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The electric field of a charged semicircle from its electric potential

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A question involving the relationship between the electric
    field E(r) and the electric potential V( r) is about computing the electric
    field on the z-axis due to a uniform line charge distribution λ
    spread out on a semicircle of radius R, lying on the first
    half of the XY plane (0 ≤ θ ≤ π). A few points on the
    semicircle are (x=R,y=0) at θ= 0, (x=0,y=R) at θ= π/2,
    and (x=-R,y=0) at θ = π (I am using the symbol π as Pi).

    2. Relevant equations

    The electric potential V( r) is easily computed giving the result
    V(z) = K*Q/Sqrt(z^2 + R^2). Q = λπR and K is the electric constant.

    3. The attempt at a solution

    The electric field is suppose to be
    obtained via E = - grad(V) = - d(V(z))/dz. But this relation leads only
    to one component of the electric field.

    Could somebody point out what is missing?

    Sergio
     
  2. jcsd
  3. Feb 8, 2013 #2

    tiny-tim

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    Welcome to PF!

    Hi Sergio! Welcome to PF! :smile:
    No, the gradient of V is the vector (∂V/∂x,∂V/∂y,∂V/∂z) :wink:
     
  4. Feb 8, 2013 #3

    haruspex

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    Re: The electric field of a charged semicircle from its electric poten

    Yes, but only knowing the potential on the z axis is not enough information to compute the field there. You also need to know how the potential will change if you move some infinitesimal off that axis. Can you compute the full potential V=V(x,y,z)?
     
  5. Feb 9, 2013 #4
    Re: The electric field of a charged semicircle from its electric poten

    All right, haruspex . I see the point now. Some how I was mislead
    by the textbook.

    After writing the potential at any point in space (x,y,z) and
    Leaving it in integral form, I then computed its derivatives
    with respect to each one of the components of the field
    point (x,y,z). Then, setting x=y=0 on each one,
    integration respect to the source variable yields the expected answer.

    Thanks,

    Sergio
     
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