Proving Gauss Law using a "bad" Gaussian surface

In summary, the student is trying to use Gauss' law to calculate the flux through a cylinder that is perpendicular to an infinite charged wire.
  • #1
math4everyone
15
0

Homework Statement


What I basically want to do is to prove Gauss Law with a cylinder perpendicular to an infinite charged wire (I know I can do this simple, but I want to do it this way)
EeTd5.png
This is what I have done so far:

hm2cR.jpg

Homework Equations


$$\Phi=\int \frac{dq}{4\pi \varepsilon_0 r^2} \hat{r} \cdot d\vec{A}$$

The Attempt at a Solution


So $$d=\frac{z}{tan(\theta)}$$ and therefore $$r^2=z^2 cot^2(\theta)+z^2$$. Now the flux through the top of the cylinder is $$\Phi=\int \frac{\lambda dx}{4\pi \varepsilon_0[z^2(1+cot^2(\theta))]} \widehat{r} \cdot d\vec{A}$$ where
$$\vec{A}=(\rho d\rho d\phi)\hat{z}$$ and $$\hat{r}=\frac{\vec{r}}{|r|}$$
So:
$$\Phi=\int \frac{\lambda dx}{4\pi \varepsilon_0[z^2(1+cot^2(\theta))]^{\frac{3}{2}}} (dcos(\phi),dsin(\phi),z) \cdot (\rho d\rho d\phi)(0,0,\hat{z})$$
But I don't know how to proceed... Maybe I can use cosine law to find $$\rho$$?
 
Physics news on Phys.org
  • #2
How do you plan to handle the flux through the sides of the cylinder?
Also: you need a clear statement of what, exactly, it is that you are proving.
 
  • Like
Likes SammyS
  • #3
Simon Bridge said:
How do you plan to handle the flux through the sides of the cylinder?
Also: you need a clear statement of what, exactly, it is that you are proving.
Using the fact that $$d\vec{A}$$ is $$\rho d\rho d\phi \hat{\rho}$$. First, I want to calculate the flux in the bottom and top of the cylinder, and I expect it will be slightly similar in the case of the sides. What I want to prove is that this flux is equal to $$\frac{q}{\varepsilon_0}$$
 
  • #4
OK good luck ... there is a reason people don't try this: it's very difficult.
It is not uncommon to end up with integrals that cannot be evaluated analytically.

The integral you are asking about needs limits ...

... and you still should make an explicit statement of what you want to prove.
Are you just rotating the standard gaussian surface for the hell of it or are you trying to find something out?
ie. why pick a cylinder in that orientation? Why not a spheroid or a cube? Why not have the line of charge pass through the cylinder at an arbitrary angle to the cylinder axis?
 

1. What is a "bad" Gaussian surface?

A "bad" Gaussian surface is a hypothetical surface that does not satisfy the requirements for using Gauss's Law. It may have an irregular shape, intersect with other objects, or have a non-uniform electric field.

2. Why is it important to use a "bad" Gaussian surface to prove Gauss's Law?

Using a "bad" Gaussian surface allows scientists to demonstrate the limitations of Gauss's Law and how it may not always accurately predict the electric field in a given scenario. It also helps to highlight the importance of using a proper Gaussian surface for accurate calculations.

3. How does using a "bad" Gaussian surface affect the calculation of electric flux?

Using a "bad" Gaussian surface can lead to incorrect calculations of electric flux, as the surface may not enclose the entire charge distribution or may intersect with other objects. This can result in a non-zero net electric flux, even though Gauss's Law states that it should always be zero.

4. Can Gauss's Law still be used to calculate the electric field when using a "bad" Gaussian surface?

No, Gauss's Law cannot be reliably used to calculate the electric field when using a "bad" Gaussian surface. This is because the irregular shape or intersecting nature of the surface can lead to inaccurate calculations of electric flux, and therefore inaccurate results for the electric field.

5. Are there any real-life scenarios where a "bad" Gaussian surface would be used?

In real-life scenarios, it is important to use a proper Gaussian surface to accurately calculate the electric field. However, a "bad" Gaussian surface can be used in thought experiments or theoretical scenarios to demonstrate the limitations of Gauss's Law and the importance of using a proper Gaussian surface for accurate results.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
373
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
384
  • Introductory Physics Homework Help
Replies
3
Views
798
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
997
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
29
Views
1K
Back
Top