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Coulombs Law help!

  1. Nov 7, 2009 #1
    Can someone tell me the variables in coulomb's law?
    much appreciated
     
  2. jcsd
  3. Nov 7, 2009 #2
    The variables in coulomb's law are the charges of the charged bodies (q1 and q2), the distance between them R, and the permissibility Eo of the medium in which the system is.
     
  4. Nov 7, 2009 #3

    Born2bwire

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    I believe you mean permittivity, but this is only in specific measurement systems, like SI. In general, we just have an abstract constant k. The constant k serves as a means of converting the law for whatever system of measurement we use. For example, in MKS, k is 1/(4\pi \epsilon_0) where \epsilon_0 is the permittivity of free space (assuming vacuum background medium). In Gaussian CGS units, k is simply 1.
     
  5. Nov 8, 2009 #4
    Uh, yeah, sorry about the misspelling of the word, english is not my mother tongue. When I said that permittivity was a variable I was thinking about how the permittivity constant epsilon changes depending on the medium in which the force between the charged object is exerted (i.e. the permittivity of the vacuum differs from the permittivity of water and therefore forces between the same two charged particles may vary between mediums, making the medium a "variable" )
     
  6. Nov 8, 2009 #5
    i think you are confused in permitivty of the mediun.
    As you must have abserved that when we palce two magnets at a distance they will repell or atrract each other and when(in case of atrraction)you push them away you will feel the force but when you place something b/w them the u will feel that by pushing them away you are exerting less force than the first case.it is b/c the of medium b/w the.the same case is with the chages.
    Coulombs's law states that:
    q1.q2
    F=---------
    4(pi)Eo r.r
    here Eo is the standrad constan.in case when there is no resistive medium which can reduce the force so we use Eo.when the medium reduce some froce it will equal to the force in vaccum which is exerted at the same magnitude of charges at the same distance devided by Er.it can be written as:
    F
    F'= ----
    Er
    =>Er=F/F'
    and also
    Er=E/Eo
    where E is permitivity of the medium.

    i m in hurry. so cant present this better.if u r not understanded then i m sorry.
     
  7. Nov 8, 2009 #6
    k thanks u guys!
     
  8. Nov 14, 2009 #7
    thanks guys you also solved my problem!.
     
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