[Physics] Confusion about energy of a continuous charge distribution

electromagnetismelectrostaticsenergy

I'm reading Griffiths's E&M book and he derives the following result for the energy required in assembling a continuous charge distribution $$W=\dfrac{\epsilon_0}{2}\left(\iiint E^2\ dV +\oint V\textbf{E}\cdot d\textbf{A} \right)$$
where $E$ is the electric field, $V$ is the potential and $\textbf{A}$ is the normal area vector. I understood the derivation but he then claims that as we extend our volume integral to all of space, the surface integral vanishes and we are left with $$W=\dfrac{\epsilon_0}{2}\iiint E^2\ dV $$
I really don't understand the reasoning behind this. I searched elsewhere and someone tried to justify this by imagining integrating over a spherical volume. The claim was that since $E$ goes like $1/r^2$, $V$ goes like $1/r$ and $A$ goes like $r^2$ so that the surface integral goes down by $1/r$. But this doesn't make much sense, since the volume integral must also go down by $1/r$? ($E^2$ goes like $1/r^4$ and the volume goes by $r^3$?). Is there a more rigorous or better explanation for the reasoning behind this?

Best Answer

In the image, I have shows the region over which we're integrating.The orange line is the area over which were doing the surface integral and the blue area is the region over which we're doing the volume integral. Assume that the charge distribution is at the centre of the sphere.

Now, Suppose you make the radius of the sphere infinitely large. When your doing the surface integral, you are always the same distance away from the centre, and hence you can neglect it because as you said, it is proportional to $\frac{1}{r}$.

But now look at the volume integral. Yes, near the outer surface the integral becomes negligible (because it's proportional to $\frac{1}{r}$), but notice that since it is a volume integral, we must also do the integration over the inner volume, which also includes near the surface, which is not negligible anymore. This is why you can ignore the surface integral and not the volume integral.

Related Solutions

Electrostatics – Why Is Energy Associated with the Electric Field Given as $U = \dfrac{\epsilon_0}{2} \int E^2 dv$?

How can the energy required for squeezing be the energy required for assembling the charges?

The energy required for squeezing the sphere from initial radius $r_0 \to (r_0-dr)$ is not the energy required to assemble the charges. He merely calculated the change in energy (or the work required) each time you squeezed the sphere by a displacement of $dr$. To find the work required for assembling the charges you'll have to integrate over all space (As in you had a sphere of Radius $\to$ infinity and squeeze it to the desired radius).

Mr. Purcell just squeezed the sphere & found the energy required to squeeze the shell against the repulsive force! ...

He found how much work is required to squeeze a sphere a little bit, and again the entire energy required to assemble the charges is to sum over all this little bit of energy from infinity to the desired radius.

It's some how like recursion; Suppose you know the energy $U$ required to assemble the charges at their current state, and now you want to squeeze them a little bit closer so you must invest extra work: $$U(r_0-dr)=U(dr)+U(r)=2U(dr)+U(r+dr)= \cdots =\text{[integration over all} \qquad U(dr) \qquad \text{from infinity to the current radius]} + U\text{(infinity)} \text{[Which is 0]}$$

If you go backwards, and let the electrons move freely to infinity (If they can do that), you'll gain back energy which is basically equal ,in value, to the work required in assembling them in the first place.

How can $4\pi\sigma$ be the field? How did he deduce it? Moreover, can anyone tell me what he is saying in the second para?

I think you have an old version of the book; The one I have has $\dfrac{\sigma}{\epsilon_0}$ instead of that. By the way, there is a "1" near the sigma, there must be an explanation at the bottom of the page I believe.

Hope this helps!

Electromagnetism – Poynting’s Theorem and Energy Conservation

Let the electromagnetic field has $u$ as its energy density (amount of energy per unit volume in the field) and let $\bf S$ represents the energy flux- the amount of energy per unit time flowing across a unit area perpendicular to the flow).

Now electromagnetic field can interact with matter and do work on them; so this energy interaction must be considered when discussing energy conservation.

The field energy inside a volume $V$ is $\displaystyle \int_V u\,\mathrm dV\;.$ Amount of energy flowing out of the volume $V$ is given by $\displaystyle \int_\Sigma \mathbf S\cdot n\,\mathrm da \;.$

Now, the work done per unit time by the field on the matter inside the volume $V$ is given by $\displaystyle \int _V Nq(\mathbf E + \mathbf v\times \mathbf B)\cdot \mathbf v\,\mathrm dV$ where $N$ is the number of particles per unit volume; this can be written as \begin{align}\int _V Nq(\mathbf E + v\times \mathbf B)\,\mathrm dV& = \int_V Nq\mathbf E\cdot v\,\mathrm dV\\ &= \int_V \mathbf E\cdot \underbrace{(Nq\mathbf v)}_\textrm{current-density}\,\mathrm dV\\ &= \int_V E\cdot \mathbf J\,\mathrm dV \end{align}

So, the continuity equation is written thus: $$\underbrace{-\frac{\partial}{\partial t}\int_V u\,\mathrm dV}_\textrm{rate of change of energy inside volume $V$}=\underbrace{\int_\Sigma \mathbf S\cdot \mathbf n\,\mathrm da}_\textrm{amount of $\mathbf{field\, energy}$ flowing out of volume $V$ per unit time} + \underbrace{\int_V \mathbf E\cdot \mathbf J\,\mathrm dV}_\textrm{work done per unit time by the field on the matter inside volume $V$} \;. $$ This definitely implies the conservation of energy and moreover, the equation which OP wrote in his query is the offshoot of this same continuity equation.

