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

In summary, 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). However, for a non-uniform charge distribution, 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
  • #1
hegtor
3
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


[tex] \oint_S \vec E \cdot \vec{dA}= \frac{Q_{enc}}{\varepsilon_0} [/tex]

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.
 
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  • #2
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.
 
  • #3
Orodruin said:
It wouldbe easier to help you if you provided sources for these statements.
For example for a sphere and for a cylinder.
 
  • #4
hegtor said:
How does this work? E(r) is not constant, right? So how can you find the integrand then?

Homework Equations


[tex] \oint_S \vec E \cdot \vec{dA}= \frac{Q_{enc}}{\varepsilon_0} [/tex]

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?
 
  • #5
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?!
 
  • #6
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
 
  • #7
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.
 

1. What is Gauss's Law and how is it used to find electric field with non-uniform charge distribution?

Gauss's Law is a fundamental law in electromagnetism that relates the electric field to the distribution of electric charge. It states that the flux of the electric field through a closed surface is equal to the enclosed charge. This law can be used to find the electric field at a point due to a non-uniform charge distribution by applying it to a small surface surrounding that point.

2. How is the Gaussian surface chosen when using Gauss's Law?

The Gaussian surface is chosen to be a closed surface that encloses the point where the electric field is to be found. It should also be chosen so that the electric field is either perpendicular or parallel to the surface, making the calculation of the flux easier.

3. Can Gauss's Law be used to find the electric field at a point due to a continuous charge distribution?

Yes, Gauss's Law can be used to find the electric field at a point due to a continuous charge distribution. The charge distribution can be divided into small elements, and the electric field at the point can be found by applying Gauss's Law to each element and then integrating over the entire distribution.

4. How does non-uniform charge distribution affect the calculation of electric field using Gauss's Law?

Non-uniform charge distribution can make the calculation of electric field using Gauss's Law more complex as it requires integration over the distribution. However, the general principle remains the same - the flux of the electric field through a closed surface is equal to the enclosed charge.

5. Are there any limitations to using Gauss's Law to find electric field with non-uniform charge distribution?

One limitation is that Gauss's Law is only applicable to static electric fields, where the charges are not moving. It also assumes that the medium surrounding the charges is a linear, isotropic, and homogeneous material. In addition, the chosen Gaussian surface must be symmetrical to simplify the calculation of the electric field.

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