Coulomb’s law presents a unique challenge to physics students struggling to understand how magnitude and direction can coexist in the same quantity–a vector. The difficulty with Coulomb’s law is that electric charge, a property of certain subatomic particles, comes in two types, arbitrarily called positive and negative. For example, the electron carries a negative electric charge, and the proton a positive electric charge. (They have the same magnitude, or amount, of charge.) Because charge can be positive or negative, Coulomb’s law often produces answers that are easily misinterpreted with regard to the direction of the force.
Most physics students pick up fairly early that in one-dimensional problems, a force vector pointing along the positive direction of the x-axis can be represented as a positive number, and a force vector pointing along the negative direction of the x-axis can be represented as a negative number. This convention leads to intuitive results when adding vectors, and is easily generalized to two- and three-dimensional problems.
But Coulomb’s law immediately creates difficulties, even for simple calculations. To wit, suppose we have a positive charge and negative charge arranged as below:
Each charge has a magnitude of 1 C, and the separation distance is one meter.
We use Coulomb’s law to calculate the force exerted on charge q1:
where k is Coulomb’s constant, and q1 and q2 are the magnitudes of the charges, and r is the distance between them. (If you’re going to build a universe, you must decide how strong the electric force will be between charged particles. You do this by setting the value of k. In our universe, and perhaps others, k has the value 9 x 109 in SI units.) Plugging in the numbers, we get
The force is so great because 1 C is a huge amount of electric charge. But what’s important here is the negative sign on the answer. Nearly all first-year physics students will interpret the negative to mean that the force points along the negative direction of the x-axis. But we know that unlike charges attract each other. Thus the force in fact points along the positive direction of the x-axis. What happened in our calculation?
The problem is not a fluke. It’s easy to see that if q2 is replaced with a positive charge, the overall force comes out positive. But the force vector in that case points along the negative x-axis since like charges repel each other.
The problem is that our simple expression for the force is not a vector equation:
calculates the magnitude of the force between the charged particles, and gives us no information about the direction. The magnitude, by definition, is always a positive number. So the proper way to use this formula is to substitute values for the charges ignoring the signs. At the end of the calculation, you simply insert a plus or minus sign to indicate the direction of the force, using the rule we all learned in kindergarten: like charges repel and unlike charges attract.
Many students (and professors) desire a more formal method of determining the direction. In other words, they want to know the general expression for Coulomb’s law wherein the force is expressed as a vector quantity, and one makes use of a simple procedure to determine the direction. (Yes, it gives you the exact same answer as the above method, but some students just aren’t comfortable with pulling the direction out of the air in the last step. They want to see the underlying rule in all its glory.)
Expressed as a vector quantity, Coulomb’s law is written
where (read “r-hat”) is a unit vector that sits on the charge that experiences the force, and points toward the other charge. Since we’re calculating the force on q1, the unit vector sits on that charge and points toward q2, as shown here:
Note the negative sign in the formula—it’s the crucial element that makes sure the direction comes out right. When using this formula, we substitute the values of the charges including the signs.
Let’s see how this works out for our example above:
The minus sign in the formula cancels the minus sign on q2 , leaving us with an overall positive force. It’s important to understand that the plus sign on the answer does not mean “along the positive direction of the x-axis.” It means “in the same direction as r-hat .” Since r-hat points toward q2 , that’s the direction of the force experienced by q1.
When we repeat the calculation for the force experienced by q2, nothing in the calculation changes except this: the r-hat unit vector now sits on q2 and points toward q1. The force still comes out positive, meaning that the force vector points in the same direction as r-hat, i.e., toward q1. (The student must be careful to not interpret the result to mean that q2 is pushed along the positive direction of the x-axis.)