Gauss Law- Conducting and Non-conducting cylindrical shells

In summary: This is incorrect thinking. The charge enclosed by the Gaussian surface in (a) varies with r whereas in (b) the enclosed charge is constant. Don't you think this should make a difference to the electric field? In fact the answer you gave for (a) is incorrect but it is correct for (b).
  • #1
Samnolan1031
4
0

Homework Statement


hnzXRls.jpg

Below is a diagram of an infinitely long non-conducting rod of radius, R, with a uniform continuous charge distribution. The uniform linear charge density of this line is lamba1. The rod is at the center of an infinitely long, conducting pipe. The linear charge density of the pipe is lamba2.
The distance from the center of the rod to the inner surface of the pipe is a and the distance between the center of the rod to the outer surface of the pipe is b. Assign r as the shortest distance from the center of the rod to an arbitrary point.
a) what is the electric field at 0<r<R
b) what is the electric field at R<r<a?
c) what is the electric field at a<r<b?
d) what is the electric field at r<b?

Homework Equations


Gauss's Law:[/B]
integral of E*da=qenc/Eo
Density of lamba 1=Q/L
Density of lamba 2=Q/V

The Attempt at a Solution


a) E*dA=qenc/Eo
E*(2pirL)=lamba1*L/Eo
E=lamba1*L/2pi*r*L*Eo
E=lamba1/2pi*r*Eo

b) the answer would the same as answer a because it is asking for the hollow space.
c) the electric field would be 0 because the pipe is conducting.
d)
E*dA= (E1+E2)/Eo
E= (E1+E2)/Eo * 1/2pir^2

Are my answers correct so far? Thank you!
 

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  • #2
Samnolan1031 said:

The Attempt at a Solution


a) E*dA=qenc/Eo
E*(2pirL)=lamba1*L/Eo
No; remember that the inner rod has uniform charge density. So what will be Q(r)?
E=lamba1*L/2pi*r*L*Eo
E=lamba1/2pi*r*Eo
The E field is not infinite at r=0 is it?
b) the answer would the same as answer a because it is asking for the hollow space.
c) the electric field would be 0 because the pipe is conducting.
d)
E*dA= (E1+E2)/Eo
E= (E1+E2)/Eo * 1/2pir^2
OK for (b) and (c) but (d) has the wrong dimensions so can't be right.
 
  • #3
Your equations are hard to read. Please
Samnolan1031 said:
) the answer would the same as answer a because it is asking for the hollow space.
This is incorrect thinking. The charge enclosed by the Gaussian surface in (a) varies with r whereas in (b) the enclosed charge is constant. Don't you think this should make a difference to the electric field? In fact the answer you gave for (a) is incorrect but it is correct for (b).
Samnolan1031 said:
d) what is the electric field at r<b?
This should be r > b otherwise it doesn't make sense as a separate question.
 
  • #4
kuruman said:
Your equations are hard to read. Please

This is incorrect thinking. The charge enclosed by the Gaussian surface in (a) varies with r whereas in (b) the enclosed charge is constant. Don't you think this should make a difference to the electric field? In fact the answer you gave for (a) is incorrect but it is correct for (b).

This should be r > b otherwise it doesn't make sense as a separate question.

Hello, thank you for your help.
In class we did a problem where the charge enclosed varies by r and these are the steps we took. In this problem, the radius of the pipe was 10cm, but the radius of the charge was 5cm. To find the charge enclosed we compare the densities of the pipe versus the density of the charge enclosed, then solved for charge enclosed.
1SRKuh7.jpg

So would I do the same thing for part a?

This was the only problem I've done where we are looking at a charge enclosed in the rod, so to me it would make sense to computationally do something similar to find the q enclosed in part a.

Thank you again for your help.
 

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  • #5
Samnolan1031 said:
So would I do the same thing for part a?
No. The example in the photograph looks like a spherical distribution is involved. Don't try to copy from another solution, just reason it out. In part (a) you have
Samnolan1031 said:
E*(2pirL)=lamba1*L/Eo
The left side of the equation is OK. The right side of the equation should be ##q_{encl.}/\epsilon_0##. So now think, (a) what is your Gaussian surface and (b) how much charge does it enclose? Draw a drawing if necessary.
 
