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Dimensions of Newton's Law of Gravitation and Coloumb's Law

  1. Nov 21, 2009 #1
    As we know, Newton's Law of Gravitation is
    {\mathbf{F}} = \frac{{Gm_1 m_2 }}
    {{r^2 }}
    and Coulomb's law is

    {\mathbf{F}} = \frac{{Qq_1 q_2 }}
    {{r^2 }}
    \] [/tex]

    We know from comparing the dimensions of the first equation that G, the gravitational constant, has the dimension
    [M^{ - 1} L^3 T^{ - 2} ]

    But for Coulomb's law, we assume Q is dimensionless. Why do we make this assumption?
  2. jcsd
  3. Nov 21, 2009 #2

    D H

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    Staff Emeritus
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    Where did you get the idea that Q (typically κ, not Q) is unitless?

    Dimensional analysis tells you what the units for κ must be:

    [tex]d(\kappa)=d(F)*d(r)^2/d(q)^2 = \frac{FL^2}{Q^2} = \frac{ML^3}{T^2Q^2}[/tex]

    where the d() extracts the dimensions of the quantity in question. This is not a dimensionless quantity. The only way it can be dimensionless is if charge is expressed in non-rational units, [tex]d(q) = \sqrt{ML^3}/T[/tex]

    See http://scienceworld.wolfram.com/physics/CoulombsConstant.html
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