What’s Wrong With Algebra-Based E&M?

It’s summer time. For me, that means I’m getting ready for summer classes. Yay! Well, at least I get paid—so that’s good, right? This year, I am teaching the physics for elementary education majors and the second semester of algebra-based physics (electricity and magnetism).

Just to be clear, there are usually two types of introductory physics at the college level. First, there is the calculus-based physics sequence. This course is for physics majors, chemistry majors, math majors…stuff like that. Of course it assumes that the students can use calculus.

The other version is the algebra-based. It does NOT use calculus. The students that take this (at least at my institution) are mostly biology, engineering technology. If you want to consider the course goals, you really need to know who is taking the course.

In order to see the problem with the algebra-based course, let me describe the second semester of the calculus-based course. For this course, I use Matter and Interactions (Chabay and Sherwood, Wiley). It’s a great textbook—here is my review of this textbook from 2014. Here is a short summary of the approach (for the second semester).

  • What is the electric field?
  • What is the magnetic field?
  • How does matter interact with the electric and magnetic fields?
  • What is the connection between electric and magnetic fields—Maxwell’s Equations.

For me, it’s all about building up to Maxwell’s Equations. Just to be clear, here are Maxwell’s Equations.

\oint \vec{E} \cdot \hat{n} dA = \frac{1}{\epsilon_0} \sum q_\text{in}

\oint \vec{B} \cdot \hat{n} dA = 0

\oint \vec{B} \cdot \vec{dl} = \mu_0 \left[ \sum I_\text{in} + \epsilon_0 \frac{d}{dt} \int \vec{E} \cdot \hat{n} dA \right]

\oint \vec{E} \cdot \vec{dl} =- \frac{d}{dt} \int \vec{B} \cdot \hat{n} dA

Of course there are many different ways to write these equations, however—one thing should be clear. You can’t really grok Maxwell’s Equations without calculus. You need to understand both derivatives, line integrals, and surface integrals.

Now for the algebra-based course. If you don’t have calculus, you can’t really get to Maxwell’s equations. Oh sure, you could do things like Gauss’s Law and Ampere’s law, but it would just be a “how do you use this equation”. Although it’s still true that Maxwell’s equations are sort of magical, without calculus they are just a game.

It’s sort of like teaching long division to 5th graders. Sure, they can learn the process of finding a division value but using the steps—but why? Why use long division when you could just use a calculator? However, if you use long division to understand the number system and division, that’s cool. But it seems that most classes just teach the “how to long divide” without going into the details.

This is exactly where most algebra-based physics textbooks end up. It becomes a giant equation salad. A bunch of equations that have no derivation. Yes, students can be “trained” to use these equations, but I really don’t see the point of that.

I should point out that there isn’t a problem in the first semester of algebra-based physics. A student can use the momentum principle or the work-energy principle without calculus. It’s not a big problem.

OK, so what am I going to do? Honestly, I don’t know. Here are some final thoughts.

  • What is the ultimate goal of this course? Why do biology and engineering technology majors take this course? The course goal will shape the course material.
  • I have two options for textbooks this semester. They both suck. OK, they don’t actually suck—but they are just a bunch of equations.
  • It would be nice to just focus on observable stuff and modeling. Do something like measure current and voltage and produce a linear function relating the two. Oh, how about repeating historical experiments to see where all this stuff comes from?

I’ll keep you updated.

MacGyver Season 3 Episode 22 Science Notes: Mason + Cable + Choices

Season 3 finale—but don’t worry, MacGyver has been renewed for a 4th season. Boom. Now for some science.

Descender Device

MacGyver needs to get down an elevator cable—to do this, he builds a descender. The basic idea is to “grab” hold of the cable to produce enough friction that it supports a human. That keeps you from falling. Of course you also want to move down, there needs to be some method to “inch” your way down. The one MacGyver builds looks like this.

Here is an early sketch for a type of descender

Of course the problem is that the elevator cable is under tension and very thick. It’s really more like a pole than a rope. That’s why the design in the episode would work better.

Two Bad MacGyver (Mason) Hacks

These aren’t bad hacks—they are hacks from the Bad MacGyver (Mason). First, there is the cable cutter. This is just a bolt cutter connected to an electric motor. That should work.

