Finding Electric Field for Spherical Charge Distribution?

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Homework Help Overview

The discussion revolves around finding the electric field generated by a spherical charge distribution with a specific charge density described by the function ρ(r) = Ke^(-br) within the region 0 < r < a. Participants are exploring the implications of using different methods to calculate the electric field based on this varying charge density.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the total charge Q by integrating the charge density over the volume and is uncertain about whether to use Q in the formula E = kQ/r^2 or to reevaluate the integral with the differential charge element dq. Some participants question the setup of the integral and the use of spherical coordinates, while others clarify the correct expression for dq.

Discussion Status

The discussion is active, with participants providing clarifications on the integration process and the correct formulation of the charge element. There is no explicit consensus on the method to use, but guidance has been offered regarding the correct expression for dq and the approach to take for the integration.

Contextual Notes

Participants are navigating the complexities of integrating a charge density that varies with radius and are addressing potential misunderstandings regarding the application of spherical coordinates in the context of this problem.

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Homework Statement


There is a charge density rho that exists in a spherical region of space defined by 0 < r < a.
[tex]\rho (r) = Ke^{-br}[/tex]
How do you find the electric field if a charge density varies as such?

The Attempt at a Solution



I found Q total = [tex]\int \int \int \rho dV[/tex]
Now I need to find E.

My real question is can I just put Q (as a function of r) into E = kQ/r^2? Or do I need to reevaluate the integral using dq = [tex]\rho r^2 sin(\theta) dr d\theta d\phi[/tex]

I get two different answers, (and I would have thought they should be the same) so which method is correct? I would have thought either would work.
 
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What do you mean? Q is the integral of the charge density over the volume. Also, I think you mean [itex]dr = rho r^2 sin(\theta) dr d\theta d\phi[/itex]. What did you do for your integral?
 
Why [tex]sin(\theta)[/tex]? rho depends only on r so [tex]dQ = 4\pi Kr^{2}e^{-br}dr[/tex]
 
Oh whoops, I shouldn't have had rho in there, and I missed it when you had it. You were right about the dq I was questioning. dq= rho *spherical jacobian (i.e. spherical integration differentials), which is what you had.

Yes, [tex]dQ = 4\pi Kr^{2}e^{-br}dr[/tex]

This is the way you want to go. I don't really understand what other way you would have gone about it.
 

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