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Electric field.

  1. Sep 7, 2007 #1
    Given is the eletrical field [tex]&\mathbf{E}& = k[y^2 \^{&\mathbf{x}&} + (2xy + z^2)\^{&\mathbf{y}&} + 2yz \^{&\mathbf{z}&}][/tex]


    I will like to find the potential so I integrate and become [tex]V(r) = -k (y^2x + xy^2 + z^2y + yz^2)[/tex] then I try to find the elektrical field again, so I differentiate this potential and become:

    [tex] \frac{\partial v }{\partial x} =- k ( y^2 + y^2 )[/tex]
    [tex]\frac{\partial v}{\partial y}=-k (2yx + 2xy + z^2 + z^2 )[/tex]
    [tex]\frac{\partial v} {\partial z}=-k (2zy + 2yz) [/tex]

    Wy find I two times the eletrical field? and not ones? Thanks.
     
  2. jcsd
  3. Sep 7, 2007 #2

    learningphysics

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    The problem is with the way you're doing the line integral...

    E.dl = k[y^2dx + (2xy+z^2)dy + 2yzdz]

    Break up the path from
    (0,0,0) to (x,y,z) into 3 paths:

    (0,0,0) to (x,0,0) here... dy and dz = 0... so you only need to worry the first part... the second and third are 0... note that y and z.
    (x,0,0) to (x,y,0) ... here only the second part (here x and z are constant)
    (x,y,0) to (x,y,z) .... third part (here x and y are constant)

    This way you'll get a different potential than the one you got.
     
  4. Sep 7, 2007 #3
    Thus the problem is path dependent? so i calculate the rot from the eletric field and that gives 0. Thus a conservative field? why not?
     
  5. Sep 7, 2007 #4

    learningphysics

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    It is a conservative field (you can check this... the curl of the field is 0 everywhere so it is conservative)... any path from (0,0,0) (or any other fixed point) to (x,y,z) will work... but you didn't use a path from (0,0,0) to (x,y,z)...

    You used:

    (0,y,z) to (x,y,z) in the first part...

    (x,0,z) to (x,y,z) in the second part

    (x,y,0) to (x,y,z) in the third part...

    But these three together don't form a path from (0,0,0) to (x,y,z).

    You need something like:

    (0,0,0) to A
    A to B
    B to C
    C to (x,y,z)

    A, B and C can be any intermediate points...
     
  6. Sep 7, 2007 #5
    sorry I don't see why my path hold not for every arbitrary (x,y,z) I will like to calculate the potential in every point in space in one time?
     
  7. Sep 7, 2007 #6

    learningphysics

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    The thing to remember is that although:

    [tex]&\mathbf{E}& = k[y^2 \^{&\mathbf{x}&} + (2xy + z^2)\^{&\mathbf{y}&} + 2yz \^{&\mathbf{z}&}][/tex]

    is path independent... the x y and z components are each path dependent...

    What I mean is suppose you get the integral of x component of the field... you choose a path from (0,0,0) to (x,y,z)... then when you do the integral of the y component, you need to use the same path for that part... and then the same path for the z part...

    For the first part, suppose you chose (0,0,0) to (0,y,0) to (0,y,z) to (x,y,z)... then in your next two parts, you should choose the same path...

    Choose the same path for each of the 3 parts... what's going to happen is everything will turn out 0 except... (0,y,z) to (x,y,z) for the first... (0,0,0) to (0,y,0) for the second... and (0,y,0) to (0,y,z) for the third...

    Let me know if this makes sense...
     
  8. Sep 7, 2007 #7

    learningphysics

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    You need the same path for the x component, y component and z component... choose 1 path... then stick to that path when doing the integration...

    From what point to what point are you calculating the line integral...
     
  9. Sep 7, 2007 #8
    oké I need to remove x y components en substitute somthing like y=kx etc ?
     
  10. Sep 7, 2007 #9

    learningphysics

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    You could... but it's unnecessary...

    Choose a simple path from (0,0,0) to (x,y,z)... let's say

    (0,0,0) to (x,0,0) along the x axis
    then (x,0,0) to (x,y,0) in the y direction
    then (x,y,0) to (x,y,z) in the z direction

    [tex]-\int{k[y^2 \^{&\mathbf{x}&}]\cdot\vec{ds}}[/tex]

    over this path... what do you get?

    use the same path for

    [tex]-\int{k[(2xy + z^2)\^{&\mathbf{y}&}]\cdot\vec{ds}}[/tex]

    And then for

    [tex]-\int{k[2yz \^{&\mathbf{z}&}]\cdot\vec{ds}}[/tex]

    summing the 3 what's your field?
     
    Last edited: Sep 7, 2007
  11. Sep 7, 2007 #10
    for the first [tex]y=0[/tex] so some ct so [tex]-k ct[/tex]

    for the second [tex] \int -k \ 2xy = -k \ \frac{2}{2}xy^2 =-k \ xy^2[/tex]

    for the last one [tex]-k \ 2 \ y \ \frac{1}{2} z^2=-k \ y \ z^2 [/tex]

    so finaly [tex]-k -kxy^2 - kyz^2 [/tex] is this oké?
     
  12. Sep 7, 2007 #11

    learningphysics

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    How do you get that?

    These two are good.
     
  13. Sep 7, 2007 #12
    the first [tex]-k \int y^2 dx [/tex] because the path over y is 0 so y=0 therfore, I think that we get [tex]- k \int 0 \ dx [/tex] and this is constant so [tex]- k \int 0 \ dx =c [/tex] is that wrong?
     
  14. Sep 7, 2007 #13

    learningphysics

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    It should really be an integral with limits... [tex]\int_{(0,0,0)}^{(x,0,0)}0dx[/tex] so the constants cancel and the result is 0.

    Also, we're taking the convention that V(0,0,0) =0
     
  15. Sep 7, 2007 #14
    Oké thanks.
     
  16. Sep 7, 2007 #15

    learningphysics

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    No prob. So V = -kxy^2 - kyz^2

    gives back the E you need...
     
  17. Sep 7, 2007 #16
    right thanks.
     
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