I know this question has been asked twice before, but i didnt find any satisfying answer there.

I learnt in my class that the energy stored in the capacitor per unit volume comes out to be

$\dfrac{\mathrm{d} U}{\mathrm{d} V} = (1/2)\epsilon E^{2}$

our teacher then said this is a general statement and is true for every configuration.

But, I thought that Internal Energy came from the interaction of charged particles. So, what does "Energy stored in electric field actually signify?"

We went on to prove the self energies of spheres using this formula which came out to be the same as if we took the work done by interaction energy while assembling the sphere. then, we used this to find the capacitance of some complex arrangements.

but, I never understood the formula itself. I would love it if someone here explains how it is derived in a general case and what it signifies. I may not understand its formal derivation since i am just a high school student but would love to have some insight on the derivation at least.


4 Answers 4


In order to move charges of equal sign that were separated by an infinite distance to some finite distance, you need to do work.

If you arrange an electric field $\vec{E}(x,y,z)$ in space by bringing in a large number of small electric charges (positive and negative) from infinitely far away to some arrangement, this takes work. For example, for a plate capacitor, you need to do positive work to collect positive charges close together on one plate, and positive work to collect negative charges on the other plate, and then negative work to bring the plates close together. The combined positive work from combining the charges on the plates will be larger what you get back from bringing the plates near each other, unless the plates touch and the charges cancel each other.

It turns out that the net work $W$ done to arrange the charges is equal to the integrated $dU/dV$ over all space: $$ W = \frac12 \epsilon_0 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} E^2(x, y, z)\, dx\,dy\,dz. $$ This equivalency of work (or energy) and the energy density of the electric field only exists in the above integral form. You cannot take an arbitrary cubic millimeter of space and somehow extract the field energy from it without affecting the field in the space around it as well.

As you have probably already done, you can verify that it works out for simple cases such as plate capacitors. Unfortunately, the general case requires mathematical knowledge (vector calculus) that you don't learn in high school. Once you have the vector-calculus toolbox and the description of electrostatics in vector calculus, the derivation is just a few lines. If you go and study physics in university, you'll learn this by the end of the first year.

I left out the effect of dielectrics ($\epsilon=\epsilon_0\epsilon_{\mathrm r}$). You can use the same reasoning if you start from an infinite dielectric medium, but it's a more tricky if $\epsilon_{\mathrm{r}}$ varies over space -- and unfortunately again too difficult for high-school level.

  • $\begingroup$ I'd like to add (as I had to struggle with such a concept in the past) that the so mentioned expression "energy stored by the fields" is just a way to say that there's a manifestation of the amount of energy somebody/something has to do, in order to maintain those fields. $\endgroup$ Commented Jul 17, 2020 at 2:34
  • $\begingroup$ Ok. Thank you for clearing my doubt. So, you are saying that this formula can be used only to derive the whole interacting energy in a system if you integrate it over all of space and that 'energy stored in some particular region of space' does not make any sense? I got confused since many questions were poorly framed in my books, for example asking for energy stored between two concentric spheres and then outside them........... $\endgroup$
    – RadP
    Commented Jul 17, 2020 at 2:53
  • $\begingroup$ "In order to move charges of equal sign that were separated by an infinite distance to some finite distance, you need to do work." Is this similar to gravitational potential energy, in reverse kinda? Like "to move two masses from finite to infinite distance" you need to do work? $\endgroup$
    – J Kusin
    Commented Sep 12, 2020 at 21:25

Let's take the 2 plates of the capacitor as planes perpendicular to the $x$ axis to simplify.

The force on a charge $q$ between them is: $F = qE$. If the charge has a volume $V$, and the charge density is $\rho$, => $q = \rho V$.

But from the Gauss law: $$\nabla.E = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} +\frac{\partial E_z}{\partial z} = \frac{\rho}{\epsilon}$$

As we assume an electric field only in the $x$ direction:

$$F = \rho V E= \epsilon \frac{\partial E}{\partial x} E V$$

The elementary work in the charge for a displacement $dx$ is $dW = F dx$

So:$$\frac{dW}{V} = \epsilon \frac{\partial E}{\partial x} E dx = \frac{\epsilon}{2}\frac{\partial (E^2)}{\partial x} dx$$

Integrating to a finite work:

$$\frac{W}{V} = \frac{\epsilon}{2}E^2$$

  • $\begingroup$ I don't think this reasoning works very well for a field in a charge-free vacuum ($\rho=0$), like inside a plate capacitor. You'd need to integrate over the charge on the plates (Dirac delta function) as well and reason about what it means to do work on a charged plane in a field that is created by that same plane. $\endgroup$ Commented Jul 17, 2020 at 8:10
  • $\begingroup$ It is only a test charge, and the energy between the plates is equivalent to the work done if it was there. It is the same idea for a gravitational potential in the vacuum. The charges on the plates plays no role. $\endgroup$ Commented Jul 17, 2020 at 13:25
  • $\begingroup$ You write that the work is $dW=\rho V E dx$, where $\rho$ is the source of the electrical field. That doesn't read like a test charge to me. $\endgroup$ Commented Jul 17, 2020 at 13:32
  • $\begingroup$ E is the total electric field, as required by Maxwell equations. It includes the field of the test charge and the field of the capacitor. Each point of the test charge is affected by the field of the capacitor and the field of the other points of the charge. $\endgroup$ Commented Jul 17, 2020 at 17:13

A2A, for the intuitive understanding of energy stored in the electric field.

I've read a book by Richard Feynman (forgot which one), where he describes a charged particle in an electric field to be likened to a ball suspended in mid air with infinitely long springs attached to it. When you shake the ball, the vibrations of the ball are carried outwards by the springs, this is electromagnetic radiation.

When you have two balls, their springs push against one another, bring them closer, and the springs are squeezed up more. The springs store the energy to push the balls apart.

Similarly, the electric field stores the energy for any electrostatic set up. Leaving the analogy, we also have an attractive force when we have particles of opposite charges, which means that the 'springs' pull in rather than push out.


Energy stored in an electric field - Means the Potential Energy (electric) in that space.

You do not even need to know volume for energy stored in electric field. It has three equations. PE = (1/2) C[V(net)^2] where C is capacity and V is 'electric potential'. I am sure you can find the other two online.

C has units of F or Farad V has units of Volt or Joules/Coulombs

  • $\begingroup$ This doesn't really answer the question. $\endgroup$
    – Chris
    Commented Sep 13, 2020 at 0:30

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