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Magnitude and direction of electric field at point P

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data

    An electric dipole is defined as two point charges separated by distance L.
    Consider point P with coordinates (x,0).
    A dipole along the x axis is separated by distance a from the origin. q- is closer to point P
    Find:

    1) Electrostatic potential due to dipole at point P
    2) Magnitude and direction of electric field E at point P in terms of dipole moment and distance x

    2. Relevant equations

    V = k[itex]\frac{q}{r}[/itex]

    3. The attempt at a solution

    1) Vtot = V+q + V-q

    = kq([itex]\frac{1}{x+a}[/itex] - [itex]\frac{1}{x-a}[/itex])

    =[itex]\frac{2kqa}{x^2}[/itex]

    2) I know that taking the divergence of [itex]\frac{2kqa}{x^2}[/itex] will give a scalar. Is this divergence going to be [itex]\frac{d}{dx}[/itex]? How do I go about getting the magnitude and direction from there?

    Thank you
     
  2. jcsd
  3. Jun 13, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The value of the gradient is the magnitude of the vector. Since both poles are on the x-axis and your given point is (x, 0), the vector will point along the x-axis to the nearest pole. That means that for x> 0 it points toward (a, 0) and for x< 0, it points toward (-a, 0).

    More generally, if the point could be any point in the xy-plane, (x, y), rather than just (x, 0), the potential function would be
    [tex]\phi(x,y)= \frac{kq}{\sqrt{(x-a)^2+ y^2}}+ \frac{kq}{\sqrt{(x+a)^2a+ y^2}}[/tex]
    and the force vector the gradient of that:
    [tex]\frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}[/tex]
     
  4. Jun 14, 2012 #3
    Ah, I see. Thank you
     
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