Gauss' Law: Enclosed Cylinder in a Hollow Shell

In summary, the problem involves a long non-conducting cylinder with a charge density of p = a*r, where a = 4.73 C/m^4 and r is in meters. It is surrounded by a hollow metallic cylindrical shell with inner radius of 0.119m and outer radius of 0.158m. The first question asks for the electric field at 0.172m from the central axis, which can be solved using the equation (a*r^3)/(3*R*ε0). However, this equation does not work for the given radius as it is outside the shell. The second question asks for the surface charge density inside the hollow cylinder, which can be solved using the equation E = (/sigma
  • #1
snakeums
2
0
A long non-conducting cylinder has a charge density p = a*r, where a = 4.73 C/m^4 and r is in meters, and a radius of 0.0437m . Concentric around it is a hollow metallic cylindrical shell with an inner radius of 0.119m and an outer radius of 0.158m.

1) What is the electric field at 0.172m from the central axis? Answer in units of N/C.
2) What is the surface charge density inside the hollow cylinder? Answer in units of C/m2.

For part 1 I've already solved a few equations for charges at various radiuses along the inside of the cylinder, between the cylinder and the shell, etc, and I have the equation of (a*r^3)/(3*R*ε0), but I know this won't work because the radius is now outside the shell. I know the shell has no net charge, conceptually, so the charge outside is negative... but I'm still not sure what my R is for this equation.

For the second part I'm almost totally lost. The E = ([tex]/sigma[/tex])/ε0) doesn't make sense to me unless I'm supposed to get E for the radius JUST inside of the shell and use that.
 
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  • #2
I'm thinking you're right about the second part, so all you technically need is the electric field. To make things simpler, you could convert the volume charge density to a linear charge density. The metallic shell doesn't devote to the electric field outside of it, so you could rely on the derived equation for outside the shell and near the inner surface of the shell.
 
  • #3
Well I tried the second part by using (a*r^3)/(3*R*ε0), where r = 0.0437 and R = 0.119 to get E. Then I multiplied by ε0 again to get the (surface charge density) but I got 0.011056 which was wrong. So I guess I'm more lost than I thought.
 

Related to Gauss' Law: Enclosed Cylinder in a Hollow Shell

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface.

2. How is Gauss' Law applied to an enclosed cylinder in a hollow shell?

In this case, the electric flux through the closed surface is equal to the charge enclosed within the cylinder divided by the permittivity of the surrounding medium.

3. What is the significance of using a cylinder and a hollow shell in this application of Gauss' Law?

Using a cylinder and a hollow shell allows us to simplify the calculations and visualize the distribution of charges and electric fields in a more manageable way.

4. Can Gauss' Law be used to determine the electric field at any point within the cylinder?

Yes, Gauss' Law can be used to determine the electric field at any point within the cylinder by considering the charge enclosed within a smaller closed surface that passes through that point.

5. Are there any limitations to using Gauss' Law in this scenario?

One limitation is that Gauss' Law assumes a uniform distribution of charges within the cylinder and a perfectly conducting hollow shell. In reality, these conditions may not always be met and may affect the accuracy of the calculations.

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