  • #6
kuruman said:
No. The example in the photograph looks like a spherical distribution is involved. Don't try to copy from another solution, just reason it out. In part (a) you have

The left side of the equation is OK. The right side of the equation should be ##q_{encl.}/\epsilon_0##. So now think, (a) what is your Gaussian surface and (b) how much charge does it enclose? Draw a drawing if necessary.

Okay thank you.

Quick question as I am working on this. The charge enclosed can only be 0 if this was a conducting rod, right? I know if this was a conducting rod, the charge enclosed would be 0, making the electric field 0. But, is there any instance where a charge enclosed can be 0 in a non conducting surface?
 
  • #7
Samnolan1031 said:
Quick question as I am working on this. The charge enclosed can only be 0 if this was a conducting rod, right?
Yes. If you had a conducting cylinder the charge will all be on its surface so the charge enclosed by a Gaussian cylinder of radius r < a would be zero. You can also have a non-conducting cylinder with charge pasted only on its surface in which case the electric field will also be zero for r < a.
 
  • #8
kuruman said:
Yes. If you had a conducting cylinder the charge will all be on its surface so the charge enclosed by a Gaussian cylinder of radius r < a would be zero. You can also have a non-conducting cylinder with charge pasted only on its surface in which case the electric field will also be zero for r < a.
Thank you, that makes a lot of sense.

Since it's inside the rod, the gaussian surface of the charge would enclose less than the total charge of the rod.
So wouldn't it be
E×2πrL= (πLr^2)/(εR^2)
Which would make E=(λ1×r)/(2πεR^2)
Is this correct?
 
  • #9
Samnolan1031 said:
Thank you, that makes a lot of sense.

Since it's inside the rod, the gaussian surface of the charge would enclose less than the total charge of the rod.
So wouldn't it be
E×2πrL= (πLr^2)/(εR^2)
Which would make E=(λ1×r)/(2πεR^2)
Is this correct?
That is correct except you have a π on the right hand side of the equation that does not belong. It has correctly disappeared from the bottom equation. Also note that the electric field is correctly zero when r = 0, which addresses the point that @rude man raised in post #2.
 

Related to Gauss Law- Conducting and Non-conducting cylindrical shells

1. What is Gauss Law and how is it used?

Gauss Law is a fundamental law in electromagnetism that describes the relationship between electric fields and charges. It states that the flux of an electric field through a closed surface is proportional to the total charge enclosed by that surface. Gauss Law is commonly used to calculate electric fields in situations where there is a high degree of symmetry.

2. What is a conducting cylindrical shell?

A conducting cylindrical shell is a hollow cylinder made of a material that allows charges to move freely, such as a metal. This means that any charge placed on the outer surface of the cylinder will distribute itself evenly on the outer surface, and there will be no electric field inside the cylinder.

3. How does Gauss Law apply to conducting cylindrical shells?

Gauss Law states that the electric field inside a conducting cylindrical shell is always zero. This is because any charge placed on the outer surface of the cylinder will distribute itself evenly on the outer surface, and there will be no net charge inside the cylinder. Therefore, the flux of the electric field through any closed surface inside the cylinder is also zero, as there is no electric field present.

4. What is a non-conducting cylindrical shell?

A non-conducting cylindrical shell is a hollow cylinder made of a material that does not allow charges to move freely, such as an insulator. This means that any charge placed on the outer surface of the cylinder will not distribute itself evenly, and there will be an electric field both inside and outside the cylinder.

5. How does Gauss Law apply to non-conducting cylindrical shells?

Gauss Law still applies to non-conducting cylindrical shells, but the electric field will be different than that of a conducting cylinder. The electric field inside a non-conducting cylinder will depend on the charge distribution on the outer surface, while the electric field outside the cylinder will depend on the total charge enclosed by the cylinder. Gauss Law can be used to calculate the electric field in these situations by considering the symmetry of the cylinder and applying the law accordingly.

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