Question

a question about Physics & Reductionism:
all physics are based in reductionism, and we construct our calculation systems in physics on it!
in EM for calculating the energy of a charge configuration we use : W=integral(epsilon0. E^2. dv).
we know that this equation is based on all charges and the field E is too!
but the energy is't! its mean that the total E is sum of are Ei (i=1, N) but the energy =(E1+E2+...+En)^2. but energy is not based on superposition principle!
what's wrong?

Answer 1

I'm not quite sure why you would expect the energy density to obey the superposition principle.

Answer 2

Just because I'm curious, could you (or someone else) clarify the problem that's been raised above?

Answer 3

First off, there's an important factor of 2 missing from W:

\[W= = \int \frac{1}{2} \epsilon_0 E^2 dV\]

Beyond that though, you can calculate the energy by squaring the total electric field from all sources. When you do, you get two types of terms: The first are just the integrals over the square of the electric field produced by each individual charge

\[W_0= = \sum_i \int \frac{1}{2} \epsilon_0 E_i^2 dV\]

Each of these terms are constant(technically infinite, but still constant in the sense that they don't depend on the locations of the charges.). Since only differences in energy are physically meaningful, we ignore constant terms in the energy in most situations.

The second set of terms are the cross-terms

\[W_0= = \sum_{i\neq j} \int \frac{1}{2} \epsilon_0 E_i\cdot E_j dV\]

When you integrate these terms, they turn into the usual potential energies
(kqq/r).

Answer 4

Those terms aren't infinite if you're dealing with a continuous charge distribution, which is the case when you use the formula
\[U =\frac{1}{2} \int \epsilon_0 E^2 dV\]
The reason it doesn't obey the superposition principle is that two distributions, when superimposed on top of one another, contribute each of their respective energies plus the energy due to their interaction.

Answer 5

Apart from the formula for the electric field energy only being applicable for continuous charge distributions, we are in total agreement. The potential energy resides in the fields.

Answer 6

I'm not arguing that the electric field energy is only applicable for a continuous charge distribution. You said yourself that the integral diverges if you're dealing with point charges. That volume integral is meaningless if a single point charge is in the bounds of the volume being integrated over. If you were considering point charges, then sure, you could use a discrete sum over the interaction energies of all of them, but not an integral like that. It's more a math quibble than anything.

Answer 7

True enough, but physically we do this kind of stuff all the time. The electric potential produced by a line charge, for example is logarithmnically divergent as the length of the line gets large, and we use a similar trick to generate a finite potential.

and there are physical objects(like electrons) that we believe to be pointlike....

Answer 8

The formula is
\[ U = \frac{1}{2} \int \epsilon_0 E^2 dV \]
We use it to calculate the energy stored in the electric field permeating a given volume of space. If that volume contains point charges, E becomes infinite at those locations. I'm not arguing that this isn't a major failing of classical electrodynamics, it is. Sometimes the description of an object as a point charge does not mesh with classical electrodynamics, this being one of those cases. Therefore, if you're talking about discrete charges, you need to use a discrete sum. Otherwise, this integral diverges. At the end of the day, U needs to equal something, like 17 or 300. A number. And it won't, if you use that integral in the presence of point charges.