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Electric Field and Potential

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Derive the electric field for the potential function V= -x*y^2+z


    2. Relevant equations

    V=-∫Eds

    3. The attempt at a solution

    x*y^2-z=∫Eds

    d/ds(x*y^2-z)= E*s

    (y^2)dx/ds+(2y*x)dy/ds-dzds=E*s

    I'm stuck at this point and I'm not really sure if I'm going in the right direction.
     
  2. jcsd
  3. Feb 16, 2012 #2

    ehild

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    The electric field is negative gradient of the potential function. Do you know what the gradient is?

    ehild
     
  4. Feb 16, 2012 #3
    All I was given is whats in the initial question. Should I just divide the s over and call it a day?
     
  5. Feb 16, 2012 #4
    On the second step what if I multiplied both sides by s then divided by s^2 to get E by itself?
     
  6. Feb 16, 2012 #5

    ehild

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    E is the electric field, and it is a vector, having x, y, z components. You get these components as negative partial derivatives of the potential function.

    Ex=-∂V/∂x; Ey=-∂V/∂y; Ez=-∂V/∂z.

    ehild
     
  7. Feb 16, 2012 #6
    So in vector form would it be <y^2,2xy,1> = E ?
     
  8. Feb 16, 2012 #7

    ehild

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    There is a minus in front of the z component.

    ehild
     
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