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Rutherford and electric force

  1. Oct 27, 2004 #1
    Rutherford and electric force....

    Rutherford thought that most of the alpha particles would pass throught the cloud of positive charge if the thomson model was correct - but why did he think this would be the case? Ive read that apparently the charge would be spread over a greater area so it would therefore be weaker, but surely electronstatic force has nothing to do with the size of an object - just its charge and distance from the other charged particle? Surely an alpha particle at distance X from gold atom would experience exactly the same sized repulsive force from the a gold atom whether this was concentrated in a nucleus or in a larger cloud? - both would have the same field as both have the same charge.

    Am I wrong?

    Thanks. :smile:
  2. jcsd
  3. Oct 27, 2004 #2


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    If I recall correctly, Thompson's plum pudding model included the electrons embedded in the positive charged matrix. So the atom was, of course, neutral, and thus exerted little force on the incoming alpha particle. In a planetary model, the alpha particle can get inside any shielding due to electrons, and interact directly with the positively charged nucleus. The rest is history.
    Reilly Atkinson
  4. Oct 27, 2004 #3
    Outside the atom there would be no difference. But the alpha particle penetrates the atom and approaches the nucleus to 10^-14 meter or so. The coulomb field is very strong there (the potential is several miljon volts there).

    If the nuclear charge would be spread out over the whole atom (10000 times larger) there would be no field close to the center. Compare it to the field of the earth and the field of a black hole with the same mass. (Well, that is always attractive, but you get the idea.)
  5. Oct 27, 2004 #4
    Both of you have given a slightly different reason - which one is right?

    On that point actually, although moving to a macromolecular scale, why does a black hole exert and greater gravitational force that a sun of the same mass? After all, gravity only relies on the mass of the 2 objects and the distance between them. (you only need to look at the equation. :wink: )

  6. Oct 27, 2004 #5

    Claude Bile

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    It is the density that is critical, not the overall charge/mass itself. Take the example of the black hole.

    While the gravitational field may be the same far from the balck hole as the sun, in the sun, once you pass through the surface of the sun, the gravitational field decreases until you get to the centre where it becomes zero. In a black hole, there is no such surface as all the mass is stored in a singularity (for want of a better word) in the centre of the black hole. This means that as you get closer to the black hole, the gravitational force will keep on rising, resulting in huge fields that make black holes as unique as they are.

    The same analogy can be applied to the case of the atom/nucleus, in Thompsons model, the electrostatic force would begin to decrease as one penetrates the atom, whereas in Rutherford's model, the electrostatic forces increase rapidly as you penetrate the atom.

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