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

  • Thread starter Pinu7
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  • #1
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As we know, Newton's Law of Gravitation is
[tex]\[
{\mathbf{F}} = \frac{{Gm_1 m_2 }}
{{r^2 }}
\]
[/tex]
and Coulomb's law is

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

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

But for Coulomb's law, we assume Q is dimensionless. Why do we make this assumption?
 

Answers and Replies

  • #2
D H
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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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