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Voltage in electric field

  1. Apr 3, 2006 #1
    Hi,

    I seem to have trouble understanding this. The way I understand it leads to an inconsistency, so there must be something I'm not getting. Please forgive me if my explanation is unclear.

    The potential difference between two points (say A and B) in an electric field is:

    -[integral from A to B] E . ds

    If s is parallel to E, then it becomes -Es
    where s is the distance between the two points.

    The potential difference between two points in the electric field of a point charge q is:

    (Ke)q * [ (1/rB) - (1/rA) ]

    where rB and rA are the distances between the respective point and q.

    Now, usually we chose V = 0 at rA = (infinty), thus the equation becomes:
    (Ke)q / rB

    However, when we consider two points A and B to be in line with a point charge q, if we let rA = (infinity) then the distance s between A and B is (infinty).

    The equation (Ke)q * [ (1/rB) - (1/rA) ] still appears fine, but what about -Es ? The situation seems to fit for that equation because s is parallel to the field line coming from the point charge. However, we end up with -E(infinity) .

    This means that:

    -E(infinity) = (Ke)q / rB

    (infinity) = (Ke)q / rB

    Which means that q must be infinite.

    There's got to be something wrong with my reasoning, can someone please tell me what it is?

    Thanks,
    Dale
     
  2. jcsd
  3. Apr 3, 2006 #2
    This is assuming that E is constant (both in direction and magnitude). In the scenario you describe, E falls off as 1/r^2.
     
  4. Apr 3, 2006 #3
    Thanks,

    that makes much more sense.

    One question though, is it actually possible to create a field that is constant everywhere in magnitude?

    Thanks,

    Dale.
     
  5. Apr 3, 2006 #4
    As for E being constant everywhere, I believe the answer is "yes" but it would be highly artificial.
    One very common example of a "uniform" field is the field between two parallel plate capacitors.
     
  6. Apr 3, 2006 #5
    If the field is uniform between the two plates, wouldn't the voltage at different points be the same everywhere?
     
  7. Apr 3, 2006 #6
    No, there is a potential difference across the plate given by V = Ed, where d is the separation of the plates.
     
  8. Apr 3, 2006 #7
    Ok thanks,

    Just to make sure I understand:

    _______________ Top plate

    ----------------- "Plane level" 1


    ----------------- "Plane level" 2

    _______________ Bottom plate

    By constant, do we mean that at Plane Level 1 and Plane Level 2, the magnitude of the field is the same? Or does it mean that the magnitudes at the planes are different, but that the magnitude is the same "across" each plane?
     
  9. Apr 3, 2006 #8
    I'm not sure what you mean by the second question. The answer is "at Plane Level 1 and Plane Level 2, the magnitude of the field is the same".

    Try this:
    http://en.wikipedia.org/wiki/Capacitor#Physics

    That the field is uniform is a consequence of Gauss' Law in integral form applied to a "pillbox" surface intersecting an infinite plane (then use superposition to get the final answer). You can most likely find this in your EM book.

    Also, bear in mind the electric field is the same at both plates, but the potential is not.
     
  10. Apr 3, 2006 #9
    Yes, I went back a bit in the book and found it. The fact the the magnitude of the field does depend on the distance from the plane of charges didn't really mean much before I started electric potential.

    Now that I think about it, it seems really shocking. The field from this charged plate will have the same effect on a charge a centimeter away as on a charge millions of kilometes away? It seems really strange...

    Thanks a lot for your help!
     
  11. Apr 3, 2006 #10
    I'm sorry, but this is really bugging me.

    So if all this is true, that means that if I hung a charged sphere from the ceiling and took a metal plate and put sufficient charge on it, and pointed it at the sphere it would move to a point where is was hanging with a certain angle to the vertical? And furthermore it would stay at that same angle no matter how far I went with the plate?
     
  12. Apr 4, 2006 #11
    Assuming the plate was infinite in extent, yes. Of course, all plates in real life are finite, so this is not a practical experiment.
    (Not to mention, it would be a difficult task to place a nonzero surface charge on a plate of infinite extend - the total charge itself would be infinite. Also, try calculating the amount of work it would require to assemble this configuration - it is also infinite.)
     
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