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Conversion of EPE to Kinetic Energy in Magnetism

  1. Dec 2, 2011 #1
    So suppose we have two objects, both with positive charge Q Coulombs and X meters apart.
    And now suppose Object 1 has a mass of m1 and Object 2 has a mass of m2.
    In other words, the charges are identical, but masses are not.

    So each of these will have an Electric Potential Energy of -QEΔX, because EPE = -QEΔd.
    Easy peasy, right? EPE = KE, just like PE = KE for a falling object! Just like mgh = (1/2)mv^2, -QEX = (1/2)mv^2. So:

    EPE = KE, so for Object 1 it's
    -QEΔX = (1/2)m1v2.

    and for Object 2 the formula is
    -QEΔX = (1/2)m2v2.

    Apparently, that's wrong and I was actually supposed to work it out this way:
    -QEΔX = (1/2)m1v2 + (1/2)m2v2

    It does make sense... but it doesn't. I mean, each object has each of their own potential energy, right? And each object loses that EPE and changes it 100% into KE, right? Isn't that what the potential energy is?

    I mean, I don't remember doing PE = KE1 + KE2 for a falling object!
    As far as I know, I've always done:
    mgh = (1/2)mv2.

    I don't remember ever making the mass of the earth share the potential energy.
    If I apply the same concept to gravity, it would be like:
    mgh = (1/2)mEarthv2 + (1/2)mobjectv2.
    What!!? I don't remember ever doing that!

    Help me out here, guys! I am greatly befuddled! :cry:
     
  2. jcsd
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