Solving Gauss' Law Problem: Comparing Electric Field and Flux Magnitudes

AI Thread Summary
The discussion focuses on applying Gauss' Law to analyze the electric field and flux in a system with multiple conducting shells and an infinite line charge. Participants compare the electric flux through cylindrical surfaces at different radii, concluding that the flux at 1.5 cm is greater than at 3.5 cm due to proximity to a positive charge, while the flux at 4.5 cm is greater than at 5.5 cm since the latter is within a conducting shell where the electric field is zero. Additionally, it is established that the electric field at 2.5 cm is zero within a conducting shell, making the field at 4.5 cm greater. Overall, the answers provided align with the principles of Gauss' Law, emphasizing the relationship between enclosed charge and electric flux. The discussion highlights the importance of understanding electric fields in relation to conducting materials and their effects on flux calculations.
TwinGemini14
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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/u.../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

----------------------------------
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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You posted an incomplete link to your diagram.
TwinGemini14 said:
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

----------------------------------
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.
What determines the net flux through a Gaussian surface per Gauss's law? How does that compare for Gaussian surfaces at the two radii in question?

----------------------------------
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

----------------------------------
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.
Good thinking.


----------------------------------
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

------------------------
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.
Right. But the flux is zero not due to Gauss's law, but due to the fact that the field is zero.
 

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