Gauss's Law long thin wire question

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SUMMARY

This discussion focuses on applying Gauss's Law to determine the electric field generated by a long thin wire with a linear charge density, denoted as λ. The participants derive the electric field magnitude, E, as E = λ/(2πrε₀), where r is the distance from the wire and ε₀ is the permittivity of free space. They also explore the Cartesian components of the electric field, concluding that E_x = (λx)/(2πr²ε₀) and E_y = (λy)/(2πr²ε₀). The conversation highlights common mistakes in applying Gauss's Law, particularly regarding the surface area of the Gaussian cylinder and the total charge enclosed.

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  • #31
Along with the attached diagram

cos theta = x/r

sin (90 - theta) = x/r

TFM
 

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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
 

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  • #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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