Finding charge density generating an Electric Field

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SUMMARY

The discussion focuses on demonstrating that the electrostatic field E = br is irrotational and determining the charge density ρ(r) that generates this electric field. The key equations utilized include ∇ x E = 0, confirming the irrotational nature of E, and ρ = ε0(div E), where div E is derived from the divergence of the vector field br. The divergence of the vector r = (x, y, z) is computed, leading to the conclusion that ρ = bε0(div r) is valid, with r interpreted as a vector rather than a scalar.

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  • Understanding of vector calculus, specifically divergence and curl operations.
  • Familiarity with electrostatics concepts, including electric fields and charge density.
  • Knowledge of Maxwell's equations, particularly the relationship between electric fields and charge density.
  • Proficiency in using mathematical notation for vector fields and operations.
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  • Study the properties of irrotational fields in electrostatics.
  • Learn about the divergence theorem and its applications in electrostatics.
  • Explore the implications of Maxwell's equations on electric fields and charge distributions.
  • Investigate the mathematical techniques for calculating divergence in three-dimensional vector fields.
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Students of physics, particularly those studying electromagnetism, as well as educators and researchers interested in electrostatic field theory and vector calculus applications.

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Homework Statement



Show that electrostatic field E = br
where b is a constant
is irrotational

Find the charge density ρ(r) which can generate this electric field

Homework Equations



∇ x E = 0 (following stoke theorem)

ρ = ε0(div E) since: div E = [itex]\frac{ρ}{ε}[/itex]

The Attempt at a Solution



The first part (I'm assuming) will suffice as ∇ x E = 0

The second part ρ = ε0(div E) = ε0(div br) = bε0(div r)

The question is whether I can compute (div r) or not. Does "r" in this particular case equal to square root (x+y+z)2? I'm assuming I can't know data given, which is why I am tempted to leave the solution as it is.

Thanks
 
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You cannot take the divergence of a scalar, so I suppose that r is actually the vector:

[itex]\vec r=(x,y,z)[/itex]
 
Ah good, that's what I was thinking. Thanks
 

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