Flux through a sphere due to a charge outside of it

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SUMMARY

The discussion centers on the application of Gauss's law to a scenario involving a point charge of +5.00 μC located at x=5.00 m, adjacent to a spherical surface of radius 4.00 m centered at the origin. According to Gauss's law, the net electric flux through the sphere is zero since it contains no enclosed charge. Despite the varying electric field strength on the near and far sides of the sphere, the total number of electric field lines entering the sphere equals the total exiting, resulting in zero net flux. The concept of electric field lines is emphasized as a useful visualization tool, despite their lack of physical existence.

PREREQUISITES
  • Understanding of Gauss's law and its mathematical formulation: integral(E dA cos(theta)) = Qenclosed / E0
  • Familiarity with electric field concepts and field line representation
  • Basic knowledge of electric charge and its effects on surrounding space
  • Ability to visualize three-dimensional charge distributions and their impact on electric fields
NEXT STEPS
  • Study the implications of Gauss's law in different charge configurations
  • Explore the concept of electric field lines and their relationship to electric field strength
  • Learn about the mathematical derivation of electric flux and its applications
  • Investigate the behavior of electric fields in non-uniform charge distributions
USEFUL FOR

Students of physics, particularly those studying electromagnetism, educators explaining Gauss's law, and anyone interested in understanding electric fields and flux in relation to charge distributions.

xSpartanCx
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Homework Statement



A point charge of +5.00 μC is located on the x-axis at x= 5.00 m , next to a spherical surface of radius x= 4.00 m centered at the origin.
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According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at x=4.00 m ) than on the far side (at x= -4.00 m ). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)?

Homework Equations



integral(E dA cos(theta)) = Qenclosed / E0

The Attempt at a Solution



I know that the flux should be zero. I drew a picture, with the charge to the right of the sphere and drew lines radially from the point. However, I can't seem to wrap my head around why the net flux is zero. If I go through adding up all the E dot dA's, does it have to do with the dA being the opposite direction of the field on the close side to the particle, so flux is negative, but on the top, bottom, and left (front and back as well) sides of the sphere, the field is in a similar direction so flux would be positive?
 
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Notice that every field line that enters the sphere will end up exiting the sphere.
 
axmls said:
Notice that every field line that enters the sphere will end up exiting the sphere.
But it exits the sphere with less magnitude than when it entered, doesn't that make a difference?
 
xSpartanCx said:
But it exits the sphere with less magnitude than when it entered, doesn't that make a difference?

Field lines are just a convenient way to picture an electric field. They don't have "intensity". The density of the field lines indicates intensity of the field. The total number of field lines is the total flux. The intensity is less on the far side of the sphere because they are spread more. But the total number exiting is the same as the total number entering.
 
xSpartanCx said:
But it exits the sphere with less magnitude than when it entered, doesn't that make a difference?

Field lines do not have actual, physical existence, but the concept is essentially that such a line is supposed to have constant strength throughout its entire length. Where there is a high density of such lines (i.e., where they are bunched up close together) the field is strong, essentially because you sum a large number of individual line strengths over a small area; where the density of such lines is low (they are "far apart") the field strength is weak. In terms of real physics this is, of course, total nonsense, but the diagrams and pictures can sometimes help one to get an intuitive feeling for the situation for various charge configurations.
 

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