# 12.3: The Coulomb Force Law

**At Grade**Created by: CK-12

The Coulomb Force Law states that any two charged particles \begin{align*} (q_1,q_2) \end{align*}

\begin{align*}\vec{F_e} = \frac{kq_1q_2}{r^2}&& \text{The Coulomb Force [1]}\\
\intertext{Where}\\
k = 8.987 \times 10^9 \ \mathrm{N \cdot m^2 \cdot C^{-2}}. &&\text{The Electric Constant}\end{align*}

This looks a lot like the Law of Universal Gravitation, which deals with attraction between objects with mass. The big difference is that while any two masses experience mutual *attraction*, two charges can either attract or repel each other, depending on whether the signs of their charges are alike:

Like gravitational (and all other) forces, Coulomb forces add as vectors. Thus to find the force on a charge from an arrangement of charges, one needs to find the vector sum of the force from each charge in the arrangement.

## Example 1

**Question**: Two negatively charged spheres (one with \begin{align*}-12\mu\mathrm{C}\end{align*}

**Answer**: Consider the diagram above; here \begin{align*} r_{s\rightarrow e} \end{align*}

\begin{align*}\frac{kq_sq_e}{r_{s\rightarrow e}^2}=\frac{kq_lq_e}{r_{e\rightarrow l}^2}\end{align*}

Now we can cancel. The charge of the electron cancels. The constant \begin{align*}k\end{align*}

\begin{align*}\frac{-3\mu\mathrm{C}}{r_{s\rightarrow e}^2}=\frac{-3\mu\mathrm{C}}{r_{e\rightarrow l}^2} \Rightarrow \frac{r_{s\rightarrow e}^2}{r_{e\rightarrow l}^2}=\frac{-12\mu\mathrm{C}}{-12\mu\mathrm{C}} \Rightarrow
\frac{r_{s\rightarrow e}}{r_{e\rightarrow l}}=\sqrt{\frac{1\mu\mathrm{C}}{4\mu\mathrm{C}}}=\frac{1}{2}\end{align*}

Given this ratio, we know that the electron is twice as far from the large charge (\begin{align*}-12\mu\mathrm{C}\end{align*}

## Example 2

## Electric Fields and Electric Forces

Gravity and the Coulomb force have a nice property in common: they can be represented by **fields**. Fields are a kind of bookkeeping tool used to keep track of forces. Take the electromagnetic force between two charges given above:

\begin{align*}\vec{F_e} = \frac{kq_1q_2}{r^2}\end{align*}

If we are interested in the acceleration of the first charge only --- due to the force from the second charge --- we can rewrite this force as the product of \begin{align*}{q_1} \end{align*}

In fact, the quantity \begin{align*}\frac{kq_2}{r^2}\end{align*}**r** from \begin{align*} q_2 \end{align*}

\begin{align*}\vec{E_f} = \frac{kq}{r^2} &&\text{[2] Electric field due to point charge } q ,\text{ distance } r \text{ away}\end{align*}

The electric field is a vector quantity, and points in the direction that a force felt by a positive charge at that point would. If we are given the electric field at some point, it is just a matter of multiplication --- as illustrated above --- to find the force any charge \begin{align*} q_0 \end{align*}

\begin{align*}\underbrace{\vec{F_e}}_{\text{Force on charge }q_0} = \underbrace{\vec{E_f}}_{\text{Field}} \underbrace{\times q_0}_{\text{Charge}} &&\text{Force on charge }q_0\text{ in an electric field}\end{align*}

Note that this is true for *all* electric fields, not just those from point charges. In general, the **electric field** at a point *is the force a* *positive test charge of magnitude 1**would feel at that point*. Any other charge will feel a force along the same line (but possibly in the other direction) in proportion to its magnitude. In other words, the electric field can be though of as “force per unit charge”.

In the case given above, the field was due to a single charge. Such a field is shown in the figure below. Notice that this a field due to a positive charge, since the field arrows are pointing outward. The field produced by a point charge will be radially symmetric i.e., the strength of the field only depends on the distance, \begin{align*} r \end{align*}

## Example 3

**Question**: Calculate the electric field a distance of \begin{align*}4.0\mathrm{mm}\end{align*}

**Answer**: To calculate the electric field we will use the equation

\begin{align*}E=\frac{kq}{r^2}\end{align*}

Before we solve for the electric field by plugging in the values, we convert all of the values to the same units.

\begin{align*}& 4.0\mathrm{mm}\times \frac{1\mathrm{m}}{1000\mathrm{mm}}=.004\mathrm{m}\\
& -2.0\mu \mathrm{C}\times \frac{1\mathrm{C}}{1000000\mu \mathrm{C}}=-2.0\times 10^{-6}\mathrm{C}\end{align*}

Now that we have consistent units we can solve the problem.

\begin{align*}E=\frac{kq}{r^2}=\frac{9\times 10^9\mathrm{Nm^2/C^2}\times -2.0\times 10^{-6}\mathrm{C}}{(.004\mathrm{m})^2}=-1.1\times 10^9\mathrm{N/C}\end{align*}

To solve for the force at the point we will use the equation

\begin{align*}F=Eq\end{align*}

We already know all of the values so all we have to do is convert all of the values to the same units and then plug in the values.

\begin{align*}& -8.0\mu \mathrm{C}\times \frac{1\mathrm{C}}{1000000\mu \mathrm{C}}=-8.0\times 10^{-6}\mathrm{C}\\
& F=Eq=-8.0\times 10^{-6}\mathrm{C} \times -1.1 \times 10^9 \mathrm{N/C}=9000\mathrm{N}\end{align*}