# Interatomic coulombic potential

1. Jun 28, 2012

### Hypatio

I am trying to calculate coulombic potentials. The equation given in wikipedia is

$$F=k_e\frac{q_{1}q_{2}}{r^2}$$

where k_e is in units of N m^2 C^-2, q are the partial charges, and r is the distance.

However, I look here and they give k_e=2.31*10^19 and remove the square of the distance. I assume that the difference is related to the units, which is in J*nm in this other resource, but if k_e is a constant then how can an equation of the form 1/r^2 give the same answer as one with the form 1/r?

Also, is the given value of k_e (2.31*10^19 J*nm) very accurate? What is the more accurate value?

2. Jun 28, 2012

### tiny-tim

Hi Hypatio!

(try using the X2 button just above the Reply box )
the force is ~ 1/r2

the potential is ~ 1/r
i don't recognise that figure

the usual constant in Coulomb's law is 8.988 x 109 N m2/C2

(or 8.987 551 787 368 176 4, see http://en.wikipedia.org/wiki/Coulomb's_law#The_law)

3. Jun 28, 2012

### nasu

The formula in the paper is for energy and not for force:
$$E=k_e\frac{q_{1}q_{2}}{r}$$
The numerical constant comes from the units: they use the numerical charge (number of electron charges) and not the charge in coulombs.
They also use the distance in nm.
So the number you see is

$$k_e\times e^2\times 10^9$$

4. Jun 28, 2012

### Hypatio

Thanks!

What is the difference between the partial charge and the 'numerical charge'? Also, what is the parameter e?

5. Jun 28, 2012

### nasu

Numerical charge is the charge in units of elementary (electron charge).
e is the value of the elementary charge (1.6x10^(-19) C).