Gauss' law for point charge inside sphere off center

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SUMMARY

The discussion centers on Gauss' Law as it applies to a point charge located off-center within a Gaussian sphere. The electric field at any point on the surface of the sphere is expressed as Qenc/(ε0 * 4 * π * r²), where Qenc is the enclosed charge and ε0 is the permittivity of free space. The conversation highlights the relationship between charge density and area in the context of electric field strength, emphasizing that while the electric field varies in strength across the surface, the total electric flux remains constant, validating Gauss' Law.

PREREQUISITES
  • Understanding of Gauss' Law and its mathematical formulation
  • Familiarity with electric field concepts and charge density
  • Knowledge of surface integrals in electromagnetism
  • Basic principles of geometry related to cones and areas
NEXT STEPS
  • Study the derivation of Gauss' Law in electrostatics
  • Explore the concept of electric flux and its applications
  • Investigate the effects of charge distribution on electric fields
  • Learn about the implications of non-uniform electric fields in physics
USEFUL FOR

Students of physics, particularly those studying electromagnetism, educators teaching Gauss' Law, and anyone interested in the mathematical foundations of electric fields and charge distributions.

gimak
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Homework Statement


If a point charge is inside a Gaussian sphere but is off center, why is its electric field still Qenc/(e0*4*pi*r^2)?

Homework Equations


surface integral of E*da=Qenc/e0

The Attempt at a Solution


If we draw cones out from the charge. the 2 surfaces from the cones' intersection has a charge/area. The bottom of the cone (circle) has areas A1 and A2. The slant of each cone is r1 and r2.

Then, the teacher said (charge density)(A1/r1^2)=(charge density)(A2/r2^2). I don't know how he jumped to this. He said this proves that the electric field is still the above.
 
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gimak said:

Homework Statement


If a point charge is inside a Gaussian sphere but is off center, why is its electric field still Qenc/(e0*4*pi*r^2)?

Homework Equations


surface integral of E*da=Qenc/e0

The Attempt at a Solution


If we draw cones out from the charge. the 2 surfaces from the cones' intersection has a charge/area. The bottom of the cone (circle) has areas A1 and A2. The slant of each cone is r1 and r2.

Then, the teacher said (charge density)(A1/r1^2)=(charge density)(A2/r2^2). I don't know how he jumped to this. He said this proves that the electric field is still the above.

The electric field is not the same. It's stronger where the charge is closer to the surface. What is the same is the total electric flux (the integral of the field through the surface). You teacher is trying to give you an argument to show Gauss's Law is true. The field gets stronger over some region but it's compensated by getting proportionately weaker over another region but in such a way that the total sum is constant.
 
Last edited:
=So for A1, let's say that the charge is closer to that than A2. Because E1=(charge density *A1)/R1^2 and E2=(charge density *A2/R2^2). We can say that magnitude E1=magnitude E2? So for A1, even though the charge is closer to that area, A1<A2. For A2, even though the charge is farther away from that, the fact that A2>A1 compensates for that?
 

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