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I What happens to the Electrical Potential the closer we are to a point charge?

  1. Nov 1, 2018 #1
    So I have been wondering:
    The potential for a point charge at the origin, is described as:
    (Using the reference point at infinity): V=1/(4πε) * q/r
    My question is, what happens to this Potential the closer we are to the point charge, and so the closer we would get, the Potential seems to go towards infinity, which does not make sense, so I would like to understand how to approach the idea since my approach gives me an illogical answer.
     
    Last edited by a moderator: Nov 2, 2018
  2. jcsd
  3. Nov 1, 2018 #2

    scottdave

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    Yes, it does go toward infinity, as you suspected. So what does this mean for another charge as it approaches the one at the origin?
     
  4. Nov 1, 2018 #3
    Its potential would be stronger the closer this one is to it? And how can a Potential go towards infinity, I feel pretty lost on the subject. Thank you for the help btw!
     
  5. Nov 1, 2018 #4

    scottdave

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    So let's agree that you cannot put zero in the denominator. Physically, the distance between two charges will always be some tiny amount.
    If the charge at the origin is positive, bringing another positive charge close to it will become increasingly difficult, as you get closer. Maybe this site will help you. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html
     
  6. Nov 1, 2018 #5

    Nugatory

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    Indeed it does not. This is the math telling you that ##F=CQ_1Q_2/r^2## for point particles only works when ##r\ne{0}##.

    If f you try to bring two charged objects together to drive the potential and the force between them to infinity, some other physics that you've been ignoring for larger values of ##r## will start to matter and Coulomb's law is no longer the whole story. For example, if the objects have non-zero sizes ##R_1## and ##R_2##, you can't reduce the distance between them to less than ##R_1+R_2## so plugging a smaller value into Coulomb's law makes no sense.
     
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