Is There Flux Through the Lateral Surface of a Cylinder?

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SUMMARY

The discussion focuses on calculating the electric flux through the lateral surface of a cylinder. It establishes that the relevant area for flux calculations is the surface area of the ends of the Gaussian surface, given by the formula πR². The enclosed charge is calculated as Q enclosed = 100 * π * (0.1)² * 8.85, resulting in 27.8 pC. It is confirmed that there is no flux through the lateral surface of the cylinder in this specific problem.

PREREQUISITES
  • Understanding of electric flux and Gauss's Law
  • Familiarity with the formula for the surface area of a cylinder
  • Knowledge of the concept of Gaussian surfaces in electrostatics
  • Basic proficiency in calculus for area calculations
NEXT STEPS
  • Study Gauss's Law and its applications in electrostatics
  • Learn about calculating electric flux through different geometries
  • Explore the implications of enclosed charge on electric fields
  • Investigate the role of Gaussian surfaces in simplifying complex electrostatic problems
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Students studying electromagnetism, physics educators, and anyone seeking to understand electric flux in cylindrical coordinates.

Fatima Hasan
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Homework Statement


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Homework Equations


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The Attempt at a Solution


Here's my work :
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Your expressions for the areas don't look correct. When calculating ##\Phi_0##, what particular area are you working with? What is the formula for this particular area ?
 
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TSny said:
Your expressions for the areas don't look correct. When calculating ##\Phi_0##, what particular area are you working with? What is the formula for this particular area ?
screenshot_14.png
 

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Last edited:
Which parts of the surface of the cylinder have nonzero flux?
 
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TSny said:
Which parts of the surface of the cylinder have nonzero flux?
The area that we're concerned will be the surface area of the ends of Gaussian surface which is equals to π / R^2
Φ = E A = Q enclosed / ε
Q enclosed = ## 100 * π * (0.1)^2 * 8.85 = 27.8 pC ##
 
Last edited:
Fatima Hasan said:
The area that we're concerned will be the surface area of the ends of Gaussian surface which is equals to π / R^2
Of course you mean π⋅R^2.
Φ = E A = Q enclosed / ε
Q enclosed = 100 * π * (0.1)^2 * 8.85 = 27.8 pC
This is correct.
 
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TSny said:
Of course you mean π⋅R^2.
This is correct.
##A = 2 \pi r h## , we use this formula to find the net flux through a cylinder , right ?
 
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Fatima Hasan said:
##A = 2 \pi r h## , we use this formula to find the net flux through a cylinder , right ?
Not in this problem. The area ##A = 2 \pi r h## is the "lateral" area of the curved surface of the cylinder, as shown below in blue
upload_2018-3-3_14-55-27.png


Is there any flux through the blue surface in the problem you are working on?
 

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TSny said:
Not in this problem. The area ##A = 2 \pi r h## is the "lateral" area of the curved surface of the cylinder, as shown below in blue
View attachment 221399

Is there any flux through the blue surface in the problem you are working on?
No
 

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