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A Charge on the Moon

  1. Nov 6, 2014 #1
    Hi All,

    Again, thanks to all the Physics Forums gurus.

    I posted a question about a year ago concerning the finite propagation speed of information and electrostatic forces between charges, which seemed confusing. I still was hoping to resolve it, so I simplified it a bit :).

    Imagine a positive charge on the moon, and an identical positive charge on earth. They faintly repel one another. Also imagine that we have a closed system (maybe a giant Dyson sphere around the system :D), so that all possible radiation from our experiment is accounted for.

    Since light takes about a second to travel between the earth and the moon, changing the location of the charge on the earth affects the charge on the moon via it's E field a second later, and vice-versa.

    Suppose you move the charges slightly closer toward each other, simultaneously, and bring them to rest, in under a second. Because the E field takes a second to propagate, each charge sees that it's partner has not moved during this transition.

    Then, suppose you repeat this action to bring the charges back to their initial position. Again, each charge experiences that it's partner has not moved during this transition.

    My question is: At first glance, the energy required to push the charges together is less that what is released when they are pushed apart. This is due to the finite speed of propagation of the E field. What is the missing fact to explain this? (Note in my previous post, I did apply the Liénard–Wiechert potential to obtain the relativistic field and force on each charge, but still came up short.)

    Thanks
     
  2. jcsd
  3. Nov 6, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    I did not calculate it, just a guess: you accelerate charges, this will emit electromagnetic radiation. And if you do this both on moon and earth at the same time, you might see less radiation than you get by moving them one by one.
     
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