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.
\(\displaystyle W = \frac{1}{2} \int \rho V~d\tau\)
\(\displaystyle d\tau\) 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: