Gauss's Law long thin wire question

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Homework Help Overview

The discussion revolves around applying Gauss's Law to a long thin wire carrying a linear charge density, λ. Participants are tasked with finding the electric field's magnitude at a certain distance from the wire and expressing the electric field components in Cartesian coordinates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Gauss's Law, particularly the choice of Gaussian surface and the calculation of electric field components. There are attempts to derive the electric field's magnitude and its components, with some participants questioning the correctness of their methods and assumptions.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's reasoning. Some guidance has been offered regarding the correct application of Gauss's Law and the interpretation of charge density. Multiple interpretations and approaches are being explored, particularly concerning the relationship between the charge and the dimensions of the Gaussian surface.

Contextual Notes

Participants are navigating potential misunderstandings about the geometry involved in the problem, including the surface area of the Gaussian cylinder and the relationship between charge density and total charge. There is also a focus on ensuring clarity in the use of coordinate systems and vector components.

  • #31
Along with the attached diagram

cos theta = x/r

sin (90 - theta) = x/r

TFM
 

Attachments

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  • #32
You're almost there. Do this. Draw a diagram, similar to what you just drew but with the x and y axes in the usual position. On that diagram show the electric field vector at some arbitrary position. Find the x and y components of that field. (You already have the value of the total field at any point.)
 
  • #33
So for the attached diagram:

E-x = Rsin theta

E-y Rcos theta

?

TFM
 

Attachments

  • #34
I presume you mean something like:
E-x = E sinθ
E-y = E cosθ

Now just rewrite sinθ & cosθ in terms of x,y, & r.
 
  • #35
So would that use the equations I made earlier of

cos θ = x/r
sin θ = y/r

giving

E-x = E x/r

E-y = E y/r

TFM
 
  • #36
TFM said:
So would that use the equations I made earlier of

cos θ = x/r
sin θ = y/r

giving

E-x = E x/r

E-y = E y/r

TFM
Good. Plug in the value of E and you're done.
 
  • #37
So:

E = lambda/(2*pi*r*epsilon-0)

Thus:

E-x = (lambda*x)/(2*pi*r^squared*epsilon-0)

and

E-y = (lambda*y)/(2*pi*r^squared*epsilon-0)

TFM
 
  • #38
finally...yes!
 
  • #39
Thats good.

Thanks Everybody :smile:

TFM
 

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