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Neutral atom and a point charge

  1. Jan 17, 2014 #1
    1. The problem statement, all variables and given/known data

    If the distance between a neutral atom and a point charge is doubled, by what factor does the force on the atom by the point charge change? (new force / old force)

    2. The attempt at a solution

    Wouldn't the answer just be (new force / old force) = 0 since there is no force between a neutral atom and a point charge? But it is wrong.
     
  2. jcsd
  3. Jan 17, 2014 #2
    Maybe they are taking the gravitational force into account? Although it is negligible, technically it does have some effect.
     
  4. Jan 17, 2014 #3

    lightgrav

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    Homework Helper

    I'll bet it is focusing on the Electric Force ... because it specifies neutral atom and "point charge".
    the old Force was zero (ignoring induced dipole in the atom being attracted to charge),
    the new Force is zero also (ignoring induced dipole)
    => that would mean the Force is the same ... multiplied by "1".

    If you're responsible for induced dipole attraction, then calculate that after finding the monopole Force is zero.
     
  5. Jan 17, 2014 #4
    Hmm I am still not understanding. So the ratio should be new force / old force = 0/1?
     
    Last edited: Jan 17, 2014
  6. Jan 17, 2014 #5
    This is in a 2nd semester elementary physics text?

    To your problem, it doesn't necessarily mean there is no net force on your test particle. Imagine this problem in one-dimension. (This will create the greatest difference in force) If your test particle was close to the atom it would recognize two distinct charges effecting it: one the electron and the other the proton. The net force would be the usual sum of them. You can, and will if your a physics major, extend this to 3-D to be more general.

    If you set the distance, [itex]D [/itex], to be the distance from the particle to the proton, then you can out the distance between the electron and the particle. (i.e. [itex]D-r_{A}[/itex], where [itex]r_{A}[/itex] is the atomic radius.) This does assume that the 'orbit' of the electron remains the same. Either way I find myself defining distance variables, which I find unsettling because I can do it a number of ways. This would be fine if you were studying advance E&M, but looking for exact answers here. Is there more information?

    Oh and you are correct about one thing, if the distance to the atom is significantly greater than the atomic radius the particle would experience no force. Basically [itex]D-r_{A}≈D[/itex]
     
  7. Jan 18, 2014 #6
    0/0≠1

    You can multiply by any number and still get zero.
     
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