# Difference between Coulomb's and E-static?

1. Aug 4, 2014

### QuestionMarks

Coming from a Chemistry background, we seem to flippantly explain away chemical phenomena using "Coulomb's Law" but are often specifically mentioning the equation for electrostatic potential energy due to their similarity. The explanatory power for us is about the same, but one is an inverse-square law and the other just proportional to 1/r. Why is this?

I recognize one is describing a force, and the other an energy, but I'm having difficulty finding any way to mathematically discover why there's the difference in denominator beyond just hand-waving or beyond stepping quite out of my knowledge set to derive it myself hah.

Also, is it appropriate for us as Chemistry instructors to use the terms interchangeably when explaining phenomena, or is this bad bad practice?

Thanks much!

2. Aug 4, 2014

### Orodruin

Staff Emeritus
Do you know how work done by a force is computed?

3. Aug 4, 2014

### QuestionMarks

Well naievly I might say w=Fd then substitute that in to internal energy assuming no term for heat, calling that something like the estatic potential and hoping the the d in work would cancel one of our d's in the inverse square... is that what youre getting at? Ive thought of something like this but felt I was making too many assumptions.

4. Aug 5, 2014

### Delta²

It is something like that, but the first law of thermodynamics has nothing to do with this.

It is by definition that the potential energy at a point A in space, where there is electric field due to a source charge Q, is equal to the work done by the force of the electric field when we move a hypothetical charge q from point A to another fixed point of reference. The fixed point of reference sometimes is taken to be a point at infinite distance from the source Q, so that the electric field is zero there.

The work done is defined as the integral $\int_{r_A}^{\infty}Fdr=\int_{r_A}^{\infty}K\frac{Qq}{r^2}dr$. If we do the math that integral equals $K\frac{Qq}{r_A}$, where $r_A$ is the distance of point A from the source Q.

We use integral in the formula for work because the force of the electric field varies inversely proportional to the square of the distance r from the source. The formula W=Fr is valid only if F is constant with respect to distance r.

Last edited: Aug 5, 2014
5. Aug 5, 2014

### Orodruin

Staff Emeritus
Let me just add that the relation between the force and the potential may also be written as
$${\bf F} = - \nabla V.$$
Constrained to the radial direction, this would be
$$F_r = - \frac{\partial V}{\partial r}.$$
Now if $V = K \frac{Qq}{r}$ we would obtain
$$F_r = -KQq \frac{d r^{-1}}{dr} = \frac{KQq }{r^2}.$$

6. Aug 8, 2014

### QuestionMarks

Alrighty, makes sense. Thanks you two!

7. Aug 8, 2014

### QuestionMarks

Wait one more question though:
Is it alright in say a high school classroom to justify the relevant phenomena to these equations by broadly talking about them as Coulomb's Law (I would guess historically one led to the other nigh simultaneously)? Subjective question perhaps but still interested in other thoughts.

8. Aug 9, 2014

### Delta²

Well it is right because as you see from our posts we prove the equivalence
(Force obeys Coulomb's law)<=>(Potential energy follows inverse distance )i.e:

1)when we have an electric field force that obeys Coulombs Law (that is the force inversely proportional to the square of the distance), then the potential energy is inversely proportional to the distance,

and vice versa that is:

2) if the potential energy is inversely proportional to the distance then the force of field follows Coulomb's law.

But maybe it would be more accurate to avoid confusion to refer to Coulomb's Law when we talk about the forces of a static E-field, and when we talk about potential energy to refer to the potential energy proportional to 1/r as a consequence of a force that follows the Coulomb's Law.

Last edited: Aug 9, 2014