Coursework question: Poisson's Equation for Electrostatics

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Homework Help Overview

The discussion revolves around the relationship between electric field and charge distribution as described by the equation \(\nabla \cdot E = \frac{\rho}{\epsilon_{o}}\). Participants are exploring whether this equation necessitates that the charge distribution occupies a specific volume, particularly in relation to Gaussian surfaces, and how the volume of charge is represented in the equation. The context is rooted in electrostatics, specifically referencing concepts from Griffiths' "Introduction to Electrodynamics".

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the implications of the equation regarding charge distribution and Gaussian surfaces. There is an exploration of the differential form of the equation versus its integral form, and how these relate to the concept of electric flux and total charge.

Discussion Status

Some participants have provided clarifications regarding the nature of the equation, noting that it holds at every point in space without requiring a Gaussian surface. Others have expressed appreciation for the explanations, indicating that the discussion has been productive in addressing misunderstandings about the material.

Contextual Notes

There is an acknowledgment that textbook explanations may sometimes lack clarity for learners, leading to ongoing questions about foundational concepts in electrostatics.

gulfcoastfella
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Homework Statement



(This isn't a homework problem; I'm just working through Griffiths' "Introduction to Electrodynamics" textbook, and can't find a very clear explanation.)

When the relationship between electric field and charge distribution are given by:

[tex]\nabla[/tex] [tex]\cdot[/tex] E = [tex]\frac{\rho}{\epsilon_{o}}[/tex]

does this require that the charge distribution fill a particular volume, like that bound by a Gaussian surface, where E is computed on the Gaussian surface? If not, how is the volume of the charge distribution accounted for in the equation? Basically, does the above equation demand that E be calculated on a Gaussian surface bounding a volume filled with the charge density [tex]\rho[/tex]?

The same question applies for Poisson's Equation...
 
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gulfcoastfella said:

Homework Statement



(This isn't a homework problem; I'm just working through Griffiths' "Introduction to Electrodynamics" textbook, and can't find a very clear explanation.)

When the relationship between electric field and charge distribution are given by:

[tex]\nabla[/tex] [tex]\cdot[/tex] E = [tex]\frac{\rho}{\epsilon_{o}}[/tex]

does this require that the charge distribution fill a particular volume, like that bound by a Gaussian surface, where E is computed on the Gaussian surface? If not, how is the volume of the charge distribution accounted for in the equation? Basically, does the above equation demand that E be calculated on a Gaussian surface bounding a volume filled with the charge density [tex]\rho[/tex]?

The same question applies for Poisson's Equation...

This is an equation which holds at every point in space. What it says is that the divergence of the E field at a point is equal to the volume charge density evaluated at that same point divided by epsilon zero. There is no gaussian surface or volume involved here.


If you choose to integrate both sides of the equation over a certain volume, then the volume integral of the left side of the equation (of the divergence) may be converted to a surface integral over the electric flux and the volume integral of the right side becomes the total charge enclosed in the volume divided by epsilon_0. This gives the integral form of Gauss' law. But the form you give above is the differential form and involves no volume, no surface.

Hope this helps
 
nrqed said:
This is an equation which holds at every point in space. What it says is that the divergence of the E field at a point is equal to the volume charge density evaluated at that same point divided by epsilon zero. There is no gaussian surface or volume involved here.


If you choose to integrate both sides of the equation over a certain volume, then the volume integral of the left side of the equation (of the divergence) may be converted to a surface integral over the electric flux and the volume integral of the right side becomes the total charge enclosed in the volume divided by epsilon_0. This gives the integral form of Gauss' law. But the form you give above is the differential form and involves no volume, no surface.

Hope this helps

That explains it perfectly. Sometimes I think textbook authors are so well versed in their craft that something appears obvious to them, while to me it requires a little bit more explaining. Your answer was spot on; thanks a bunch!
 
gulfcoastfella said:
That explains it perfectly. Sometimes I think textbook authors are so well versed in their craft that something appears obvious to them, while to me it requires a little bit more explaining. Your answer was spot on; thanks a bunch!
:cool:
You are very welcome! Believe me, this feeling (that some details that would make things much more clear are left out) remains present as you progress in your studies and read more and more advanced books. It never goes away. after a few years, you still get stuck on details but now it's in books on quantum field theory or advanced condensed matter physics. But if you then look back at lower level books you used to struggle with, you often think that was clearly explained, why did I have so much trouble back then! . This is when you realize that you have learned a lot and absorbed a lot of ideas and techniques. And years later, the quantum field theory/condensed matter textbooks will seem clear but you will be struggling with understanding research papers! So it's a constant struggle to figure out what other people mean! But it's fun!

best luck
 

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