A mathematical doubt in the expression for the energy of a continuous Charge Distribution.

I'm studying Introduction to Electrodynamics by David J.Griffiths and here I'm gonna present his method of calculating the the energy stored (or work done in assembling) of continuous charge distribution.
$$\displaystyle W = \frac{1}{2} \int \rho V~d\tau$$
$$\displaystyle d\tau$$ stands for volume element.
By Gauss' Law $$\displaystyle \frac{\rho}{\epsilon_0} = \mathbf{\nabla} \cdot \mathbf{E}$$ so, $$\displaystyle W= \frac{\epsilon_0}{2} \int (\mathbf{\nabla}\cdot\mathbf{E})V~~d\tau$$
Now, using integration by parts and considering first function $$\displaystyle (\mathbf{\nabla} \cdot \mathbf{E})$$ and second function $$\displaystyle V$$ we can write things like this
$$\displaystyle W = \frac{\epsilon_0}{2} \int (\mathbf{\nabla}\cdot\mathbf{E}~~d\tau - \int \mathbf{\nabla}V ~~\int(\mathbf{\nabla}\cdot\mathbf{E})~d\tau ~~~~~~~~d\tau$$
Now, by The Fundamental Theorem For Divergences we have $$\displaystyle \int (\mathbf{\nabla}\cdot\mathbf{E})~~d\tau ~~~~~~= ~~~~~~~\oint_S \mathbf{E} \cdot d\mathbf{a}$$ and we can write the above equation as
$$\displaystyle W = \frac{\epsilon_0}{2} V \oint_S \mathbf{E}\cdot d\mathbf{a} - \int \mathbf{\nabla}V \oint_S \mathbf{E} \cdot d\mathbf{a}~~~~~~~d\tau ~~~~~~~~~~~~~~~~~ (1)$$
But in the book the équation (1) is written as $$\displaystyle W = \frac{\epsilon_0}{2} \left [ - \int \mathbf{E}\cdot(\nabla V) ~~d\tau + \oint_S V\mathbf{E}\cdot d \mathbf{a} \right ]$$
The problem is how he has converted $$\displaystyle \int \nabla V \oint_S \mathbf{E} \cdot d\mathbf{a}~~~~~d\tau$$ into $$\displaystyle \int \mathbf{E} \cdot (\nabla V) ~~~~~~d\tau$$.
I mean how converted that surface integral to get $$\displaystyle \mathbf{E}$$. I'm attaching an image of the book so that you can his arguments.

Thank you. Any help will be much appreciated.

Attachments

• 112.8 KB Views: 2
Last edited:

topsquark

Forum Staff
All right. You are having some problems with your integration by parts. I often do too, so let's do this a slightly different way.

Let's look at
$$\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau$$

Note that $$\displaystyle \vec{ \nabla } \cdot \left ( V \vec{E} \right ) = V \left ( \vec{ \nabla } \cdot \vec{E} \right ) + \vec{E} \cdot \left ( \vec{ \nabla } V \right )$$

So the integral becomes
$$\displaystyle W = \dfrac{ \epsilon _0}{2} \int \left ( \vec{ \nabla } \cdot \left ( V \vec{E} \right ) - \vec{E} \cdot \left ( \vec{ \nabla } V \right ) \right ) ~ d \tau$$

By Gauss' Law $$\displaystyle \int _{vol} \vec{ \nabla } \cdot \left (V \vec{E} \right ) ~d \tau = \oint _{ \sigma } V \vec{E} \cdot d \vec{a}$$

So we finally get
$$\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau = \dfrac{ \epsilon _0}{2} \oint _{ \sigma } V \vec{E} \cdot d \vec{a} - \dfrac{ \epsilon _0}{2} \int \vec{E} \cdot \left ( \vec{ \nabla } V \right ) d \tau$$

-Dan

All right. You are having some problems with your integration by parts. I often do too, so let's do this a slightly different way.

