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The charge interaction as a force

  1. Jul 24, 2015 #1
    I think this thing came to my mind some times after I studied Electromagnetism when I studied Physics; otherwise, I would have asked it to my professor. (Incidentally, I'm still studying and I'm involved in Electromagnetism, but for a totally different course :wideeyed:)

    We know from Coulomb's Law that, if we have two charges, they attract or repel each other and the result is a force: (sorry, but writing formulas, even simple, is a pain in the... :rolleyes:)

    F=(k*q1*q2)/r²

    While Newton's Second Law of Dynamics states that:

    F=m*a

    If we compare the members of the equations (sorry, I don't know if it's the right way to say that), we have:

    a=(k*q1*q2)/(m*r²)

    What struck me once and what I want to point out is:
    we have an interaction between charges, but its result applies to masses. Why doesn't it apply to charges themselves? Why charge interaction is a force?

    Let's suppose we have a particle with charge, but no mass: what would it happen then if it interacted with another particle of the same type? There would be an interaction, but it would apply to what?
    (Ok, perhaps I know the answer: no particles with charge e no mass (for some very complex reason) can exist)
     
  2. jcsd
  3. Jul 24, 2015 #2

    Student100

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    You basically answered your own question, it's not that massless electrical charged particles couldn't theoretically exist, but experimentation has shown they don't. Under the standard model, massless electrically charged particles would make our world quite unstable, and be easy to detect. They may have existed at one time under SM, but I'm not real sure about this and will defer to others with more experience. As far as interactions, they wouldn't interact in normal ways with other electrically charged particles.
     
  4. Jul 24, 2015 #3

    Drakkith

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    You're confusing the F = MA equation to mean that a force acts 'on a mass'. The equation simply gives us a relationship between the net force on an object, the acceleration of the object under that net force, and the mass of the object. Remember that 'a charge' means 'an electrically charged object'. This object could be a single fundamental particle or a massive object composed of trillions upon trillions of particles. The equation still applies to objects under the influence of gravity, regardless of whether they have charge or not.
     
  5. Jul 24, 2015 #4

    lightgrav

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    It is the object property mass that fixes the object into our set of reference frames (called "inertial frames").
    ... this means that massless objects would not need to remain in our inertial frame (or anybody's inertial frame)
    we have observed objects whose intrinsic mass m =0, and the implication holds, for light.
    ... if an object does not recognize the property mass, that object (probably) cannot conjure a mechanism to remain in an inertial frame
    such things might appear (to us, ponderous bodies) to "pop in and out" of our locality
    (it would "still exist", if they are a conserved quantity, but not anywhere near us ; a=∞ would be possible, and maybe v=∞ also.)

    It is the object property charge that has electric Force applied to it (by the Electric Field).
    The Electric Field is "well-defined" only in inertial reference frames
    ... or frames that are accelerating gently enough that they can be treated as "momentarily inertial"
    (with this velocity for a little while, before being boosted to a new velocity and inertial frame, for the "next little while", etc).
    ... charged objects carry with them an Electric Field, which imposes an Energy density to the space (Volume) it permeates.
    We (ponderous bodies) interpret this Field Energy (/c2) as mass, and notice that they do carry inertia (m≠0).

    I can't think of any mechanism to have a charge without an Electric Field, that would still be influenced by other charges' E-fields.
    ... maybe I'm just not creative enough ... but (by para.1) how would we recognize them anyway? (my lab is inertial ... sorry, lack of $ ;-)
     
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