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Electric potential integral

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    An electric field is given by E = 2xi-3y2j N/C. Find the change in potential from the position rA = i - 2j m to rB = 2i + i + 3k m.

    2. Relevant equations

    VB - VA = -[itex]\int_A^B[/itex] E[itex]\cdot[/itex]ds

    3. The attempt at a solution

    ΔV= -[itex]\int_1^2[/itex]2x dx - [itex]\int[/itex]-3y^2 dy

    The second integral is supposed to be from -2 to 1

    And when I calculate this I get -10 V when the answer is +6 V

    I also noticed that if I simply invert the limits on the integral I get +6 V, is it just a coincidence or have I calculated the integral wrong?
     
  2. jcsd
  3. Oct 6, 2011 #2

    ehild

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    You can have a sign error somewhere. Show your work.


    ehild
     
  4. Oct 6, 2011 #3
    Well i have noticed that i'm very prone to making rudimentary mistakes often but I don't believe that to be the case here.

    When calculating the integral I do as follows:

    - [itex]\left[ x^2 \right]_{1}^{2}[/itex] - [itex]\left[ -y^3 \right]_{-2}^{1}[/itex] =
    -(4-1) - (-1+8)= -3 - 7 = -10 V
     
  5. Oct 6, 2011 #4

    ehild

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    You made the sign error at y^3. At the lower bound it is -(-(-(-8)))

    -x^2-(-y^3)=-x^2+y^3.


    -(4-1)+(1^3-(-2)^3)=-3+(1+8)=6

    ehild
     
  6. Oct 6, 2011 #5
    Oh I see now. My homework would be alot loss painful if I could avoid stuff like this but it doesn't matter how thorough I am, I often miss things anyways. I think I might have a mild form of dyscalculia or something..

    But thank you for your assistance!
     
    Last edited: Oct 6, 2011
  7. Oct 6, 2011 #6

    ehild

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    Get rid of the lot of minuses as soon as possible and use parentheses. No dyscalculia. I got the correct result at the fourth attempt.:rofl:
     
  8. Oct 6, 2011 #7
    Hehe well it's good to know that i'm not the only one who struggles with stuff like that :smile:
     
  9. Oct 6, 2011 #8

    SammyS

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    What is the z coordinate for rA ?
     
  10. Oct 7, 2011 #9
    There isn't a z coordinate for rA but as I understand that doesn't matter anyways since the electric field has no component in the z direction.
     
  11. Oct 7, 2011 #10

    ehild

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    That is the z coordinate is 0. :smile:

    ehild
     
  12. Oct 7, 2011 #11

    SammyS

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    Yup , so rB - rA has a z component.
     
  13. Oct 7, 2011 #12

    ehild

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    It has, but E has not.

    ehild
     
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