Explaining Electric Flux with Enclosed Charges

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SUMMARY

The discussion centers on the concept of electric flux and its relationship with enclosed charges, specifically using Gauss's Law. Electric flux is defined as the integral of the electric field (E) over an area (A), expressed mathematically as flux = E dot A. It is established that the total electric flux through a closed surface is proportional to the charge enclosed, regardless of the position of the charge within the surface, as demonstrated by the example of a sphere with a charge at its center versus a charge near the edge.

PREREQUISITES
  • Understanding of electric fields and their properties
  • Familiarity with Gauss's Law
  • Knowledge of vector calculus, specifically dot products
  • Basic concepts of charge distribution
NEXT STEPS
  • Study Gauss's Law in detail, including its mathematical derivation
  • Explore the concept of electric field lines and their relation to electric flux
  • Learn about different charge distributions and their effects on electric fields
  • Investigate the implications of electric flux in various geometrical configurations
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of electromagnetism, particularly those studying electric fields and flux in relation to enclosed charges.

brushman
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I under stand that electric flux is proportional to the charged enclosed but I don't get why.

Homework Statement


Consider the sphere with the charged enclosed:
[PLAIN]http://electron9.phys.utk.edu/phys136d/modules/m4/images/Image985a.gif

As you can see the area and electric field vectors are always parallel.

Now compare that situation to one exactly the same except the charge is moved towards the edge of the sphere (but still inside).

In the first situation the area and electric field are parallel to each other so will have the max value. In the second situation the angle is not always 0 degrees.

How can situation two have the same electric flux when it has smaller angles?

Homework Equations


flux = E dot A



Thanks
 
Last edited by a moderator:
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hi brushman! :smile:

the flux is ∫∫ E.dA

that is the same as ∫∫∫ ∇.E dxdydz

and ∇.E = 0 in empty space …

so draw a small imaginary sphere with the charge as its centre … the flux through the surfaces of both spheres must be the same (since there's no charge between them) :wink:
 

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