[Physics 2 w/calc Uni] Cylinder inside of a cylindrical shell

In summary: ENDMENT: In summary, the problem involves a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are nonconducting and thin and have uniform surface charge densities on their outer surfaces. The radial component E of the electric field versus radial distance r from the common axis is given in Figure (b). The graph shows a discontinuity at r = 3.5 cm, indicating that the outer surface has a diameter of 3.5 cm and contains opposite charge of a linear charge density that is greater in magnitude than that of the inner cylinder. The question asks for the linear charge density of the shell, which is simply the amount of charge per unit length of the cylinder. To solve the problem, you can ignore
  • #1
jonathanlv7
26
1
Figure (a) shows a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are nonconducting and thin and have uniform surface charge densities on their outer surfaces. Figure (b) gives the radial component E of the electric field versus radial distance r from the common axis. The vertical axis scale is set by Es = 4.8 × 103 N/C. What is the linear charge density of the shell?

The problem also comes with this graphic (#13 on this page http://www.phys.ufl.edu/courses/phy2049/old_exams/2010f/exam1sol.pdf) --- yes I know the problems are different but the graph is identical

I have tried to solve this problem multiple times and every time it's wrong. I think my issue lies in the following questions I have in this problem's setup.

1. It says "narrow charged solid cylinder" is this thing 1 dimensional? 2 dimensional? I'm not sure because in the hint I got it said to use the equation for an infinite LINE of charge.

2. If it is 2 dimensional, then how is it also solid? A solid cylinder would have electric field lines coming out of the caps of the cylinder as well as the curved part. In the solutions I have seen they completely ignore this.

3. In the graph, what is being represented? When the value of E goes from positive to negative, what just happened? Did r exceed the radius of the inner cylinder? Or the outer cylinder?

4. The question asks for a linear charge density, how do I find a linear charge density of an object that is more than 1 dimensional? If I have a cylinder, it has to have a surface charge density, right? I found the following equation to convert from surface charge density to linear charge density, but I'm not sure if it's correct for this application - lambda = [ (surfacechargedensity)*area ]/L

5. Also, it says that the two objects have charge on their OUTER shells. Does this imply that there are only electric field lines spawning on the outside of the objects? Pointing out towards infinity (or in if negative)?

If someone could clarify these questions for me I'm confident I will be able to take it from there.

Thanks!
 
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  • #2
jonathanlv7 said:
Figure (a) shows a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are nonconducting and thin and have uniform surface charge densities on their outer surfaces. Figure (b) gives the radial component E of the electric field versus radial distance r from the common axis. The vertical axis scale is set by Es = 4.8 × 103 N/C. What is the linear charge density of the shell?

The problem also comes with this graphic (#13 on this page http://www.phys.ufl.edu/courses/phy2049/old_exams/2010f/exam1sol.pdf) --- yes I know the problems are different but the graph is identical

I have tried to solve this problem multiple times and every time it's wrong. I think my issue lies in the following questions I have in this problem's setup.

1. It says "narrow charged solid cylinder" is this thing 1 dimensional? 2 dimensional? I'm not sure because in the hint I got it said to use the equation for an infinite LINE of charge.
It is 3 dimensional but you can ignore the diameter of the cylinder. This allows you to ignore any edge effects, so it approximates a line charge. Otherwise, you would not be able to determine the gaussian surface for applying Gauss' law.

2. If it is 2 dimensional, then how is it also solid? A solid cylinder would have electric field lines coming out of the caps of the cylinder as well as the curved part. In the solutions I have seen they completely ignore this.
You can ignore the edge effects.

3. In the graph, what is being represented? When the value of E goes from positive to negative, what just happened? Did r exceed the radius of the inner cylinder? Or the outer cylinder?
There appears to be a discontinuity when r = 3.5 cm. Since the charge on the outer shell was on the outer surface, this means that the outer surface has a diameter of 3.5 cm and must contain opposite charge of a linear charge density that is greater in magnitude than that of the inner cylinder.

4. The question asks for a linear charge density, how do I find a linear charge density of an object that is more than 1 dimensional? If I have a cylinder, it has to have a surface charge density, right? I found the following equation to convert from surface charge density to linear charge density, but I'm not sure if it's correct for this application - lambda = [ (surfacechargedensity)*area ]/L
Linear charge density is simply the amount of charge per unit length of the cylinder. You don't have to worry about the distribution around the cylinder surface - just the amount per unit length.

5. Also, it says that the two objects have charge on their OUTER shells. Does this imply that there are only electric field lines spawning on the outside of the objects? Pointing out towards infinity (or in if negative)?
No. But using Gauss' law you can ignore the inward field lines. IF you choose the right gaussian surface (a surface that, by reasons of symmetry, has the same field strength over the entire surface) it becomes apparent that the contributions from all but the enclosed charge will cancel out.

AM
 
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1. What is the formula for finding the volume of a cylinder inside of a cylindrical shell?

The formula for finding the volume of a cylinder inside of a cylindrical shell is V = πh(R2 - r2), where h is the height of the cylinder, R is the radius of the cylindrical shell, and r is the radius of the smaller cylinder inside.

2. How do I calculate the surface area of a cylinder inside of a cylindrical shell?

To calculate the surface area of a cylinder inside of a cylindrical shell, you can use the formula SA = 2πrh + 2π(R2 - r2), where r is the radius of the smaller cylinder, R is the radius of the cylindrical shell, and h is the height of the cylinder.

3. Can the volume of the smaller cylinder be greater than the volume of the cylindrical shell?

No, the volume of the smaller cylinder cannot be greater than the volume of the cylindrical shell. This is because the smaller cylinder is contained within the larger cylindrical shell, so its volume must be smaller than or equal to the volume of the shell.

4. How can I solve for the height of the cylinder inside of a cylindrical shell?

To solve for the height of the cylinder inside of a cylindrical shell, you can use the formula h = V/π(R2 - r2), where V is the volume of the smaller cylinder and R and r are the radii of the cylindrical shell and smaller cylinder, respectively.

5. Is there a difference between a cylinder inside of a cylindrical shell and a hollow cylinder?

Yes, there is a difference between a cylinder inside of a cylindrical shell and a hollow cylinder. A cylinder inside of a cylindrical shell has a smaller cylinder inside of a larger cylindrical shell, while a hollow cylinder is a cylinder with a hollow space inside. The calculations for volume and surface area are different for these two shapes.

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