Nonconducting concentric cylinders

  • Thread starter Thread starter J2012
  • Start date Start date
  • Tags Tags
    Cylinders
Click For Summary
The discussion focuses on determining the electric field around a solid nonconducting cylinder and its surrounding concentric shell using Gauss's law. For the regions defined by the radii (0 < R < R1, R1 < R < R2, R2 < R < R3, and R > R3), participants are analyzing how to calculate the electric field, considering the effects of the charge density and the geometry of the cylinders. There is confusion regarding whether to treat the electric field as being directed towards or away from the surfaces and how the surrounding shell influences the electric field calculations. Participants are also discussing the implications of the charge density and the assumption that the shell is nonconducting. Overall, the thread emphasizes understanding the electric field behavior in different regions around the cylinders.
J2012
Messages
6
Reaction score
0

Homework Statement



A very long solid nonconducting cylinger of radius R1 is uniformly charged with a charge density of volume charge densty pE. It is surrounded by a concentric cylinder tube of inner radius R2 and outer radius R3.

Determine the electric field as a function of the distance R from the center of the cylinders for:
a) 0 < R < R1
b) R1 < R < R2
c) R2 < R < R3
d) R > R3
e) If the charge density pE = 15 uC/m3 and R1 = 1/2R2 = 1/3R3 = 5.0 cm, plot E as a function of R from R = 0 and R = 20.0 cm. Assume the cylinders are very long compared to R3

Image looks like one in this thread: https://www.physicsforums.com/showthread.php?t=390934

Homework Equations



Gauss's law

Flux = E * A = pE / e_0 * [Volume of cylinder]

The Attempt at a Solution



So part a). R is between zero and the radius of the inner cylinder.

flux = pE/e_0 [pi*R^2*l]

flux = E * Curved Area = E * 2*pi*R1*l

pE/e_0 [pi*R^2*l] = E * 2 * pi * R1 * l

E = (pE * R^2)/ e_0 * 2R1

Am I on the right track? I did this as if I was calculating the electric field outside the cylinder. Should I have done it for the electric field inside the cylinder?
 
Physics news on Phys.org
Also, how do I know if the electric field is directed towards the circular surface of the cylinders or to the curved part?
 
Okay, so I think I would calculate a) as if it was a radius enclosed in the small cylinder. I'm assume the E is perpendicular to the curved surface. I'm not sure how to determine if the field is perpendicular to the curved surface or the flat circles.

b) I would calculate it as a radius outside the small cylinder.

When calculating a) and b) does the shell surrounding the cylinder have any affect on the E of the solid cylinder?

c) I am a bit confused on this part - The radius is inside the cylindrical shell that is surrounding the solid cylinder. So am I calculating the electric field within the shell? How do I calculate that?

d) The radius surrounds the entire structure (shell and solid cylinder). Same with part c) how am I supposed to calculate this?
 
Normally, for this kind of problem, it is assumed that you are calculating the field very far from either end of the cylinder, so that the electric field (where it is non-zero) is purely in the radial direction. (The radial direction is perpendicular to the common axis of the cylinders.)

I would guess that the concentric cylinder tube of inner radius R2 and outer radius R3 is made of a conducting material. I suppose it could also be made of the same material (with the same charge density) as the inner cylinder.

To answer your very last question: The interior of the inner cylinder has r values in the range; 0 < r < R1, so you need to find the field here also.
 
The shell is shown to be nonconducting and equal in charge density.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

Replies
26
Views
1K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
1K