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Electric Field Lines Can't Cross Each Other?

  1. Feb 2, 2013 #1
    Hello,

    This query isn't concerned with solving a problem, but only with details appendant to the problem.

    The question is as follows: A uniformly charged ring of radius 10.0 cm has a total charge of 64.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

    (a) 1.00 cm


    (b) 5.00 cm


    (c) 30.0 cm


    (d) 100 cm

    A commentary on the problem says that, if you were to place a charged particle in the center of this ring, the force due to the electric fields at that point would be zero. My question is, is the force zero because the electric fields can't point radially towards the center because they could cross? My next question would then be, why can't electric field lines cross?
     
  2. jcsd
  3. Feb 2, 2013 #2

    Nugatory

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    Staff: Mentor

    There's nothing wrong with electric field lines pointing radially inward - that's what you see around a charged point particle (if you think they're pointing radially outwards instead, try again with the opposite charge on the particle).

    However, it is true that the field lines can never cross. This is because at any given point, the direction of the electric field is determined by the gradient of the potential at that point, and that gradient points in only one direction. If the lines crossed, that would mean that at the crossing point the gradient would have to point in two direction at once, and that's impossible.
     
  4. Feb 2, 2013 #3
    So, the reason why the net force on the particle at the center of a charged ring is because the forces are pointing in opposite directions but with equal magnitude, thus cancelling out?
     
  5. Feb 2, 2013 #4

    Nugatory

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    Staff: Mentor

    Yes. And at that point, the gradient of the potential is zero.
     
  6. Feb 2, 2013 #5
    May I ask, what exactly do you mean by gradient?
     
  7. Feb 2, 2013 #6

    Nugatory

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    Staff: Mentor

    If [itex]\phi[/itex] is a scalar function like the electrical potential, then its gradient, written as [itex]\nabla\phi[/itex], is the vector [itex]\frac{\partial \phi}{\partial x}\hat{x}+\frac{\partial \phi}{\partial y}\hat{y}+\frac{\partial \phi}{\partial z}\hat{z}[/itex].

    Intuitively, if [itex]\phi[/itex] is the height above sea level of any point (x, y), then the gradient [itex]\nabla\phi[/itex] would be a vector pointing in the direction of the steepest slope at that point. (this would be a two dimensional example, so no z coordinate). Zero gradient means flat ground.

    The electrical field at any point is the gradient of the potential.

    Wikipedia has more detail.
     
  8. Feb 2, 2013 #7
    While what Nugatory wrote is correct, it is unnecessarily complicated. Suppose electric field lines could cross, what would happen if one tried to measure the electric field at the point where they crossed? It makes no sense, because electric field measurements (just like any other measurement of a single well-defined quantity) can only return a single value. Since one arrives at a contradiction (that one should measure two values when one only measures one) by assuming that electric field lines can cross, they must not be able to cross. It is just a matter of the field being well-defined, if one knows that the field points in some direction and has some magnitude, then logically it can't be pointing anywhere else or have any other magnitude (at the location and time at which it is known).
     
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