# Why do we say that in Coulomb's law the force is proportional to $\frac{1}{r^{2}}$ and not $\frac{1}{r^{3}}$?

I am going over Coulomb's law and there is something that is a bit confusing for me:

According to Coulomb's law, if I have a charge $q_{1}$ at a point $\vec{r_{1}}$ and a charge $q_{2}$ at a point $\vec{r_{2}}$ then the force that the first charge applies to the second charge is given by $$F_{1,2}=\frac{Kq_{1}q_{2}}{|\vec{r_{2}}-\vec{r_{1}}|^{2}}(\vec{r_{2}}-\vec{r_{1}})=\frac{Kq_{1}q_{2}}{|\vec{r_{2}}-\vec{r_{1}}|^{3}}\hat{(\vec{r_{2}}-\vec{r_{1}})}$$

I see that in the first expression is does look like the force is proportional to $\frac{1}{r^{2}}$, but I don't understand why it isn't the second expression that matters, giving us that the force is proportional to $\frac{1}{r^{3}}$:

Say both charges are on the $x-$axis, then $\hat{(\vec{r_{2}}-\vec{r_{1}})}=\hat{x}$. Increasing the distance between the charges to two times what is by moving the second charge on the $x$- axis was, would decrease the force $2^{3}=8$ times, since $\hat{(\vec{r_{2}}-\vec{r_{1}})}$ is still $\hat{x}$.

What is the mistake in my reasoning ?

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A different question on $1/r^2$ versus $1/r^3$: physics.stackexchange.com/q/73028/2451 – Qmechanic Aug 5 '13 at 18:44

The mistake you made is in the way you stated Coloumb's law.

It's either

$$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}}$$ OR $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}}$$

but definitely NOT

$$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}}$$

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The notion behind probing the variance of one physical quantity with respect to another is simple.

Simplify your equation, before you conclude.

So when you write the vector equation like this :

$$\overrightarrow{F} = K \frac{q_{1} q_{2}}{r^{3}} \overrightarrow{r}$$

You are actually writing a shortened version of the vector equation. The complete vector equation is like this :

$$|\overrightarrow{F}| (\widehat{F}) = K \frac{q_{1} q_{2}}{r^{3}} |\overrightarrow{r}| (\widehat{r})$$

Now do note that from this equation, it is clear that the magnitude of coulombic force is directly proportional to one power of the magnitude of displacement and inversely proportional to the third power of displacement.

Hence simplifying all this, the conclusion remains that coulombic force is inversely proportional to the square of the displacement.

These confusion stem from the fact that the equation you used was not in its most simple form. When equations are presented in their most simplified format, they bring out the true colors. Cheers :)

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