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# 12.3: The Coulomb Force Law

Created by: CK-12

The Coulomb Force Law states that any two charged particles $(q_1,q_2)$ --- with charge measured in units of Coulombs --- at a distance $r$ from each other will experience a force of repulsion or attraction along the line joining them equal to:

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

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 $-12\mu\mathrm{C}$; the other with $-3\mu\mathrm{C}$) are $3\mathrm{m}$ apart. Where could you place an electron so that it will be suspended in space between them with a net force of zero (for this problem we will ignore the force of repulsion between the two charges because they are held in place)?

Answer: Consider the diagram above; here $r_{s\rightarrow e}$ is the distance between the electron and the small charge, while $\vec{F}_{s\rightarrow e}$ is the force the electron feels due to it. For the electron to be balanced in between the two charges, the forces of repulsion caused by the two charges on the electron would have to be balanced. To do this, we will set the equation for the force exerted by two charges on each other equal and solve for a distance ratio. We will denote the difference between the charges through the subscripts “s” for the smaller charge, “$e$” for the electron, and “$l$” for the larger charge.

$\frac{kq_sq_e}{r_{s\rightarrow e}^2}=\frac{kq_lq_e}{r_{e\rightarrow l}^2}$

Now we can cancel. The charge of the electron cancels. The constant $k$ also cancels. We can then replace the large and small charges with the numbers. This leaves us with the distances. We can then manipulate the equation to produce a ratio of the distances.

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

Given this ratio, we know that the electron is twice as far from the large charge ($-12\mu\mathrm{C}$) as from the small charge ($-12\mu\mathrm{C}$). Given that the distance between the small and large charges is $3\mathrm{m}$, we can determine that the electron must be located $2\mathrm{m}$ away from the large charge and $1\mathrm{m}$ away from the smaller charge.

## 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:

$\vec{F_e} = \frac{kq_1q_2}{r^2}$

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 ${q_1}$ and $\frac{kq_2}{r^2}$. The first part of this product only depends on properties of the object we're interested in (the first charge), and the second part can be thought of as a property of the point in space where that object is.

In fact, the quantity $\frac{kq_2}{r^2}$ captures everything about the electromagnetic force on any object possible at a distance r from $q_2$. If we had replaced $q_1$ with a different charge, $q_3$, we would simply multiply $q_3$ by $\frac{kq_2}{r^2}$ to find the new force on the new charge. Such a quantity, $\frac{kq_2}{r^2}$ here, is referred to as the electric field from charge $q_2$ at that point: in this case, it is the electric field due to a single charge:

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

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 $q_0$ would feel at that point:

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

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, $r$, from the charge, not the direction; the lengths of the arrows represent the strength of the field.

## Example 3

Question: Calculate the electric field a distance of $4.0\mathrm{mm}$ away from a $-2.0\mu \mathrm{C}$ charge. Then, calculate the force on a $-8.0\mu \mathrm{C}$ charge placed at this point.

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

$E=\frac{kq}{r^2}$

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

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

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

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

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

$F=Eq$

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.

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

Feb 23, 2012

Aug 01, 2014