# Coulomb's law -- vector character

#### Prerequisites

In our page on Coulomb's law, we worked out how the magnitude of the electric force between two charges depend on the charges and on the distance between them. If we have more than two charges to consider, we'll have to add electric forces, and since forces are vectors, we'll have to add them as vectors. So we'll have to consider the character of Coulomb's law in vector form.

If we look at the basic picture of two charges and the directions of the forces they feel, we see that the force each charge exerts on the other acts along the line joining them. So we need to find a vector representation of this direction.

One way to do it is to think about the vector starting on one charge and ending on the other. This is a **displacement** **vector, **$\overrightarrow{r}_{Q \rightarrow q}$. (See the page on Kinematic Variables.) If the position of charge $Q$ is the vector $\overrightarrow{r}_Q$ and the position of charge $q$ is the vector $\overrightarrow{r}_q$, then what we want is the displacement vector from $Q$ to $q$.

$$\overrightarrow{r}_Q + \overrightarrow{r}_{Q \rightarrow q} = \overrightarrow{r}_q$$

$$\overrightarrow{r}_{Q \rightarrow q} = \overrightarrow{r}_q - \overrightarrow{r}_Q$$

If we just want the **direction**, not the actual displacement, we can divide by the length of that displacement vector and get a vector of unit length (without any units) in the right direction:

$$\hat{r}_{Q \rightarrow q} = \frac{\overrightarrow{r}_{Q \rightarrow q}}{r_{Q \rightarrow q}}$$

Careful! This is a very hard equation to read! We've used the same symbol three times — $r_{Q \rightarrow q}$ — but with different arrow markers on top of it to show the different meanings.

- The one on the right on top with the arrow on top of it ($\overrightarrow{r}_{Q \rightarrow q}$) means the displacement vector starting at $Q$ and ending at $q$.
- The one on the right on bottom with no arrow on it ($r_{Q \rightarrow q}$ means the magnitude of that displacement — the distance between them.
- The one on the left with the hat ($\hat{r}_{Q \rightarrow q}$)means the vector direction from $Q$ to $q$ but with the distance between them divided out to give a pure direction with length 1 (with no units).

So with this notation, we can put the vector character into the Coulomb law for the force between two charges:

$$\overrightarrow{F}^E_{Q \rightarrow q} = \frac{k_CqQ}{r^2_{qQ}} \hat{r}_{Q \rightarrow q}$$

and changing the charges gives

$$\overrightarrow{F}^E_{q \rightarrow Q} = \frac{k_CqQ}{r^2_{Qq}} \hat{r}_{q \rightarrow Q} = -\overrightarrow{F}^E_{Q \rightarrow q} $$

giving us agreement with N3, since the direction from $Q$ to $q$ is opposite the direction from $q$ to $Q$:

$$\hat{r}_{q \rightarrow Q} = - \hat{r}_{Q \rightarrow q}$$

The second equation shows us the forces exerted by q on Q: r^{2} is the square of the magnitude of the distance vector between q and Q so it does not change sign. Since the direction from Q to q is opposite to the direction from q to Q, our unit vectors just get a minus sign and it is clear Newton's 3rd law holds.

This is starting to look messy! There are a lot of symbols and a lot of subscripts. But if you keep in mind a physical picture of the two charges and the forces they exert on each other, you ought to be able to see how these equations are coding for that physical picture. Remember that subscripts and superscripts can be your friends and make this equation easier to read! Some of the subscripts identify the object that is exerting the force and the object on which the forces is exerted. The subscript on the constant k indicates that this is related to Coulomb's law. Finally the superscript on F tells us that we are dealing with electrical forces.

## Negative charges

We've done all our considerations of Coulomb's law as if all the charges were positive. But once we go to the vector form we can let the charges be negative and all the directions will work out fine. If we make one of the charges negative, we can just associate that minus sign with the unit vector direction. Making a vector negative just flips its direction. If we switch just one charge then the directions of both forces will flip and the forces will be attractive. If we flip both, (-1)^{2} = +1 so we are back where we started with repulsive forces.

Joe Redish & Wolfgang Losert 10/11/12

#### Follow-ons

Last Modified: March 11, 2020