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littlecat

Charge q1 is at origin and charge q3 is at (a,a). Why is unit vector r1,3 written as r1,3 = cosθi + sinθ j = sqrt2/2 (i + j)?

  • one year ago
  • one year ago

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  1. littlecat
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    Visualization of my question

    • one year ago
  2. Xishem
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    Writing...\[\vec{r}_{13} =\cos \theta \hat i + \sin \theta \hat j\]is the same thing as writing...\[\vec r_{13}=r_x\hat i +r_y\hat j\]Can you see where this relationship is derived now?

    • one year ago
  3. littlecat
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    hmm.. if it is the same, why was it not written as \[r⃗ 13=iˆ+jˆ\] ? I attempted to use polar co-ordinates which gave me the answer r^ = a (i^ + j^), this eventually leads to a wrong answer in this superposition question

    • one year ago
  4. littlecat
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    also kudos to your superb equation typing skills :)

    • one year ago
  5. Xishem
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    i-hat and j-hat only give an indication of the direction of the vector. By writing...\[\vec r_{13}=\hat i+\hat j\]You are implying that \[\vec r_{13}=1\hat i+1\hat j\]Which is not the case. The i-hat is just a unit vector, only implying direction. If you project the vector r_13 onto the x- and y- axes and break it down into its component vectors you can see where the expression it gives is derived from.

    • one year ago
  6. Xishem
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    As far as where the far RHS of the equation is derived from... Let me see what I can figure out.

    • one year ago
  7. littlecat
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    cos θ = sqrt2/2 = sin θ ... but i'm still confused about the LHS!

    • one year ago
  8. Xishem
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    Alright. I don't understand how the middle part of the equation was derived. It seems to be missing a factor of r.|dw:1345972919614:dw|I would think that it would be, then...\[\vec r_{13}=\vec r_{13}\cos \theta \hat i+\vec r_{13}\sin \theta \hat j\]I'm not sure where that r_13 factor is lost.

    • one year ago
  9. littlecat
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    the figure shows an isosceles triangle.. why is the projection of the hypothenuse r1,3 not equal to the length a...?

    • one year ago
  10. Xishem
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    The projection would be equal to a.\[r_x=a \hat i; r_y=a\hat j\]I believe.

    • one year ago
  11. littlecat
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    yes! then r_13 = a(i^ + j^) ? but the answer shows r_13 = sqrt2/2 (i^ + j^) :E

    • one year ago
  12. littlecat
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    the factor of r may have been lost due to the fact that it is a unit vector... i'm not sure myself

    • one year ago
  13. Xishem
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    Oh! It is a unit vector, so that would explain why the factor is lost... And for the other part, hmm.

    • one year ago
  14. Xishem
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    I think the issue is that you can't express a unit vector in terms of a constant. Since r is a unit vector and not a standard vector, the following aren't true.\[\hat r \neq a\hat r_x+a\hat r_y\]

    • one year ago
  15. Xishem
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    If r were a standard vector, it would be true, but since it's a unit vector, things work out differently. My understanding of unit vectors is shaky, so, that's about as much as I can tell from it.

    • one year ago
  16. littlecat
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    OH that actually helps :) I'll think of it as basic polar co-ordinates x = rcosθ = r sqrt2/2 y = rsinθ = r sqrt 2/2 without the factor r but keeping sqrt2/2 since it is referenced to θ THANKS!

    • one year ago
  17. Xishem
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    Yep. I think that the constant is worked into the unit vector somehow. Unit vectors are really something I should review :P.

    • one year ago
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