}{\mathrm{m}^3\cdot\mathrm{kg}}\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{C}\cdot\mathrm{s}^2}\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{C}\cdot\mathrm{s}^2}=\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{m}^3\cdot\mathrm{s}^2}=\frac{\mathrm{J}}{\mathrm{m}^3}$

Joules per cubic meter — energy per unit of volume, just as we’d expect for an energy density.

February 14, 2012 -

1. [...] time we calculated the energy of the electric field. Now let’s repeat with the magnetic field, and let’s try to be a little more careful [...]

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2. [...] in sight! On the left, we’re taking the derivative of the combined energy densities of the electric and magnetic [...]

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3. dV has been subscripted in the formula about halfway down.

Comment by Hunt | February 18, 2012 | Reply

4. About the hand-wavy part: but the surface area goes to infinity, so is there reason to believe that it’s true?

Comment by Hunt | February 18, 2012 | Reply

5. Thanks for catching that; I fixed a few places it had gotten dropped entirely as well.

As for the hand-wavy bit, basically if you get far enough from a compact charge distribution it looks like a point charge, so $E$ falls off as the square of the distance, $\lvert E\rvert^2$ falls off as the fourth power, and the surface area only goes up as the square of the distance.

Comment by John Armstrong | February 18, 2012 | Reply