Gauss's Law insulating cylinder concentric shell

In summary, The conversation is about a physics problem that involves calculating the electric field for a charged cylinder. The conversation includes a discussion of the correct equations to use, potential errors in the solution, and suggestions for posting clear and readable work in the future. The final comment mentions a common issue with rotated attachments.
  • #1
ilovetswift
4
0

Homework Statement


photo.jpg


photo2.jpg



Homework Equations





The Attempt at a Solution


I'm not sure if I am doing part a,b,e, and f correctly
 
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  • #2
You'd get more responses if you were to post attachments that are the right way up! Even after rotation, the handwritten parts are pretty hard to read.
For a), the denominator is hard to read, but it looks like .1π(.01)2, which would be right.
For b), you seem to have used 'a' in place of the given variable r, and used r as the .01m radius. The cylinder length is ten times the radius, so I would think you are supposed to ignore end effects. If I ignore that term in the denominator and simplify what's left, it seems to give a value which increases with distance from the central cylinder, which cannot be right.
Pls take the trouble to post your working in LaTex, or write very clearly (no still-visible rubbings out!). It would also help if you were to define your variables.
 
  • #3
How do you post the attachments that are up right away? Also, on my pc they do not need to be rotated for whatever reason when they uploaded it turned them sorry.

a) i have written .1 pi (.01)^2 in the denominator
Thank you, and dually noted
 
  • #4
ilovetswift said:
How do you post the attachments that are up right away?
Sorry, I don't know what the magic is there. I've never run into the problem, but I see you're not alone. Some come out 180 degrees out. Sounds like a landscape/portrait mix-up.
 
  • #5
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I would like to clarify that Gauss's Law is a fundamental law in electromagnetism 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 net charge enclosed by that surface divided by the permittivity of free space.

Regarding the specific scenario of an insulating cylinder and a concentric shell, Gauss's Law can be applied to determine the electric field inside and outside the cylinder and shell. The total charge enclosed by the surface of the cylinder and shell would determine the electric flux through that surface, and by using the appropriate symmetry considerations, the electric field can be calculated at any point inside or outside the cylinder and shell.

In terms of the specific homework questions, it would be helpful to provide more context and information about the problem, such as the geometry and dimensions of the cylinder and shell, and any given values or conditions. Without this information, it is difficult to determine if the approach for parts a, b, e, and f is correct. However, in general, the key steps would involve identifying the appropriate Gaussian surface, determining the total charge enclosed by that surface, and using symmetry considerations to determine the electric field. It is important to also consider any given boundary conditions or constraints that may affect the calculation.
 

Related to Gauss's Law insulating cylinder concentric shell

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is named after the German mathematician and physicist Carl Friedrich Gauss.

2. What is an insulating cylinder?

An insulating cylinder is a three-dimensional shape that is made of an insulating material, meaning it does not conduct electricity. It can come in various sizes and shapes, but its main characteristic is that it does not allow the flow of electric charge through it.

3. What is a concentric shell?

A concentric shell is a spherical or cylindrical shell that has the same center as another spherical or cylindrical object. In the case of Gauss's Law, it refers to a conducting shell that surrounds an insulating cylinder, with both objects sharing the same center.

4. How is Gauss's Law applied to an insulating cylinder and concentric shell?

In this scenario, Gauss's Law states that the electric flux through the surface of the insulating cylinder is equal to the charge enclosed by the cylinder divided by the permittivity of free space. This is because the electric field inside the insulating cylinder is zero, and the electric field inside the concentric shell is constant.

5. What are some real-world applications of Gauss's Law with insulating cylinders and concentric shells?

Gauss's Law with insulating cylinders and concentric shells is commonly used in the design of capacitors, which are electronic devices used to store electric charge. It is also used in understanding the behavior of electric fields in insulating materials, such as dielectrics, which are commonly used in electronics and electrical insulation.

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