Gauss Law and the wrong Gaussian surface.

In summary, the conversation discusses the problem of calculating the electric field inside an infinite length cylinder with a non-uniform charge density. The use of a Gaussian cylinder and sphere as the Gaussian surface is considered, with potential solutions and difficulties discussed. The need to convert r to spherical coordinates and use Poisson's equation is mentioned.
  • #1
Scherejg
5
0

Homework Statement



The problem was to calculate the electric field inside an infinite length cylinder (with radius R) with a non uniform charge density. The charge density depended on r. Its easy enough to solve using a gaussian cylinder with r less than R. But what if I wanted to complicate things and use a gaussian sphere inside the cylinder with r < R?

Homework Equations



∫E[itex]\bullet[/itex]dA = q / ε° Gauss' Law

ρ = ρ°(1 - r/R) This is charge density distribution

q = ρ[itex]\bullet[/itex]dV

V= 4/3 π r^3

A= 4 π r^2

The Attempt at a Solution



Since the electric field is not the same everywhere, it can't be removed from the first integral. It is however constant over dθ when r and d[itex]\Phi[/itex] are held constant.
q =∫ ρ*dV = ∫ ρ°(1- r/R) * A* dr
q = ∫ρ°(1-r/R) * 4 π r^2 * dr
q is easy enough to solve. So the problem lies in ∫E*dA
I had a few ideas about this, First one: convert the problem into spherical coordinates and solve it that way. Second one: The electric field is a vector quantity, so could I say the total electric field at a point r is equal to the sum of the partial derivatives of the electric field at that point? In that case, I would need to find an expression for dE/d[itex]\Phi[/itex]. Any guidance on this would be appreciated. Thanks!
 
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  • #2
As I understand r is the distance from axis, so you can not find the total enclosed charge inside a sphere by simply integrating ρ = ρ°(1 - r/R) you first need to convert r to spherical coordinates. After doing that you can try to solve equations for electric field but probably it will not be easy because taking a sphere as gaussian surface actually makes thing worse than finding electric field using Poisson's equation.
 

1. What is Gauss Law and how is it related to electricity and magnetism?

Gauss Law is a fundamental law in electromagnetism that relates the electric field and electric charge. It states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. This law helps us understand the behavior of electric fields and the relationship between electric charges and their surrounding fields.

2. Why is it important to choose the right Gaussian surface when applying Gauss Law?

The choice of Gaussian surface is crucial in applying Gauss Law because it determines the enclosed charge. If the wrong Gaussian surface is chosen, the electric field and enclosed charge will be incorrectly calculated, leading to incorrect results. The correct Gaussian surface should enclose all of the electric charge being considered and should be symmetrical to simplify calculations.

3. Can Gauss Law be applied to non-uniform electric fields?

Yes, Gauss Law can be applied to both uniform and non-uniform electric fields. However, the calculations can become more complicated for non-uniform fields, as the electric field strength will vary at different points on the Gaussian surface. In these cases, it is important to choose a small enough surface element to accurately calculate the electric flux.

4. How does Gauss Law relate to Coulomb's Law?

Gauss Law and Coulomb's Law are closely related, as they both describe the relationship between electric charge and electric field. However, Gauss Law is more general and can be applied to any closed surface, while Coulomb's Law is specifically for the electric field created by a point charge.

5. What happens if the electric charge is located outside of the Gaussian surface?

If the electric charge is located outside of the Gaussian surface, it will not contribute to the electric flux through the surface. Therefore, it will not be included in the calculations for Gauss Law. The only charges that should be considered are those enclosed by the Gaussian surface.

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