AN implementation of gauss's law

In summary, the electric field is Gaussian in nature and is symmetrical around the center of the charged object.
  • #1
impendingChaos
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A long solid nonconducting cylinder or radius R1 is uniformly charged with a charge density (p). It is surrounded by a concentric cylindrical tube or innder radius R2 and outer radius R3, it also has uniform charge density, p. Before I can go on i need to find the electric field as a function of the distance r from the center for (a) 0<r<R1 (b) R1<r<R2 (c) R2<r<R3 and (d) r>R3)

The integration is screwing me up, thanks
Kael.
 
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  • #2
To state my current attempt:
for part (a) i know
p=Q/4*pi*permittivity of free space*R1^2*L
V(enclosed)=pi*r^2*L
so p*V(enclosed)= the ratio (r^2/R1^2)Q which is the total Q enclosed
 
  • #3
You shouldn't need to do any integration. Hint: Write Gauss's law, which will give you the total flux through a surface in terms of the enclosed charge. (The enclosed charge is just the charge density times the volume.) Use a cylindrical Gaussian surface, of course, to take advantage of symmetry.
 
  • #4
Ok ok I'm starting to see it, so i would have
(permittivity of free space= Eo)

int(E) x dA = Q(enc)/Eo
EA= (r^2/R1^2)Q/Eo
then dividing both sides by the area would give E?
and if so would I need to use the left side of the equation divided by the area of the face of the cylinder plus the equation divided by the area of the length of the cylinder?
 
  • #5
impendingChaos said:
Ok ok I'm starting to see it, so i would have
(permittivity of free space= Eo)

int(E) x dA = Q(enc)/Eo
OK.
EA= (r^2/R1^2)Q/Eo
Almost there. Get rid of Q. You are given charge density, not Q. (As I suggested earlier, express the enclosed charge as charge density times the volume.)

Express area and volume in terms of r.
 
  • #6
hmm, well i thought i did that since if
charge density = Q/(4pi*L*R1^2)
and the volume is pi*L*r^2
then multiplying the two would give (Q*r^2)/(4*R1^2)
ok, now I take this Q(enc) and put it Gauss to obtain:
EA=(Q*r^2)/(4*R1^2*Eo)
but you said Q is not present?

In any case I continued from this by divinding the right side by A which would give me my electric field. Now, do I need to do this for the ends AND the length of the tube or are the ends considered negligible since it is "very long." I have solved it and obtained the correct answer but only when excluding the ends of this tube such that
E=pr/2Eo (this is the stated result)
So now, assuming the math was correct, is there a reason I should be considering the ends to be zero?
 
  • #7
impendingChaos said:
but you said Q is not present?
Well, you tell me. If you aren't given Q, how can your answer be in terms of it? You are given the charge density though. :wink:

In any case I continued from this by divinding the right side by A which would give me my electric field. Now, do I need to do this for the ends AND the length of the tube or are the ends considered negligible since it is "very long." I have solved it and obtained the correct answer but only when excluding the ends of this tube such that
E=pr/2Eo (this is the stated result)
So now, assuming the math was correct, is there a reason I should be considering the ends to be zero?
You are presumably finding the field at positions far enough from the ends so that any non-uniformity of field can be neglected. Your Gaussian surface is a cylindrical section in the middle of the long rod.

When you are finding the electric flux through the Gaussian surface, you are multiplying the area times the component of E perpendicular to that surface. In which case, what would be the flux through the end pieces of that Gaussian cylinder?
 
  • #8
Zero! :rofl:
 
  • #9
Exactamundo!
 
  • #10
Thanks alot, I think I can get parts b through c from here!
 

1. What is Gauss's Law and why is it important?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the net electric charge enclosed within that surface. It is important because it provides a mathematical relationship between electric fields and electric charges, allowing for the calculation of electric fields in various situations.

2. How is Gauss's Law implemented in a scientific setting?

In a scientific setting, Gauss's Law is typically implemented using mathematical equations and calculations. These calculations involve determining the electric flux through a closed surface and using it to find the electric field at a particular point.

3. What are the assumptions made in an implementation of Gauss's Law?

The main assumption made in an implementation of Gauss's Law is that the electric field is constant over the surface being considered. This is known as the "constant field assumption." Other assumptions may include that the electric field is perpendicular to the surface and that the surface is closed.

4. What are some real-world applications of Gauss's Law?

Gauss's Law has many real-world applications, including the design of electrical circuits, the calculation of electric fields in various situations, and the understanding of the behavior of charged particles in electric fields. It is also used in the design of antennas and other electronic devices.

5. Are there limitations to an implementation of Gauss's Law?

Yes, there are limitations to an implementation of Gauss's Law. One limitation is that it only applies to static electric fields and cannot be used to calculate the behavior of changing electric fields. Additionally, it assumes that the electric field is constant over the surface being considered, which may not always be the case in real-world situations.

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