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What is this nature of Electrostatic Field ?

  1. Sep 26, 2013 #1
    Hi all,

    Work done to move a charge between two points in electrostatic field is independent of path taken.

    what is in this electrostatic force that is causing this? I mean how is it making it path independent?

    Does it has to do with the inverse-square nature of the field? or Its radially outward pointing field?

    Secondly, is the relationship between conservativeness of field and path independence bidirectional?

    I am having really a tough time understanding its nature intuitively.Almost 2 weeks gone. Still couldn't make.kind request to help me get out of this.

    Last edited: Sep 26, 2013
  2. jcsd
  3. Sep 26, 2013 #2


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    The curl of the static electric field is zero - this implies that the work done is path-independent.

    These are called "conservative forces". Classical gravity is another conservative force.
  4. Sep 26, 2013 #3
    To add to UltrafastPED's answer

    When a field has zero curl everywhere, you can use Stoke's theorem to deduce that its circulation around any closed loop is also zero. This is equivalent to saying the energy to go from point A to point B doesn't depend on the path, because you can define your closed loop to be A -> B -> A. If the circulation around this closed loop is zero regardless of the path, then the energy used in going from A to B must be exactly the negative of the energy used in going back from B to A.

    It's an intuitive concept which is served far better by diagrams. Youtube Stoke's theorem for the idea.
  5. Sep 26, 2013 #4


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    You should be a bit more careful! From [itex]\vec{\nabla} \times \vec{V}=0[/itex] in a region you can only conclude that in any simply connected region within that region the vector field is the gradient of a scalar potential.

    The most simple example is the "potential vortex"
    [tex]\vec{V}=\frac{\Gamma}{2 \pi (x^2+y^2)} (-y,x,0)^T[/tex]
    or in standard cylinder coordinates [itex](r,\varphi,z)[/itex]
    [tex]\vec{V}=\frac{\Gamma}{2 \pi r} \vec{e}_{\varphi}.[/tex]
    It's easy to see that
    [tex]\vec{\nabla} \times \vec{V}=0 \quad \text{for} \quad x^2+y^2 \neq 0,[/tex]
    but that for any closed curve [itex]C[/itex] circling around the [itex]z[/itex] axis once gives
    [tex]\int_{C} \mathrm{d} \vec{x} \cdot \vec{V}=\Gamma \neq 0.[/tex]
  6. Sep 27, 2013 #5

    I looked at stroke's theorem. It relates line integral of field along a path(closed) to curl of field over surface enclosed by that closed path.

    But my question is fundamental. Consider only a path which is not closed between A to B.Let this path be P1. The line integral of Electro-Static Force between such A and B is not zero. Consider another Path P2 between A and B. In this case also Line Integral is same as for Path P1.

    What is in this electro-static force that makes 2 line integrals around 2 different paths between A and B equal? What is its nature?

    My question is more fundamental to physics than mathematical.
    Are these 2 different line integrals becoming equal because of
    1. The radial outward direction of field from source
    2. Inverse square law of the electric force

    These are my so so guesses.

    Thanks for a note on that.

  7. Sep 27, 2013 #6
    Taken together, the two different paths between A and B form a closed loop. To see this, go along P1 from A to B, then go along P2 from B to A. If the integral along this loop is zero, then that means that the integral of P1 from A to B is equal and opposite to the integral of P2 from B to A. Going the other direction on P2 just means a change of sign, so the integral along the two paths are equal.
  8. Sep 27, 2013 #7
    If we can assume that the elementary charges produce radial symmetric and constant fields, then it is clear that going from radial distance 1 meter to 2 meters the work is exactly opposite of going from 2 meters to 1 meter. Also you can move around a circle of constant radius without doing any work. So always a fixed amount of work has to be done to transition between two different radial distances from the charge. Therefore all points can be labeled with a certain potential energy. And if you have a lot of charge it is just the sum of all the small charges and similarly the force and the potential adds up.
  9. Sep 27, 2013 #8
    Yeah my closed loop was A-B-A since you end up back at A.
  10. Sep 27, 2013 #9


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    [itex]{{\bf dr}\cdot(\bf r}/r^3)=-d(1/r)[/itex], so when it is integrated around any closed loop it gives zero.
    From this it follows that the integral from A to B is path independent.
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