Find electric field strength expression for a specfic region

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Homework Help Overview

The problem involves finding an expression for the electric field strength within a thick, spherical shell characterized by an inner radius \(a\) and an outer radius \(b\), which carries a uniform volume charge density \(\rho\). The focus is on the region where \(a < r < b\).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Gauss's Law and the determination of the enclosed charge within the spherical shell. There are attempts to express the enclosed charge in terms of the charge density and the relevant dimensions, with some confusion regarding the use of area versus volume in calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have attempted calculations and expressed uncertainty about the correct application of volume and area in the context of Gauss's Law. Guidance has been offered regarding the need to consider the volume of the shell and how to express the enclosed charge correctly.

Contextual Notes

Participants note the importance of correctly identifying the volume of the spherical shell and the relationship between the inner and outer radii. There is a mention of constraints related to the problem statement, which specifies expressing the answer in terms of \(r\), \(a\), \(b\), \(\rho\), and \(\epsilon_0\).

  • #31
gneill said:
Right. And similarly, the volume of a sphere of radius a is (4/3)πr3.

The sphere of radius a happens to be the volume of the empty cavity within the shell. So the volume of the shell from a out to r is equal to the volume of the solid sphere of radius r minus the volume of the sphere of radius a (you're removing the volume of the cavity from that of a solid sphere).
So ((4/3)π(r3-a3))/(4πε0))ρ.
 
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  • #32
jlmccart03 said:
So ((4/3)π(r3-a3))/(4πε0))ρ.
Close. You're missing the r2 in the denominator that belongs to the Gaussian surface area. Then you can cancel the 4π's that are common to the numerator and denominator.
 
  • #33
gneill said:
Close. You're missing the r2 in the denominator that belongs to the Gaussian surface area. Then you can cancel the 4π's that are common to the numerator and denominator.
Ok so that would mean it becomes (r3-a3)/(3ε0r2))ρ?
 
  • #34
Yes, that looks good.
 
  • #35
gneill said:
Yes, that looks good.
Phew! Thanks for the help even though I got really lost, but I really appreciate you working through the roadblocks I was having! Thank you!
 
  • #36
You're welcome :smile:
 

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