osanseviero
  • osanseviero
Electric field question
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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osanseviero
  • osanseviero
There is an electrical field. There is a line with constant density. We know L (how long is the line). At a distance a from the middle, there is a point. Which is the field there? |dw:1441428125288:dw|
UnkleRhaukus
  • UnkleRhaukus
is \(L\gg a\)?
osanseviero
  • osanseviero
Yep

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osanseviero
  • osanseviero
It is not infinite though, so I don't think we can use Gauss here
osanseviero
  • osanseviero
Actually it does not say anything about the relationship between a and L
UnkleRhaukus
  • UnkleRhaukus
i don't know how to deal with the fringing effects the occur at the the end point of the line, i do know how to get the E field if the line if infinite, (good approximation if a<
UnkleRhaukus
  • UnkleRhaukus
hmm i maybe if a is a point above the MIDDLE of the line , the fringing effect from both sides cancel
osanseviero
  • osanseviero
It is in the middle
osanseviero
  • osanseviero
So I would guess that in horizontal all of them cancel
UnkleRhaukus
  • UnkleRhaukus
|dw:1441430062457:dw|
osanseviero
  • osanseviero
So I would have to use Gauss?
UnkleRhaukus
  • UnkleRhaukus
yeah use this cylinder as your gaussian pill box
osanseviero
  • osanseviero
Doing that I have E(2pi*a*L) = Qin / epsilon. But they told me doing this was not correct
UnkleRhaukus
  • UnkleRhaukus
is Qin the total charge of the line?
UnkleRhaukus
  • UnkleRhaukus
i think you have to express the charge in terms of a linear charge density \(\lambda\) \[\lambda = q/L\]
osanseviero
  • osanseviero
E = lambda/(2 * pi * a * Eo) (Eo is the epsilon)
osanseviero
  • osanseviero
But still...how would you solve this without Gauss?
UnkleRhaukus
  • UnkleRhaukus
you mean like using Coulombs law?
osanseviero
  • osanseviero
Yep. Coulomb + Electric field
osanseviero
  • osanseviero
And charge distributions
osanseviero
  • osanseviero
Found an answer, thanks a lot

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