The other one is the hydrochloric acid in the basement of a building around support pillars. So, would this work? Well, hydrochloric acid does indeed dissolve concrete and cement—it’s not super fast though. Everyone likes to think of acid as being that kind in the movie Alien. It’s not like that.

Of course a pillar isn’t just cement. It has steel rods in there too. But acid will eat through steel as well—again, it just takes a while. But you don’t have to completely dissolve the pillars to cause destruction. Just making them weak could do the job.

Oh, it’s a good thing the hydrochloric acid is in plastic barrels. It would melt steel barrels.

Atwood Machine

MacGyver’s plan is to connect one elevator to the one next to it. When the cable is cut, the two elevators will create an Atwood machine. This is of course a real physics problem.

The idea is to have two different masses connected by a string. This string then runs over a pulley. If the masses are different, the two masses will accelerate (one up and one down) with a constant acceleration. The key is that this acceleration will be much smaller than the acceleration of a free falling object. That’s a good thing since waaaaay back in the day, it was very difficult to measure the motion of an object with a large acceleration.

I think I will save the physics of an “Atwood Machine Problem” (no one really calls it that) for a later post. Instead, here is my calculation.

But wait! There’s more! This calculation would give you the tension in the cable, but once there is a tension the cable would stretch. How much the cable stretches depends on:

  • Tension
  • Cable length
  • Cable diameter
  • Type of material

So you see that the stretch really depends on two things—the material and the size of the cable. For the material dependence on stretch, we call this Young’s Modulus.

There is one more thing—maximum tension before a cable breaks. This also depends on the type of material and the shape of the cable. Here is a sample calculation.

Too bad MacGyver never got a chance to put these calculations into practice. Of course it’s Mason’s fault.

Recover Serial Number

It is possible to recover a serial number that’s scratched off a metal. Essentially, when the number is stamped into the metal there is a more than just a surface effect. The deeper metal is also changed in some way. Using acid, it’s possible show these differences and find this number. Yes, this is real.

For a circuit chip, the serial number is not likely to be stamped—it will be printed. Still, it’s entirely plausible that you could still recover some type of artifact.

Evolution of a Physics Lab

When I think about the physics labs I teach, I realize things have changed over the past 18 years. The way that I run introductory labs is different than when I first started. Here is a review of my lab philosophy over the years.

I’m going to leave off the labs I taught as a graduate student since I wasn’t really in charge of the lab design.

Phase 1: Mostly Traditional – But With Computers

Really, when you first start off with a tenure track position you have to go with the flow. You can’t jump in and start doing crazy stuff. There are too many other things to focus on (grants, papers, projects…). So, for me—I just took the departmental physics lab manual and started with that. It was pretty traditional.

But I quickly set out on my own. I stopped using the lab manual and made my own labs. Oh, they were still pretty traditional in the format of:

  • Here is some physics theory.
  • Here are detailed instructions on how to collect data.
  • Here are detailed instructions on how to analyze the data.

However, my labs had data acquisition stuff to make it cooler. I found some money to put new (at the time new) iMacs in the room and used Vernier Logger Pro with sensors and stuff. Wait, I actually have a picture of this room from 2003.

Check that out. Those are some classic iMacs. Those suckers were in use for at least 10 years.

There was another important aspect of this “phase 1 lab”. I wanted to have the students work on the following:

  • Physics concepts
  • Data analysis
  • Error analysis (uncertainty)
  • Technical writing and communication
  • Experimental design.

Note: you can not do this many things. It’s either a 2 or 3 hour lab. At most you could focus on two of these things.

In terms of writing, I think I was making excellent progress on this front. I was working on an idea about peer evaluation of writing. The basic idea is that students evaluate other students writings as a way of helping everyone write better. I still think this is a good idea, but I moved on (because of many issues and other things to work on).

Phase 2: Make Pre Lab Great Again

If you have taught labs, you know that students aren’t always properly prepared. Most faculty know this. They might spend the first 30 minutes of lab time with a lecture to cover the important points. But this still doesn’t work. It’s hard for the students to pay attention and to fully grok the lab. They end up just asking questions about stuff you just told them.

OK—I can fix this. I will just make super awesome lab materials and post it online. Note only that, I will include videos and everything. Students will look at this and then we can just rock and roll during lab.

