AN implementation of gauss's law

  • #1
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.
 

Answers and Replies

  • #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
Doc Al
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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
Doc Al
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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
Doc Al
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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?
 
  • #9
Doc Al
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Exactamundo!
 
  • #10
Thanks alot, I think I can get parts b through c from here!
 

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