How Cooulomb's law follows from Gauss' law?

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    Gauss Gauss' law Law
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Discussion Overview

The discussion revolves around the relationship between Coulomb's law and Gauss's law, focusing on how one can be derived from the other. Participants explore theoretical aspects, mathematical reasoning, and conceptual clarifications related to electric fields and symmetry in physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest reviewing concepts such as the inverse-square law and flux to understand the compatibility of Gauss's law with Coulomb's law.
  • One participant discusses the symmetry of the electric field around a point charge, emphasizing that the vector must be directed along the axis connecting the charge and the point of evaluation.
  • Another participant elaborates on the dependence of the electric field's magnitude on the distance between the charge and the observation point, noting that it must be independent of the coordinate system used.
  • There is a discussion about the relationship between electric field strength and the total charge, with an analogy to light intensity from a bulb to illustrate the inverse-square relationship.
  • One participant clarifies that the goal is to prove Gauss's law from Coulomb's law, highlighting the importance of symmetry in determining the direction and magnitude of the electric field.
  • A mathematical expression of Gauss's law is presented, along with a specific solution for the electric field due to a point charge.

Areas of Agreement / Disagreement

Participants express differing views on the derivation process, with some focusing on proving Gauss's law from Coulomb's law while others emphasize the need to establish the electric field's characteristics first. The discussion remains unresolved regarding the best approach to connect these two laws.

Contextual Notes

Participants note that the proof from Coulomb's law to Gauss's law is considered more challenging, and there are mentions of omitted steps in educational contexts that may lead to confusion.

skyliner34
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Can anyone explain this to me?
 
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take a look at the Wikipedia articles on inverse-square law, flux (the physical quantity you're interested in is really "flux density"), and then Gauss's law.

if the enclosing surface is a sphere surrounding the point charge, with the point at center, it's pretty easy to see how Gauss's law is directly compatible with the inverse-square relationship.
 
First you must think what are the possible vector fields caused by a single point source (charge).
All information you have is the location of 2 points: the source and the point where you want to evaluate the vector. The problem has cylindric symmetry around the axis connecting the two points, so vector MUST be directed parallel to this axis: either straigth away or towards the source.
The magnitude of the vector must be independent of the choise of coordinate system, so it can only depend on the distance between the two points. Under this circumstances the integral of the electric field over a sphere is not difficult to evaluate.
 
Lojzek said:
First you must think what are the possible vector fields caused by a single point source (charge).
All information you have is the location of 2 points: the source and the point where you want to evaluate the vector. The problem has cylindric symmetry around the axis connecting the two points, so vector MUST be directed parallel to this axis: either straigth away or towards the source.
The magnitude of the vector must be independent of the choise of coordinate system, so it can only depend on the distance between the two points..

there are a lot of different scaler functions of that distance.

Under this circumstances the integral of the electric field over a sphere is not difficult to evaluate.

the reason why that surface integral evaluates to a constant that is proportional to the enclosed point charge is because the magnitude of that field vector is inversly proportional to the square of the distance and proportional to the charge. why it's proportional to the amount of charge is because if you had two different charges sitting virtually next to each other, the total force you get from both should come from adding the force you get due to each charge. why it's inverse-square with distance is because of this concept of flux and flux density and where we model the field as being proportional to the flux density (or, with judicious choice of units, the same as flux density). and because the surface area of a sphere is [itex]4 \pi r^2[/itex].

think of a 100 watt light bulb surrounded by a bunch of concentric spheres all centered on the 100 watt light bulb. intensity of radiant energy is how much radiating power crosses a unit area that is held perpendicular to the flow of this radiant power. so if you were 10 meters out, that 100 watts is distributed over 4[itex]\pi[/itex]102 square meters. if you're 20 meters out, the same 100 watts is distributed over 4[itex]\pi[/itex]202 square meters. and the intensity is 100 watts divided by the area which is an inverse-square relationship.

Gauss's law does not work unless it's an inverse-square relationship with distance and proportional to the amount of "stuff" at the point source that we're measuring distance from.
 
Last edited:
Of course I meant that the dependence E(r) must be determined from Gauss law.

I only solved the first part of the problem: we are not proving Gauss law from Coulomb's. We are proving the opposite, which is more difficult. For this proof we must first use symmetry to show that the electric field vector is directed away from the source and it's magnitude depends only on the distance (when I was taught about Coulomb's law I felt cheated because this first part was omitted and the proof Coulomb->Gauss was sold as Gauss->Coulomb proof).
 
Gauss's Law is

[tex]\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}[/tex]

One can verify that given
[tex]\rho = q\delta^3(\vec{r})[/tex]
A solution for E is
[tex]\vec{E} = \frac{q}{4\pi\epsilon_0 r^2}[/tex]
 

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