Flux through a sphere due to a charge outside of it

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Homework Help Overview

The discussion revolves around the concept of electric flux through a spherical surface due to an external point charge, specifically a +5.00 μC charge located at a distance from the sphere. Participants are exploring the implications of Gauss's law, which states that the net flux through a closed surface is proportional to the charge enclosed, leading to questions about the behavior of electric field lines and flux in this scenario.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to reconcile the idea that the net flux is zero with the observation that the electric field strength varies across the sphere's surface. Questions are raised about how the flux entering and exiting the sphere can be equal despite differences in field strength.

Discussion Status

The discussion is ongoing, with participants providing insights about field lines and their relationship to electric flux. Some have suggested that while the magnitude of the field changes, the total number of field lines entering and exiting remains constant, indicating a productive exploration of the topic.

Contextual Notes

There is a focus on the conceptual understanding of electric field lines and flux, with some participants expressing confusion about the implications of varying field strengths on the net flux. The discussion reflects a mix of interpretations regarding the physical representation of electric fields.

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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