Let's look at
$$\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau$$

Note that $$\displaystyle \vec{ \nabla } \cdot \left ( V \vec{E} \right ) = V \left ( \vec{ \nabla } \cdot \vec{E} \right ) + \vec{E} \cdot \left ( \vec{ \nabla } V \right )$$

So the integral becomes
$$\displaystyle W = \dfrac{ \epsilon _0}{2} \int \left ( \vec{ \nabla } \cdot \left ( V \vec{E} \right ) - \vec{E} \cdot \left ( \vec{ \nabla } V \right ) \right ) ~ d \tau$$

By Gauss' Law $$\displaystyle \int _{vol} \vec{ \nabla } \cdot \left (V \vec{E} \right ) ~d \tau = \oint _{ \sigma } V \vec{E} \cdot d \vec{a}$$

So we finally get
$$\displaystyle W = \dfrac{\epsilon _0}{2} \int V \left ( \vec{ \nabla } \cdot \vec{E} \right ) d \tau = \dfrac{ \epsilon _0}{2} \oint _{ \sigma } V \vec{E} \cdot d \vec{a} - \dfrac{ \epsilon _0}{2} \int \vec{E} \cdot \left ( \vec{ \nabla } V \right ) d \tau$$

-Dan
$$\displaystyle \vec{ \nabla } \cdot \left ( V \vec{E} \right ) = V \left ( \vec{ \nabla } \cdot \vec{E} \right ) + \vec{E} \cdot \left ( \vec{ \nabla } V \right )$$

topsquark

Forum Staff
It's simply the product rule as used by the divergence. Consider the 1-D case:

$$\displaystyle \dfrac{d}{dx} ( f(x) g(x) ) = g(x) \dfrac{df}{dx} + f(x) \dfrac{dg}{dx}$$

We can look at this in terms of components. Let $$\displaystyle \partial _i$$ be the derivative in the ith direction (just like x, y, z except I'm using an i instead.)

This all looks a lot worse than it is. Don't try to inhale the formulas all at once...take it one step at a time.

The gradiant operator is $$\displaystyle \vec{ \nabla } = \hat{x} \partial _x + \hat{y} \partial _y + \hat{z} \partial _z = \sum_i \hat{i} \partial _i$$. (Where $$\displaystyle \hat{i}$$ is the unit vector in the "ith" direction.)
So the gradiant of a scalar B is
$$\displaystyle \vec{ \nabla } B = \sum_i \hat{i} \partial _i B$$.

The divergence of $$\displaystyle \vec{A}$$ would be writen as
$$\displaystyle \vec{ \nabla } \cdot \vec{A} = \left ( \sum _i \hat{i} \partial _i \right ) \cdot \left ( \sum_i \hat{i} A_i \right ) = \sum_i \partial _i A_i$$

So putting it together:
$$\displaystyle \vec { \nabla } \cdot \left ( V \vec{E} \right ) = \left ( \sum_i \hat{i} \partial _i \right ) \cdot \left ( \sum_i \hat{i} V E_i \right ) = \left ( \sum_i \hat{i} \partial _i V \right ) \cdot \left ( \sum_i \hat{i} E_i \right ) + V \left ( \sum_i \partial _i E_i \right )$$

$$\displaystyle = \left ( \vec{ \nabla } V \right ) \cdot \vec{E} + V \left ( \vec{ \nabla } \cdot \vec{E} \right )$$.

Here's what I could find on the internet. Focus on the sub-section "Cartesian Coordinates" (in the "Definition in Coordinates" section) and the "Properties" section. However I'd advise looking in a Calculus III textbook.

-Dan

Last edited:

Sir your little note that it’s just a product rule has solved all my problem at once. I didn’t even look at the link.

topsquark

topsquark

Forum Staff
Sir your little note that it’s just a product rule has solved all my problem at once. I didn’t even look at the link.
You mean I had to type all of that and for nothing?! You ungrateful little..

Just kidding. I'm glad it helped.

-Dan

Last edited: