Why Does Gauss's Law Use Cylinder Circumference Instead of Area?

In summary, the conversation discusses the use of Gauss's Law and the calculation of electric flux on a cylindrical surface. It is noted that the circumference of the cylinder is used in the calculation instead of the full area, which is explained by the fact that the electric field lines do not pass through the ends of the cylinder. The understanding is clarified and gratitude is expressed to "Doc Al".
  • #1
CollegeStudent
109
0

Homework Statement



So I was looking at this site reading up again on Gauss's Law and was a tad confused

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html

1st thing on the page...

Homework Equations


The Attempt at a Solution



It states that the electric flux will just be the Electric field times the area of the cylinder...

However in the calculation, the circumference of the cylinder is used...why is that?
 
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  • #2
CollegeStudent said:
However in the calculation, the circumference of the cylinder is used...why is that?
The circumference times the length L gives the surface area.
 
  • #3
Oh okay! I was taking into account the full area but since the Electric field lines will never cross through the 2 ends of the cylinder the area needed will just be the circumference times the length..

Thank you Doc Al!
 

What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface.

How is Gauss's Law used in physics?

Gauss's Law is used to calculate the strength and direction of the electric field created by a distribution of charges. It is also used to determine the net charge contained within a closed surface.

What is the mathematical formula for Gauss's Law?

The mathematical formula for Gauss's Law is ∮E⋅dA = Q/ε₀, where E is the electric field, dA is a differential area element, Q is the total charge enclosed by the surface, and ε₀ is the permittivity of free space.

How do you apply Gauss's Law to solve problems?

To apply Gauss's Law, you must first choose a closed surface that encloses the charge distribution of interest. Then, you need to calculate the electric flux through that surface using the formula ∮E⋅dA. Finally, you equate that value to Q/ε₀ and solve for the electric field at any point on the surface.

What are some real-life applications of Gauss's Law?

Gauss's Law has many real-life applications, including determining the electric field strength inside a charged capacitor, calculating the force between two charged objects, and understanding the behavior of lightning. It is also used in the design and analysis of electrical circuits and electronic devices.

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