Gauss's law to solve for electric field.

In summary, the conversation discusses the use of Gauss's law to find the electric field generated by an infinitely long line charge inside a cylindrical Gaussian surface. The experimental data suggests the existence of an electric field, but the calculated electric field is always zero due to the cancellation of positive and negative charges. The conversation also raises questions about the validity of the coaxial model used and the location of where the electric field was measured.
  • #1
playoff
80
1
An infinitely long line charge with a linear density +λ inside an infinitely long cylinder of radius R and area density -λ/(2pi*R).

So if I set up a cylindrical Gaussian surface with length L, the positive charge inside the surface would be λL and negative charge inside the surface would be area density multiplied by area so -λ/(2pi*R)*(2pi*RL) => -λL, cancelling each other out no matter what.

But the problem is that electric field has to exist, because I am supposed to compare this derived electric field with experimental data.

Did I miss out on something?
 
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  • #2
What does the experimental data tell you?
 
  • #3
Well there definitely seems to be an existing electric field according to my data. I should have added this: I am supposed to find the size of lambda that would generate the electric field recorded from the lab. But since electrical field is always 0, there can't be a lambda that would fit my experimental data.

Could someone evaluate whether I have used Gauss's law correctly and came up with a correct answer?
 
  • #4
If the coaxial model you are using is a good fit to the experiment, then the electric field should be very small outside the coax.
That is why I asked you what the data is telling you. Is the electric field almost zero?

It is unlikely that you actually have an infinitely long bit of coax in the lab though ... so you will not get an exactly zero field.

The answer to the question as you wrote it down is, as you have pointed out, there is no value of lambda that can give the measured field.
If you have missed something, it was in something you have not told us yet.
 
  • #5

Hey, how did you get that lambda into your post? I mean the Greek letter.
 
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  • #6
Are you sure you were'nt supposed to find the E field between the inner wire and the outer shield?
 
  • #7
You mean ##\small{\lambda}##? Same way as always - LaTeX.
But it looks like the first post may have used the "insert unicode" trick.
 
  • #8
rude man said:
Are you sure you were'nt supposed to find the E field between the inner wire and the outer shield?

I would like to second this question. Where exactly did you measure the field?
 
  • #9
It's my bet for the missing information - yes.
 

1. What is Gauss's law?

Gauss's law is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space.

2. How is Gauss's law used to solve for electric field?

Gauss's law can be used to solve for the electric field in situations where there is symmetry in the distribution of charges. By using a Gaussian surface that follows the symmetry of the charge distribution, the electric field can be found by taking the dot product of the electric field and the surface normal and setting it equal to the total charge enclosed by the surface divided by the permittivity of free space.

3. What is a Gaussian surface?

A Gaussian surface is an imaginary surface that is chosen to follow the symmetry of the charge distribution in a given scenario. It can be any shape, as long as it is closed and encloses the charges of interest. This surface is used in Gauss's law to simplify the calculation of the electric field.

4. Can Gauss's law be used to solve for electric field in all situations?

No, Gauss's law can only be used in situations where there is symmetry in the distribution of charges. In cases where the charge distribution is not symmetrical, other methods such as Coulomb's law or integration must be used to calculate the electric field.

5. How does Gauss's law relate to electric flux?

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. This means that the electric flux is directly proportional to the charge enclosed and is inversely proportional to the permittivity of free space.

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