Gauss' Law With conducting shells

In summary, we discussed how to approach problems involving Gauss' Law and conducting shells. We also clarified the answers to the three questions posed in the conversation, emphasizing the importance of considering the electric field, charge density, and distance in calculations.
  • #1
TwinGemini14
34
0
An infinite line charge lies on the z-axis with l = 2 µC/m. Coxaial with that line charge are: an infinite conducting shell (with no net charge) with thickness 1 cm and with inner radius 2 cm and outer radius 3 cm, an infinite shell with a radius of 4 cm and with a net charge of -5 µC/m, and another infinite conducting shell (with no net charge) with a thickness of 1 cm and with an inner radius of 5 cm and outer radius of 6 cm. A cross sectional view of this setup is shown below:

http://i662.photobucket.com/albums/uu347/TwinGemini14/elecshell.gif
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Gauss' Law = |(E . DA) = Qenclosed / (Epsilon-not)
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1) Compare the magnitude of the electrical flux through a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 1.5 cm to that of a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 3.5 cm.

A) Flux1.5 cm > Flux3.5 cm
B) Flux1.5 cm = Flux3.5 cm
C) Flux1.5 cm < Flux3.5 cm

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My logic is this. At 1.5cm the field is closer to the surface with a +2u C/m. Near 3.5, it is near a surface with a net charge of -5u C/m. So the magnitude of the e field is probably greater near the -5u C/m surface. According to Gauss' Law, since the magnitude of the e field is greater and the area is greater (3.5cm radius > 1.5cm radius), then the answer should be C.

So I said the answer is C.

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2) Compare the magnitude of the electric field at 2.5 cm from the z-axis and 4.5 cm from the z-axis.

A) E2.5 cm > E4.5 cm
B) E2.5 cm = E4.5 cm
C) E2.5 cm < E4.5 cm

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Here, E2.5 cm is zero because it is within the conducting shell. Since the E field is not zero at 4.5 cm, the magnitude must be greater.

So I said the answer is C.


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3) Compare the magnitude (i.e., the absolute value) of the electrical flux through a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 4.5 cm to that of a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 5.5cm.

A) Flux4.5 cm > Flux5.5 cm
B) Flux4.5 cm = Flux5.5 cm
C) Flux4.5 cm < Flux5.5 cm

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Since 5.5 cm is within a conducting shell, it's e field = 0. So then its flux must also be zero due to Gauss' Law. So the magnitude of the flux is greater in 4.5cm.

So I said the answer is A.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Can somebody please help me with this problem and review my answers because I am not sure about it. How do I go about this problem in a more precise fashion? Thanks for the help in advance!
 
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  • #2


Hi there,

I am glad to see that you are using Gauss' Law to approach these problems. Your logic and reasoning are correct for the most part, but there are a few things that I would like to clarify and expand upon.

1) You are correct in saying that the magnitude of the electric field is greater near the surface with a net charge of -5 µC/m compared to the surface with a charge density of +2 µC/m. However, the magnitude of the electric field is not just dependent on the charge density, but also on the distance from the source. In this case, the distance from the z-axis is 1.5 cm for the first surface and 3.5 cm for the second surface. Since the electric field decreases with distance, the electric field at 3.5 cm will be smaller than the electric field at 1.5 cm. Therefore, the flux through the surface with a radius of 3.5 cm will be smaller than the flux through the surface with a radius of 1.5 cm.

So the correct answer is actually A) Flux1.5 cm > Flux3.5 cm.

2) Your reasoning for this question is correct. Since the electric field is zero at 2.5 cm from the z-axis due to the conducting shell, and non-zero at 4.5 cm from the z-axis, the magnitude of the electric field must be greater at 4.5 cm. So the correct answer is C) E2.5 cm < E4.5 cm.

3) You are correct in saying that the electric field is zero at 5.5 cm from the z-axis due to the conducting shell. However, the electric field is also zero at 4.5 cm from the z-axis due to the inner conducting shell. So both surfaces will have the same flux through them, which is zero. Therefore, the correct answer is B) Flux4.5 cm = Flux5.5 cm.

In order to approach these problems more precisely, it is important to remember the relationship between electric field, charge density, and distance. The electric field decreases with distance and is directly proportional to the charge density. Also, when dealing with conducting shells, it is important to consider the electric field both inside and outside of the shell.

I hope this helps clarify your understanding of these concepts. Good luck with your studies!
 
  • #3


I would like to provide a more detailed explanation for each of the questions:

1) The magnitude of electric flux is directly proportional to the strength of the electric field and the area of the surface. In this case, the electric field is stronger near the -5 µC/m surface compared to the +2 µC/m surface. Additionally, the surface with a radius of 3.5 cm has a larger area compared to the surface with a radius of 1.5 cm. Therefore, the magnitude of electric flux through the cylindrical surface with a radius of 3.5 cm will be greater than that of the surface with a radius of 1.5 cm. Hence, the correct answer is C) Flux1.5 cm < Flux3.5 cm.

2) The electric field is zero at 2.5 cm from the z-axis because it is within the conducting shell. However, at 4.5 cm from the z-axis, the electric field is not zero due to the presence of the net charge -5 µC/m. Therefore, the magnitude of the electric field at 4.5 cm will be greater than that at 2.5 cm. Hence, the correct answer is C) E2.5 cm < E4.5 cm.

3) The electric field is zero at 5.5 cm from the z-axis because it is within the conducting shell. According to Gauss' Law, the electric flux through a closed surface is directly proportional to the net charge enclosed by that surface. Since there is no net charge enclosed by the cylindrical surface with a radius of 5.5 cm, the electric flux through this surface will be zero. On the other hand, the cylindrical surface with a radius of 4.5 cm encloses a net charge of -5 µC/m, which will result in a non-zero electric flux. Hence, the correct answer is A) Flux4.5 cm > Flux5.5 cm.
 

1. What is Gauss' Law With conducting shells?

Gauss' Law With conducting shells is a principle in electrostatics that states the electric flux through a closed surface created by a charge outside the surface is equal to the charge enclosed by the surface divided by the permittivity of free space.

2. How does Gauss' Law apply to conducting shells?

Gauss' Law can be applied to conducting shells by considering the electric field outside and inside the shell separately. The electric field outside is the same as it would be for a point charge, while the electric field inside is zero. This results in the electric flux through the conducting shell being equal to the charge enclosed by the shell.

3. What is the significance of conducting shells in Gauss' Law?

Conducting shells are important in Gauss' Law because they allow for the simplification of calculations. Since the electric field inside a conducting shell is zero, the electric flux through the shell is solely determined by the charge enclosed. This allows for easier calculations and analysis of electric fields.

4. How do you determine the direction of the electric field using Gauss' Law with conducting shells?

The direction of the electric field can be determined by considering the direction of the electric flux. If the electric flux is positive, the electric field points away from the enclosed charge, while a negative flux indicates an electric field pointing towards the enclosed charge.

5. Can Gauss' Law be applied to non-spherical conducting shells?

Yes, Gauss' Law can be applied to non-spherical conducting shells as long as the charge is evenly distributed and the electric field can be determined for each point on the surface. The electric flux can then be calculated and used to determine the enclosed charge.

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