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

Click For Summary
SUMMARY

The discussion focuses on applying Gauss' Law to compare electric field and flux magnitudes in a system involving an infinite line charge and multiple conducting shells. The line charge has a linear charge density of 2 µC/m, while the conducting shells have varying radii and net charges. Participants concluded that the electric field at 2.5 cm from the z-axis is zero due to the conducting shell, while the electric field at 4.5 cm is non-zero, leading to the conclusion that the flux through a 4.5 cm radius is greater than that through a 5.5 cm radius, which is zero. The answers provided by users were confirmed as correct based on the principles of Gauss' Law.

PREREQUISITES
  • Understanding of Gauss' Law and its mathematical formulation.
  • Familiarity with electric fields and flux concepts in electrostatics.
  • Knowledge of the behavior of conducting materials in electric fields.
  • Ability to analyze cylindrical symmetry in electrostatic problems.
NEXT STEPS
  • Study the derivation and applications of Gauss' Law in various geometries.
  • Explore the concept of electric field lines and their relationship to charge distributions.
  • Learn about the properties of conductors in electrostatic equilibrium.
  • Investigate the implications of charge density variations on electric field strength.
USEFUL FOR

Students of physics, electrical engineers, and anyone interested in understanding electrostatics and the application of Gauss' Law in complex charge configurations.

TwinGemini14
Messages
31
Reaction score
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/u.../elecshell.gif
----
Gauss' Law = |(E . DA) = Qenclosed / (Epsilon-not)
----

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.

----------------------------------
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.


----------------------------------
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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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!
 
Last edited by a moderator:
Physics news on Phys.org
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.
 

Similar threads

Electrostatics: Gauss' Law Problem Finding the Flux through a Conducting Spherical Shell
  • · Replies 9 ·
Replies
9
Views
2K
Gauss' law problem -- Should the field due to both the inside and outside charges be taken into account?
  • · Replies 28 ·
Replies
28
Views
1K
Gauss' Law: Conductor, Insulator, Electric Field
  • · Replies 7 ·
Replies
7
Views
1K
I Thought I Understood Gauss' Law (Sheets)
  • · Replies 26 ·
Replies
26
Views
3K
No Electric Charges if No Electric Field in Region
  • · Replies 22 ·
Replies
22
Views
3K
Gauss' law and Faraday cage problem
  • · Replies 10 ·
Replies
10
Views
2K
Electric field external to Conducting Hollow Sphere with charge inside
  • · Replies 23 ·
Replies
23
Views
4K
Calculating Electric Field Using Gauss's Law for a Spherical Charge Distribution
  • · Replies 4 ·
Replies
4
Views
4K
Solve Gauss' Law Homework: Charge of Insulating Sphere & Conductor Shell
  • · Replies 20 ·
Replies
20
Views
3K
Gauss' law: find the electric field
  • · Replies 5 ·
Replies
5
Views
2K