Physics Asked by An student on April 22, 2021
In Jackson’s electrodynamics Page (153), it is given that the total potential is the sum of potential due to free charge and potential due to dipole. Therefore,
begin{align*}
phi(x) & = frac{1}{4pivarepsilon_0}intfrac{rho(vec{x}’)}{|vec{x} – vec{x}’|}d^3x’
+ frac{1}{4pivarepsilon_0}int{vec{P}(vec{x}’)}cdotvec{nabla}’frac{1}{|vec{x} – vec{x}’|}d^3x’
& = frac{1}{4pivarepsilon_0}left[intfrac{rho(vec{x}’)}{|vec{x} – vec{x}’|}d^3x’
+ intvec{nabla}’cdotleft[frac{{vec{P}(vec{x}’)}}{|vec{x} – vec{x}’|} right]d^3x’
– intfrac{vec{nabla}’cdot vec{P}(vec{x}’)}{|vec{x} – vec{x}’|}d^3x’ right]
end{align*}
Using the Gauss’s divergence theorem, the 2nd term becomes
begin{equation*}
intvec{nabla}’cdotleft[frac{{vec{P}(vec{x}’)}}{|vec{x} – vec{x}’|} right]d^3x’ = intfrac{vec{P}(vec{x}’)cdot vec{ds}’}{|vec{x} – vec{x}’|}
end{equation*}
This term has been neglected while reaching Eq. (4.32) in the book. How we can do that?
Jackson is assuming that the medium fills all of space, so that the boundary term is taken at infinity, and it is assumed to vanish.
Whether it does actually vanish, i.e. whether a suitable surface-to-infinity limit exists that will take that integral to zero, is a separate (and often fairly subtle) question, and the answer may well be negative. However, I think it is justified in Jackson's usage in that section, since he is primarily concerned with the behaviour of the bulk, with the surface-charge effects embodied by that integral getting deferred to later sections of the book.
To be honest, I find this aspect of the presentation to be lacking sufficient clarity. However, you can clearly see that the bulk of §4.3 is treating the medium as infinite since its last paragraph draws a line of demarcation and talks about what happens if the medium is not infinite.
Generically, though, that term doesn't vanish, and there are important polarization surface charge effects (say, the charge marked 'bound surface charge density' in this Wikipedia page) at any region where the dielectric susceptibility changes discontinuously.
Answered by Emilio Pisanty on April 22, 2021
Jackson is talking about potential due to dipoles associated with some selected region of material medium. Such potential contains contribution due to all charges of dipoles implied, including those that are outside the boundary of the region. At the same time, it does not have contributions due to charges inside the boundary that come from other dipoles that are outside the boundary. This means Jackson's $Phi$ as he defines it and expresses in 4.30 is not necessarily potential due to all charges inside the boundary surface, but differs by some contribution that can be associated with the boundary surface. It is common to talk about surface polarization charges, with surface charge density $mathbf Pcdot mathbf n$.
The term that you found Jackson removed is exactly the potential due to these surface polarization charges.
There are two ways to justify the removal.
If the boundary is entirely outside the polarized body, then the term is zero and can be removed without any change in value of potential.
But if the boundary is at least partially inside the polarized body, when the term is removed, the resulting expression is no longer what Jackson defined $Phi$ to be, but a slightly different thing: it is the potential of all dipoles that have at least some charged part inside the boundary. A microscopic distinction, but with macroscopic difference in potential.
Personally, I think the most useful and clear case is the first one, where the potential is unambiguously potential of the whole polarized body and the microscopic distinction in the two ways of counting what belongs to the integral does not arise.
Answered by Ján Lalinský on April 22, 2021
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