Energy stored in a space with electric field is $\tfrac12 \varepsilon E^2$ per unit volume. Why?

Question

I'm a high school student and my book says that once it has been established that a region containing electric field $E$ has energy $\tfrac12 \varepsilon E^2$ per unit volume, the result can be used for any electric field whether it is due to a capacitor or otherwise. I understand how it is true for a parallel plate capacitor. But it's hard for me to accept that it is true for all possible geometries or settings of capacitors. Can anyone please explain how this is ought to be true. A derivation process for spherical capacitor would be very much appreciated. Or simply just an intuitive approach would be very helpful.

Answer 1

A general demonstration would require using advanced concepts in Maxwell's theory, so I will follow your suggestion of giving a demonstration for a spherical capacitor.

Consider a hollow sphere of radius $R$ with a charge $Q$ uniformly distributed on the surface. This can be regarded as a spherical capacitor, with the second conductor of infinte radius, so the capacity is just $C=4\pi\varepsilon_0 R$, and the energy stored in the capacitor is $U=\frac{Q^2}{2C} = \frac{Q^2}{8\pi\varepsilon_0R}$.

Now, the electric field generated by this configuration is just $0$ inside the sphere, and follows Coulomb's law outside, so it has a modulus $E=\frac{Q}{4\pi\varepsilon_0 r^2}$ on a generic point at distance $r>R$ from the center. As you can see, this is not uniform, as it depends on $r$, so different points have a different energy density; however, all the points on spherical shells of radius $r$ have the same energy densities. If you want to compute the total energy, you have to

1. compute the energy on a given spherical shell of radius $r$

2. sum up the contributions of all the spherical shells.

For point 1., you can just compute the volume $dV$ of a spherical shell of width $dr$ as volume of outer sphere minus volume of inner sphere: $$ dV = \frac{4}{3}\pi (r+dr)^3 - \frac{4}{3}\pi r^3 \approx 4\pi r^2 dr $$ where I have neglected terms like $dr^2$ or $dr^3$ which are very very small if $dr$ is small (our shell is in fact very thin!). The energy on this shell is finally $$ dU = \frac{\varepsilon_0}{2}E(r)^2 dV = \frac{\varepsilon_0}{2} \frac{Q^2}{16\pi^2\varepsilon_0^2 r^4} 4\pi r^2dr = \frac{Q^2}{8\pi \varepsilon_0 r^2}dr $$

For point 2 you need to know how to deal with integrals. They are essentially used to perform a sum of a given function for all the continuous values of the independent variable ($r$ in our case takes all the real values from $R$ to $\infty$). So the total energy is just $$ U = \int dU = \int_R^{\infty} \frac{Q^2}{8\pi\varepsilon_0 r^2} dr. $$ I assume you know how to deal with this, but if not, let me know and I can give more details. Solving this we get: $$ U=\frac{Q^2}{8\pi\varepsilon_0} \left[- \frac{1}{r} \right]_R^{\infty} = \frac{Q^2}{8\pi\varepsilon_0R}. $$ which is what we found from simple arguments above!