Gauss's Law to find E with non-uniform charge distribution

Click For Summary
SUMMARY

This discussion focuses on applying Gauss's Law to determine the electric field E(r) in scenarios involving non-uniform charge distributions. The participants clarify that while E(r) can be constant over a Gaussian surface for uniform charge distributions, this is not the case for non-uniform distributions. The conversation highlights that if the charge density exhibits spherical symmetry, E(r) remains constant, allowing for the application of Gauss's Law. For arbitrary charge distributions, the Poisson equation can be utilized to derive the electric potential and subsequently the electric field.

PREREQUISITES
  • Understanding of Gauss's Law: \(\oint_S \vec E \cdot \vec{dA}= \frac{Q_{enc}}{\varepsilon_0}\)
  • Knowledge of electric field concepts and charge density
  • Familiarity with spherical symmetry in charge distributions
  • Basic understanding of the Poisson equation
NEXT STEPS
  • Study the application of Gauss's Law in non-uniform charge distributions
  • Learn about the Poisson equation and its role in electrostatics
  • Explore examples of electric fields in various symmetrical charge configurations
  • Investigate numerical methods for calculating electric fields in arbitrary charge distributions
USEFUL FOR

Students of electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields and Gauss's Law in complex charge scenarios.

hegtor
Messages
3
Reaction score
0

Homework Statement



Hello,

this is more of a conceptual question than a concrete homework assignment question. I'm learning about Gauss's law and the Prof did an exercise on a sphere with uniform charge distribution, where he found E(r). The trick was, that E(r) was constant over the Gaussian surface he picked. Normally you can't find the integrand of an unknown function if you know the value of the integral, but you can if the function is constant. That enabled him to find E(r).

When he finished he said you cannot deduce E(r) for a non-uniform charge distribution, because E(r) would not be constant so you couldn't pull it out of the integral. I did some research and found exercises where one finds E(r) for a non-uniform charge distribution on a sphere and there are even some posts in this forum on that.

How does this work? E(r) is not constant, right? So how can you find the integrand then?

Homework Equations


\oint_S \vec E \cdot \vec{dA}= \frac{Q_{enc}}{\varepsilon_0}

The Attempt at a Solution



My first idea was to construct a Gaussian surface so that E(r) would be constant on it throughout. I mean this should be possible, but the shape of this surface could be weird. This could get arbitrarily complicated though, and one couldn't compute the area of it easily if it was a weird shape...
So I'm stuck and hope you can help me look in the right way.
 
Last edited:
Physics news on Phys.org
hegtor said:
I did some research and found exercises where one finds E(r) for a non-uniform charge distribution on a sphere and there are even some posts in this forum on that.
It wouldbe easier to help you if you provided sources for these statements.
 
Orodruin said:
It wouldbe easier to help you if you provided sources for these statements.
For example for a sphere and for a cylinder.
 
hegtor said:
How does this work? E(r) is not constant, right? So how can you find the integrand then?

Homework Equations


\oint_S \vec E \cdot \vec{dA}= \frac{Q_{enc}}{\varepsilon_0}

The Attempt at a Solution



My first idea was to construct a Gaussian surface so that E(r) would be constant on it throughout. I mean this should be possible, but the shape of this surface could be weird. This could get arbitrarily complicated though, and one couldn't compute the area of it easily if it was a weird shape...
So I'm stuck and hope you can help me look in the right way.
E(r) means that E is function of r, the distance from a certain central point. If the charge density has spherical symmetry, depending only on r, the electric field also has spherical symmetry, its magnitude is constant over a sphere of radius r, and can be determined with the help of Gauss' Law.
As an example, assume that the charge density is of the form ρ=B/r2. Can you determine the charge enclosed in a sphere of radius R?
 
ehild said:
E(r) means that E is function of r, the distance from a certain central point. If the charge density has spherical symmetry, depending only on r, the electric field also has spherical symmetry, its magnitude is constant over a sphere of radius r, and can be determined with the help of Gauss' Law.
As an example, assume that the charge density is of the form ρ=B/r2. Can you determine the charge enclosed in a sphere of radius R?
That makes sense!
So as long as the charge density has spherical symmetry, E(r) is constant over a sphere of some radius r from the center which I "cleverly" choose to be my Gaussian surface. Because E(r) is constant there I can pull it in front of the integral and solve for it.

But what happens for a arbitrarily weird charge density generally? Can you solve for E with Gauss's Law?
And what's with other shapes than spheres, some weird shape which you don't know the area of? Does Gauss's Law help here finding E?

As I understood up until now you always need some symmetry present otherwise Gauss's Law won't help in finding E?!
 
hegtor said:
As I understood up until now you always need some symmetry present otherwise Gauss's Law won't help in finding E?!
Yes, the integral form of Gauss' Law can be used to find E when there is symmetry in the charge density.
In principle, you can use the Poisson equation to find the electric potential V(x,y,z) in case of arbitrary charge density ρ(x,y,z), and get E as the negative gradient of the potential function.
http://farside.ph.utexas.edu/teaching/em/lectures/node31.html
 
You can in principle also vectorially sum the forces due to all the charges in your sphere on a unit test charge but that is similarly practically infeasible for arbitrary charge distributions.
 

Similar threads

Replies
6
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
23
Views
4K
Replies
10
Views
4K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 18 ·
Replies
18
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K