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

1. Nov 21, 2009

### Pinu7

As we know, Newton's Law of Gravitation is
$$${\mathbf{F}} = \frac{{Gm_1 m_2 }} {{r^2 }}$$$
and Coulomb's law is

$$${\mathbf{F}} = \frac{{Qq_1 q_2 }} {{r^2 }}$$$

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. Nov 21, 2009

### D H

Staff Emeritus
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:

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

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, $$d(q) = \sqrt{ML^3}/T$$

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