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Determining if a field is electrostatic or not

  1. Nov 2, 2011 #1
    If i have an arbitrary vector field, say F = constant / (x^2 + y^2 +z^2)
    I want to determine if it is electrostatic or not. For this I am thinking I should first determine if it is a conservative field, but not 100% sure, any guidance would be appreciated


  2. jcsd
  3. Nov 2, 2011 #2


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    An electrostatic field has curl F=0, but the F you right is not a vector, unless "constant" is a vector.
  4. Nov 3, 2011 #3
    Yea i think its a trick question, the F is bold in the example I'm looking at, but then there are no vector directions on the right side of the = so in that case, you can't take the curl can you?
  5. Nov 4, 2011 #4
    I think the OP refers to the modulus of a vector.Notice x^2+y^2+z^2 equals r^2.
    My guess is that F is the modulus of a gravitational or electrical field. In such a case we can write the vector field using the direction cosines and, upon taking the curl of that vector, we find it's conservative.
  6. Nov 4, 2011 #5


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    If you have only the modulus of a field, you cannot determine whether it's electrostatic or not.

    Of course, any time-independent vector field, which is curl free, can be an electrostatic field. The source (charge distribution) is given by

    [tex]\rho=\vec{\nabla} \cdot \vec{E},[/tex]

    where I'm using Heaviside-Lorentz (rationalized Gauß) units.
  7. Nov 4, 2011 #6
    Unless there's more hidden in the problem than stated, F = constant / (x2 + y2 +z2) is just F = constant / r2. You have not stated the direction of this vector, but I assume it is radial:
    F = Fr = constant / r2
    This is just the static field produced by a point source if the field is central and inverse-square-law in nature. For instance this would be the electrostatic field due to a point charge, or the gravitational field due to a point mass.
  8. Nov 4, 2011 #7


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    The radius vector, [itex]\vec{e}_r=\vec{r}/r[/itex] (with [itex]r=|\vec{r}|[/itex]) is not a constant. Of course, the Coulomb field is potential field,

    [tex]\vec{E}=\frac{q}{4 \pi r^2} \vec{e}_r=-\vec{\nabla} \frac{q}{4 \pi r},[/tex]

    and thus an electrostatic field. The charge distribution is

    [tex]\vec{\nabla} \cdot \vec{E}=-\Delta \frac{q}{4 \pi r}=q \delta^{(3)}(\vec{r}).[/tex]
  9. Nov 4, 2011 #8
    Actually, I think the question is a bit more simple. It is asked whether or not a field is electrostatic. In other terms, if it is time-independent. By looking at the expression, there is no time variable in there. So there we have the "static" part. Now is it electrostatic? Well, from Faraday:
    [tex]\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}[/tex]

    Where the right side becomes zero (time-independent). Thus, in electrostatic, we would need the electric field E to be conservative also.
  10. Nov 5, 2011 #9


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    Sure, any conservative time-independent field can be an electrostatic field. I've only shown this for the Coulomb field.
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