Is There Flux Through the Lateral Surface of a Cylinder?

AI Thread Summary
The discussion centers on whether there is flux through the lateral surface of a cylinder. Participants clarify that the relevant area for calculating flux is the surface area of the ends of the Gaussian surface, which is πR². The equation for electric flux, Φ = EA = Q enclosed / ε, is confirmed, with a specific example calculating Q enclosed as 27.8 pC. It is emphasized that the lateral area formula (A = 2πrh) does not apply in this context, as there is no flux through the lateral surface. Ultimately, the conclusion is that there is no flux through the lateral surface of the cylinder in this problem.
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 ?
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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 ##
 
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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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