Here is the normal routine in introductory physics class.
Start off with the electric field due to point charges. You can find the total electric field at some location by first determining the vector value of the electric field due to individual charges. Then, using the superposition principle, the total electric field is just the vector sum of these fields. It’s not always trivial since the electric field is a vector.
After that, you introduce the idea of the electric potential (with respect to infinity). Again, you find the electric potential due to multiple points and then use superposition to find the total electric potential. But in this case, it’s much easier to add them since the potential is a scalar quantity — you don’t need to worry about direction of vectors and stuff.
So, let’s say you have an electric dipole consisting of two equal, but opposite charges. Could you find the electric potential at some point and then use this potential to find the value of the electric field? Yup. That’s exactly what I’m going to do.
Defining Electric Potential and Electric Field
I’m going to start with the definition of the change in electric potential. It’s really the definition of work (physics work) per unit charge. The work is a path integral of a force and in this case that force is the electric force on a charge. That gives the following definition of the change in potential between points A and B.
Since the change in potential is a path integral of the electric field, the electric field must be a derivative of the electric potential. But just a plain derivative wouldn’t give you a scalar value. That’s where the del operator comes in. It’s essentially a 3-dimensional operator where the components are derivatives.
With that, the electric field at some point would be:
But you can’t just take the derivative of the electric potential if you have just a single value. It would be like trying to find the slope of a single point — you just can’t do it. So, there are two options. First, you could know the electric potential as a function (of x, y, and z) then you can just take the derivatives like a normal human being. Second, if you just have numerical values for the electric potential then you can calculate a numerical derivative. This second method is going to be my plan.
Finding the Electric Potential
If you want to use the electric potential, you first have to get some values for the electric potential. So, let’s suppose I have 2 electric charges on the x-axis. Charge 1 is at the origin with a value of 6 nC and the other point is at x = 0.01 meters with a charge of -2 nC. I just picked these values mostly at random. Here is a diagram of the situation.
Now, let’s find the numerical value of the electric potential at another point on the x-axis — just to keep this in one dimension. I’m going to use the location of x = 0.02 meters to find the potential.
The first step is to use the expression for the electric potential due to a single point charge with respect to infinity.
In this equation, k is the Coulomb constant (9 x 10^-9 Nm²/C²) and r is the distance from the charge to the location that you want to find the electric potential. Since there are two charges, you just need to find this potential twice and then add them together.
That’s just a warm up. Next I can use stuff like this to find the electric field.
Calculating the Electric Field from the Electric Potential
I need to take a numerical derivative. So, here’s the plan — this is just in the x-direction to make it simpler. Find the electric potential at some point (call this point x0). Then move forward some small x-value (call this dx) and find the new electric potential. Finally, move back an amount dx and find the potential again. The x-component of the electric field will then be:
Let’s do this. Of course it’s easiest to use something like python for your calculator. It’s not too complicated, check it out. Oh, here is a link to the code also.
Notice that I use the change in electric potential to calculate the electric field (x-component). However, I also calculated the electric field by using superposition of the electric field due to the two point charges. You can see that the values are quite close — that’s a win in my book.
But if you can find the x-component of the electric field at one point — you can find it at every point. It’s just the same calculation repeated at other values of x. Here is what that looks like (and here is the code).
Notice that I’m plotting both the electric field due to superposition of the field due to individual points AND the field calculated from the electric potential. They are pretty much right on top of each other (so that’s a good thing).
But check this out. At what point on the x-axis is the x-component of the electric field equal to zero? It’s somewhere around x = 0.0236 meters (from the graph). Is the electric potential ALSO zero there? Nope. Remember that the electric field depends on how the electric potential changes — not on the value of the potential itself. It’s possible (and very common) for the field to be zero and the potential to be some non-zero value.
How about you find the vector value of the electric field at some location that is NOT on the x-axis? This means that you need to find the numerical derivative in both the x- and the y-directions. It will be fun.