How Can Surface Charge Density Equal bcos(θ)?

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Homework Help Overview

The discussion revolves around the interpretation of surface charge density defined on a hemisphere with a radius denoted as b. The original poster questions how the surface charge density can be expressed as Rcos(θ) = bcos(θ) while maintaining appropriate units, particularly in relation to the coordinate z.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the surface charge density and the coordinates involved, particularly questioning the units of measurement for z and b. There is discussion about potential typos in the original statement regarding units.

Discussion Status

Some participants have provided clarifications regarding the units of charge density and have suggested alternative formulations for the surface charge density. There is an ongoing examination of the definitions and relationships between the variables involved, with no explicit consensus reached yet.

Contextual Notes

Participants are addressing potential misunderstandings regarding the dimensional analysis of the variables involved, particularly focusing on the implications of using different units for charge density in a surface context versus a volume context.

jeff1evesque
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Homework Statement



A surface is defined by a hemisphere of radius b, centered on the x-y plane. The surface charged density is given by [tex]\rho_s(z) = z (\frac{Coul}{m^3})[/tex].

Homework Equations


[tex]\rho_s(z) = z = Rcos(\theta) = bcos(\theta) (\frac{Coul}{m^3})[/tex].


3. Question
My question is how can the surface charge density equal to [tex]Rcos(\theta) = bcos(\theta)[/tex]? That is a measure of the distance from the origin to the surface [element], and thus only [to my knowledge] have units of radius b, or meters [tex]\neq (\frac{Coul}{m^3})[/tex].

Thanks,


JL
 
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z has no units, it's just a coordinate in the coordinate system, and what they are saying is that the charge density at points with coordinates (x,y,z) is equal to z C/m2. (You wrote C/m3 which I'm assuming is a typo since we're talking about a surface, not a 3D region.)
 
dx said:
z has no units, it's just a coordinate in the coordinate system, and what they are saying is that the charge density at points with coordinates (x,y,z) is equal to z C/m2. (You wrote C/m3 which I'm assuming is a typo since we're talking about a surface, not a 3D region.)

Sorry about the typo. But what about the variable b which has units of meters? How does that fit into the interpretation of surface charges?

Thank you.
 
dx said:
... the charge density at points with coordinates (x,y,z) is equal to z C/m2. (You wrote C/m3 which I'm assuming is a typo since we're talking about a surface, not a 3D region.)
No typo. One unit of length in z cancels one unit of length in the denom. of C/m^3.

jeff1evesque said:
[tex]\rho_s(z) = z = Rcos(\theta) = bcos(\theta) (\frac{Coul}{m^3})[/tex].
This is misleading (i.e. wrong). The first inequality should be confusing. I would suggest:
[tex] \rho_s(z)=\rho_0z\rightarrow{}\rho_0Rcos(\theta)=bcos(\theta) (C/m^3)[/tex]
where [itex]\rho_0[/itex] is some unknown constant that has units of charge-per-volume and R and b have units of length.
Or, better yet,
[tex] \rho_s(z)\rightarrow\rho_0cos(\theta)[/tex]
where [itex]\rho_0[/itex] is some unknown constant with units of C/m^2.
 
turin said:
No typo. One unit of length in z cancels one unit of length in the denom. of C/m^3.

This is misleading (i.e. wrong). The first inequality should be confusing. I would suggest:
[tex] \rho_s(z)=\rho_0z\rightarrow{}\rho_0Rcos(\theta)=bcos(\theta) (C/m^3)[/tex]
where [itex]\rho_0[/itex] is some unknown constant that has units of charge-per-volume and R and b have units of length.
Or, better yet,
[tex] \rho_s(z)\rightarrow\rho_0cos(\theta)[/tex]
where [itex]\rho_0[/itex] is some unknown constant with units of C/m^2.

That makes much more sense, thank you.
 

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