I was recently studying Electrostatics from the book Introduction to Electrodynamics by David J Griffiths whose chapter 2 is dedicated to the subject. I am stuck at a point in that book from where I cannot move on. The concept is about the energy in a continuous charge distribution. The book says in the page 93 that the work done to assemble a continuous charge distribution is

$$ \begin{equation} W ={1\over 2} \int{\rho Vd\tau} \end{equation} $$

where the $\rho$ is the volume charge density and the $V$ is the potential of the volume element $d\tau$.

Now Griffiths wants to eliminate the $\rho$ and $V$.

So he first recognizes the $\rho$ as

$$ \begin{equation} \rho = ({\vec{\nabla} \cdot \vec{E}})\epsilon_{0} \end{equation} $$

from Gauss’s law so that

$$ \begin{equation} W = {\epsilon_{0}\over 2}\int{(\vec{\nabla}\cdot\vec{E}) V d\tau} \end{equation} $$

Next he performs integration by parts and gets

$$ \begin{equation} W = {\epsilon_0 \over 2}[\oint V\vec{E}\cdot d\vec{a} - \int{\vec{E}\cdot \vec{\nabla}V}d\tau] \end{equation} $$

Next, he recognizes $V = -\vec{\nabla}\cdot\vec{E}$ and gets the following integral

$$ \begin{equation} W = {\epsilon_0\over 2}[\int{E^2 d\tau + \oint{V \vec{E}\cdot d\vec{a}}}] \end{equation} $$

At this point he makes an argument.

That is, what is the volume that we should integrate upon?

As the first equation suggests this should be the volume where the charge is located but he says that we can also take any volume we care to take because then the extra volume will contribute nothing since the charge is zero there. But at this point me makes a statement that literally confused the hell out of me.

He says if we increase the volume then the volume integral in the last equation will increase since the integrand is positive and since the work should remain the same as the work for just the volume confining the charge the surface integral must decrease. He also says since the $\vec{E}$ goes like $1/r^2$ and $V$ goes lke $1/r$ and the surface increases like $r^2$ the entire integral must decrease like $1/r$.

But all I ask is shouldn’t the same argument be applied to the volume integral?

The $E^2$ goes like $1/r^4$ and the volume $d\tau$ goes like $r^3$ shouldn’t then the entire volume integral go like $1/r$?

It seems like I am missing some thing here. Can you guys help me?

  • $\begingroup$ Note that on the surface, the tendency of $E$ to go like $1/r^2$ applies everywhere, but in the volume it doesn't... $\endgroup$ Commented Apr 24 at 5:18

2 Answers 2


For the surface integral, only the integrand's value at the boundary of the surface matters. On the other hand, for the volume integral, you add however diminishing contributions you get from each volume element


The argument in Griffith's textbook goes as follows.

At the level of the formula $$\begin{equation} W ={1\over 2} \int{\rho Vd\tau} \tag{1} \end{equation} $$ we could choose any volume containing the region where the charge density is different from zero without changing the value of the integral. Let's start with a volume $V_1$ defined as the region's volume where $\rho \neq 0$. By using Gauss' law, we can transform the integral in equation ($1$) into the sum of a volume integral and a surface integral, as in $$ \begin{equation} W = {\epsilon_0\over 2}\left[\int_{V_1}{E^2 d\tau + \oint_{\partial V_1}{V \vec{E}\cdot d\vec{a}}}\right]. \tag{2} \end{equation} $$

Now, let's consider another volume $V_2>V_1$, containing the region of volume $V_1$. Clearly, $$ {\epsilon_0\over 2}\left[\int_{V_2}{E^2 d\tau + \oint_{\partial V_2}{V \vec{E}\cdot d\vec{a}}}\right]-{\epsilon_0\over 2}\left[\int_{V_1}{E^2 d\tau + \oint_{\partial V_1}{V \vec{E}\cdot d\vec{a}}}\right] = 0\tag{3} $$ because in the the volume $V_2-V_1$ we have $\rho=0$. We can use equation ($3)$ to write the integral over the surface $\partial V_1$ in equation ($2$) as $$ \oint_{\partial V_1}{V \vec{E}\cdot d\vec{a}}={\epsilon_0\over 2}\left[\int_{V_2-V_1}{E^2 d\tau + \oint_{\partial V_2}{V \vec{E}\cdot d\vec{a}}}\right]. $$ The analysis of the asymptotic behavior of the field and potential tells us that in the limit of an infinite volume $V_2$, the first integral of the previous formula has a finite value (being the integral of a positive function), but the surface integral vanishes (regular surface area grows as $r^2$).

Therefore, the surface contribution of the border of volume $V_1$ can be expressed as a volume integral over the complement to the whole space of $V_1$, ${\mathbb R}^3-V_1$ of the positive function $E^2$. Once combined with equation $(1)$, we get the usual formula containing only the volume integral over ${\mathbb R}^3$ of the squared electric field.

  • $\begingroup$ Can you please clarify the last two paragraphs a little bit? I still don't seem to have grasped the idea of why the first integral will have a finite value. After all should the $E^2$ not go like $1/r^4$? Where as the volume like $r^3$ so the entire contribution of the volume integral go like $1/r$? $\endgroup$ Commented Apr 24 at 15:43
  • $\begingroup$ @MayukhNath An integral over an infinite volume of a positive function can only give a positive finite number (if the function decays fast enough) or $+\infty$, if the function doesn't decay fast enough $\endgroup$ Commented Apr 24 at 20:53

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