Energy density and Poynting Vector:

We can write the differential form of the continuity equation above as:

$$-\frac{\partial u}{\partial t}= \textrm{div}\,\mathbf S + \mathbf E\cdot \mathbf J$$ (since, all the derivation is done in Cartesian coordinates, then $\textrm{div}\equiv \mathbf \nabla$).

Or, we can write $$\mathbf E\cdot \mathbf J= -\frac{\partial u}{\partial t}-\mathbf E\cdot \mathbf J\tag 1$$

In order to find out what $u$ and $\bf S$ actually are, it is assumed that they solely depend on the fields $\bf E$ and $\bf B\;.$

Now, using $$\mathbf J = \epsilon_oc^2\,\mathbf \nabla\times\mathbf B- \epsilon_0\frac{\partial\mathbf E}{\partial t}\,,$$ we can write $\mathbf E\cdot \mathbf J$ as $$\mathbf E\cdot \mathbf J= \epsilon_oc^2\,\mathbf E\cdot (\mathbf \nabla\times\mathbf B)- \epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\;.$$

Now, \begin{align}\mathbf E\cdot (\mathbf \nabla\times\mathbf B)&= (\mathbf \nabla\times\mathbf B)\cdot \mathbf E\\ &= \mathbf \nabla\cdot (\mathbf B\times\mathbf E)+ \mathbf B\cdot (\mathbf \nabla \times \mathbf E)\;.\end{align}

Therefore, \begin{align}\mathbf E\cdot \mathbf J&=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\\ &=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E)+ \epsilon_oc^2 \mathbf B\cdot \left(-\frac{\partial\mathbf B}{\partial t}\right)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E) - \frac{\partial}{\partial t}\left(\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B+ \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E\right) \tag 2\end{align}

Comparing $(1)$ and $(2)\,,$ we get $$u = \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E+\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B\\ \mathbf S=\epsilon_oc^2\,\mathbf E\times\mathbf B \;.$$

This vector $\bf S\,,$ that is the energy flux vector we considered at the beginning of the making up of the continuity equation, is called the Poynting Vector.

The whole derivation is based on the continuity equation, which is the mathematical expression of conservation of energy.

OP can conclude the equation he wrote in the question by converting the differential form of equation $(2)$ into integral form; using the definition of $u$ and $\bf S$ derived above and lastly using Gauss's theorem.

References:

$\bullet$ Lectures on Physics by Feynman, Leighton, Sands.

you're saying that energy conservation is needed as an independent equation, in order to deduce that the Poynting vector gives energy flux. I don't think this is true. I think the energy continuity equation can be shown as a consequence of Maxwell's equations.

No! Never, ever I've made such statement. I've answered to your query I know the books say it represents energy flux, but how do you prove it represents energy flux?; that's it. What I've done is to clearly present the continuity equation before OP and have tried to clearly interpret each term and how they together imply conservation of energy. The continuity equation can indeed be derived from Maxwell's equation and this is what I've done to define $u$ and $\bf S\;.$ I'm really pretty upset on the allegation OP made.

But this is not a proof of conservation of energy.

It's indeed a proof of conservation of energy.

Contrast this with conservation of charge within electromagnetism. Conservation of charge is a consequence of Maxwell's equations, it does not have to be assumed independently.

Okay, let's fix this thing. Conservation of charge implies this continuity equation(this can be derivable from Maxwell's equations):

$$-\frac{\partial \rho_V}{\partial t}= \mathbf \nabla \mathbf J\;.$$

This suggests conservation of energy would look like:

$$-\frac{\partial u}{\partial t}= \mathbf \nabla \mathbf S\;.$$

But that is incomplete as it is the total energy and not just field energy which is conserved; the interaction of field with matter has to be taken into consideration.

is energy conservation a consequence of Maxwell's equations or something assumed independently from it?

Yes. The continuity equation meant for conservation of energy is derived from Maxwell's equation; that's how the definition of $u$ and $\bf S$ have been computed. OP must be confused to see in my answer first the continuity equation as if I'm saying continuity equation should be considered independently in order to consider conservation of energy. No! That was intentionally done in order to conceive that the Poynting Theorem indeed implies conservation of energy.