The Value of Physics, and Coulomb’s Law

~Sadly, I do not have any beers with me today. However, I do have a 12 pack of snapple and some Uji Gyokuro, so tea will be my poison of choice today~

For those of you who don’t know, I’m in the final year of an undergraduate degree in physics. The last core course I have remaining is Electricity and Magnetism, although I also have to fulfill major elective requirements, so ultimately, I’m taking four physics courses this semester. As of now, I have no intention of pursuing a career in physics unless I get into one of the select few graduate programs I’m interested in.

So the first question is “why?” Why waste my time and money working towards a degree in a field that won’t ultimately make me money. To that question, I have two answers:

First, I thoroughly enjoy physics. The subject matter is interesting.

Second, physics is a unique degree in that, like similar subjects such as mathematics, philosophy, and computer science, physics trains the skills of critical thinking and problem solving.

The first response is self-explanatory, and completely subjective. I won’t touch on that for now; however, the second response explains the true value of a physics degree. You see, when a physics major finishes college, and decides to go into a field such as finance in order to make money, the physics major starts at a huge disadvantage when compared to finance majors. This logic makes sense, because a finance major has been training for four years in order to understand the workings of finance, and how to be successful in finance, while a physics major has been talking about why the lattice structure of sodium chloride doesn’t collapse and destroy the universe. For some potential employers, it’s hard to see the crossover.

The thing is, there is crossover. While there are no direct applications of Coulomb’s Law in finance, there is direct application of the critical thinking and problem solving skills learned in physics. Give that recent physics graduate a year, and he can be just as competent in finance as the finance major. Give that recent physics graduate five years, and more often than not that physics major will be better at finance than the finance major will be.

Why is this? Without going into too much detail, the problems encountered in physics are much harder than the problems encountered in most other majors. Physics requires more in the way of critical thinking and intuition and less in the way of tedious work and memorization. The physics major isn’t spending his time in university typing long papers with skills largely learned in previous education, he is spending his time solving some of the world’s historically most complex problems using math and logic. Physics is analogous, in a way, to training in as an MMA fighter, whereas a lot of other specialized majors, such as finance, are like training as a boxer. A boxer has no chance against an MMA fighter in an MMA fight, whereas an MMA fighter fighting a boxer in a boxing match could probably put up a good fight. Additionally, that MMA fighter will certainly pick up boxing much more quickly than an average non-fighter, and will likely pick up boxing a lot faster than a boxer will pick up other martial arts disciplines.

With this in mind, I’d like to introduce something I did in my Electricity and Magnetism class that showcases the problem solving skills I have learned in physics. The original question is as follows:


This question is asking to find the Electric Field of an infinite cylinder with a charge density dependent on the distance from the axis running down the center of the cylinder. For convenience, I have drawn the cylinder, where increased shading signifies a greater density of charge.

Some background on Electric Fields. Electric Fields are essentially the ‘gravity’ that affects charged particles, such as electrons. Electric Fields are created by the stationary charged particles. The theory is that charged particles repel like charges and attract opposite charges, and the magnitude of the attraction is dependent on the distance between the two charged particles. As such, an electron in your computer effects your computer a lot more than an electron in space does. The electric field, in contrast to the attraction of charged particles, is the intrinsic distortion in spacetime caused by these particles.

Visualize the following; two stationary, oppositely charged particles, across from one another. What we expect to happen is that these two charges attract and move in a straight line towards one another. Now visualize this; one particle alone in spacetime; this particle would attract another particle if there were another particle to attract. However, there is not another particle to attract. So instead of an attraction, a field can be modeled, which says how a particle will be attracted at every point in space due to the initial particle.

Now, the most primitive way of calculating the Electric Field comes from what is called Coulomb’s Law, a law that imitates the Law of Gravitation. Coulomb’s Law states that the size of the Electric Field depends on how the ‘electric’ part of the electromagnetic force interacts with space, how big each respective charge is, and the distance between the charges, and that the direction of the Electric Field depends on how the charges are oriented in space. In a question such as the question provided, the Electric Field is expected to be the addition of each Electric Field created by each charge in the charge distribution. To make solving this question easier, we assume that the charge distribution is infinite. This assumption means that the charge distribution is going to be symmetric about the height no matter where in space we want to find the Electric Field, and that the Electric Field is going to ultimately cancel in all directions but one.

In order to solve this question, we use a technique called integration. Integration assumes that our cylinder can be broken into an infinite amount of infinitesimally small points, then sums up each of these very small points. Integrating this problem will give some very messy math that will yield a relatively simple answer due to the symmetry of the problem. I could have done this.

However, I’m lazy. In this problem, I noticed the following:


What this says is that the derivative of charge with respect to the z axis is 0. What this means in English is that the density of charge is independent of what height that charge is on the cylinder. Due to this, we have the ability to cut this cylinder into a bunch of disks, each with the same thickness, dz. We know that, since dq/dz=0, each of these disks has the same amount of total charge Q.


We notice that if we make the width of each disk, dz, infinitesimally small, then we are left with an infinite amount of disks, each with the same total charge on them. If we look at one of these disks from above, we see a charge distribution where the disk has radial symmetry. This means that dq/dθ=0, or that the density of charge is independent of the angle made with the center of the circle.


And so, we can do the following.


We apply what’s called the Shell Theorem. The Shell Theorem is a theorem used by Newton when computing gravitation. The Shell Theorem is a proof stating that for any body mass distribution with spherical (or in this case, radial, because our slice is two dimensional rather than three) symmetry, that body can be reasonably approximated as a point in the center of mass with a value of the total mass, M. Since charge is to the Electric Field what mass is to gravity in Classical Physics, we can apply the Shell Theorem to our charge distribution in the same way that we would apply the Shell Theorem to a mass distribution. Since this is a two dimensional surface, the total charge on the disk can be found by the surface integral, which is found by integrating the charge density times the differential unit of area; the charge density is given in the initial problem, and we select cylindrical coordinates due to ease of use for the differential unit of area. The differential unit of area in cylindrical coordinates is the radius x the differential unit of the radius x the differential unit of the angle theta, and our bounds are from 0 to a for the differential unit of radius, and from 0 to 2pi for the differential unit of theta, since we are integrating around a circle. This integral gives us a total charge, Q, for the disk.


Since each disk has the same total charge Q, we can use the Shell Theorem on each disk to approximate each disk as a point. Since there is an infinite amount of points, we are left with a line of charge, and since each point has the same charge Q, this line of charge has a constant charge density, in contrast to the non-constant charge density of the initial problem. The integration of Coulomb’s Law for an infinite line of constant charge is trivial; for consolidation, I just wrote down the answer. The answer is dependent on the constant charge density lambda; plugging in for lambda gives our final answer, which is the correct answer.

This method solves the entire question in under 5 minutes with barely any math, and shows how the knowledge obtained during a degree in Physics can have direct results on critical thinking and problem solving.


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