I am an undergraduate physics major. I've taken an introductory E&M course, worked through the electrostatics chapters of Griffiths, and have just started looking at some sections of Jackson.
In section 1.8 of the 3rd edition, Jackson uses Green's first and second identities to show how to get to a solution for electric potential. I will quote the result below:
$$ \int_V [-4\pi\Phi(x')\delta(x-x')+\frac{\rho(x')}{R}]d^3x' = \oint_S [\Phi \frac{\partial}{\partial{n'}}\frac{1}{R} – \frac{1}{R}\frac{\partial{\Phi}}{\partial{n'}}]da'. $$
To my understanding, if x is inside V, i.e., if x only goes to x' (where charge is located), then the first volume integral term evaluates to $-4\pi\Phi(x)$. Then, you move the second volume term to the right hand side and divide by the constant. Thus, a solution for electric potential within V:
$$ \Phi(x)= \frac{1}{4\pi\epsilon_0}\int_V\frac{\rho(x')}{R}]d^3x' + \frac{1}{4\pi}\oint_S [-\Phi \frac{\partial}{\partial{n'}}\frac{1}{R} + \frac{1}{R}\frac{\partial{\Phi}}{\partial{n'}}]da'. $$
Next, Jackson discusses how this result changes when you consider potential outside of S. If my reasoning above is accurate, then if x is always outside of V, the volume integral term with the dirac delta distribution will yield 0. Thus, that term vanishes. What's left is:
$$ 0= \frac{1}{4\pi\epsilon_0}\int_V\frac{\rho(x')}{R}]d^3x' + \frac{1}{4\pi}\oint_S [-\Phi \frac{\partial}{\partial{n'}}\frac{1}{R} + \frac{1}{R}\frac{\partial{\Phi}}{\partial{n'}}]da'. $$
- Is my understanding of the potential inside V accurate and complete?
- What does the above relationship have to do with being "consistent with the interpretation of the surface integral as being the potential due to a surface-charge density $\sigma = …$ and a dipole layer $D=…$."
Any help would be greatly appreciated.
Best Answer
Seems fine to me.
The boundary terms in the expression above can be interpreted as being caused by a surface charge density or a dipole layer on the boundary $S$. He is referring to (in the German 2nd ed. I have on my shelf) (1.22) and (1.27), which are $$ \left(\vec{E}_1 - \vec{E}_2\right) \cdot \vec{n} = 4 \pi \sigma $$ (i.e. the normal part of the jump of the electric field gives the surface charge density) and $$ \phi_1 - \phi_2 = 4 \pi D $$ (i.e. the jump of the potential gives the surface dipole density) (note that $D$ here is not the dielectric displacement, but the density of dipoles in the surface!). Now, in the expression above the normal part of the jump of the electric field across the surface would be $\frac{\partial\phi}{\partial n'}$ and the jump of the potential across the surface is just $\phi$ when the potential vanishes outside the volume $V$. This is what is meant by the sentence you are referring to.