Nope. That doesn’t work. It doesn’t matter how great the video teaches the concept if students never watch it. In fact, I would find many students watching the video IN LAB. This drives me crazy—mostly because I hate hearing my own voice.

I tried online pre-lab quizzes. That didn’t work. They would just do the bare minimum to get the stuff done before class. It was just a pain in the rear.

Oh, what about pre-lab quizzes in class? Again, those are more trouble than they are worth.

Phase 3: Play and Compete

This one works fairly well. Forget about the pre lab stuff. Drop the lecture at the beginning of lab too. Give the students stuff to play with and see if they can come up with their own questions.

Here is an example in the realm of 1-D collisions.

  • Show students the tracks, carts, and different bumper options.
  • Tell them “keep the track level”, but otherwise just play with it.
  • Students love the magnetic bumpers. Many of them will try collisions between different mass carts.
  • After they have played, suggest they try to calculate the kinetic energy and the momentum of the carts.
  • Let them come up with their own methods for calculating velocities (I give some options).

That works fairly well. Some students don’t do too much, but for the students that find cool stuff it works great.

Here is another example with a competition. Again, no pre-lab.

  • Show students an inertial balance (oscillates back and forth).
  • Let them play with it.
  • Now for the challenge. Can you use this to find the mass of 4 unknown masses? The quiz at the end of the lab is just finding the mass. Your score is based on your accuracy.

This works fairly well—but not every lab can be in the form of a contest. However, students love to compete and it’s fun.

Phase 4: Free-for-all

This is where I am at now. I don’t expect students to prepare for lab because I will just be disappointed. The labs are a combination of all types of lab. Sometimes they are just verifying an equation. Sometimes they get to build stuff. I don’t expect the lab to match up with the lecture course (because apparently that doesn’t matter).

Sometimes labs still suck, but sometimes they are awesome. I will keep changing my labs until everything is perfect.

Oh, here is a more recent picture of the lab.

Numerical Calculation of an Electric Field of a Half-Ring

Suppose there is a charge distribution that is half a circle with uniform charge. How do you find the electric field due to this half-ring? Here is a picture.

If you wanted to find the electric field at the origin (center of the half-ring), you could do this analytically. If you want to find the electric field somewhere else, you need to do a numerical calculation.

Here is the plan for the numerical calculation.

  • Break the ring into N pieces (where N can be whatever number makes you happy).
  • Treat each of these N pieces as though they were point charges.
  • Calculate the electric field due to each of these pieces and add them all up.
  • The end.

Maybe this updated picture will be useful.

Let’s say the total charge is 5 nC and the ring radius is 0.01 meters. We can find the electric field anywhere, but how about at < 0.03 ,0.04, 0 > meters.

I’m going to break this ring into pieces and let the angle θ determine the location of the piece. That means I will need the change in angular position from one point to the next. The total circle will go from θ = π/2 to 3π/2. The change in angle will be:

d\theta = \frac{\pi}{2N}

I know it’s wrong, but I will just put the first piece right on the y-axis and then space out the rest. Here is what that looks like for N = 7.

Here is the code.

That works. Oh, and here is the link to the code. Go ahead and try changing some stuff. See what happens if you put N = 20.

But there is a problem. If I make these charge balls, I need to also calculate the electric field due to each ball. I was going to make a list (a python list) to put all these balls in, but I don’t think I need it.

Here is my updated code.

With the output of:

I think this is working, but let me go over some of the deets.

  • Line 13: you need to know the charge of each piece—this depends on the number of pieces.
  • Line 12: We need to add up the total electric field from each piece. This means that we need to start with a zero electric field.
  • Line 15: I named the point charges so I can reference them. But here you can see that with this method, there is only one charge—it just moves.
  • Line 16: calculate r from a piece to observation location.
  • Line 17: electric field due to a point charge.

Homework

I’m stopping here. You can do the rest as homework.

  • How do you know this answer is correct? Hint: put the observation location at the origin.
  • How many pieces do you need to get a valid answer?
  • Make a plot of E vs. distance along the x-axis. This graph should show E approaching zero magnitude as you get farther away.
  • What about electric potential with respect to infinity? Oh yeah. That’s a good one.
  • Display the electric field as an arrow at different locations.

Elastic Collisions in 1D

I need a nice model to predict the final velocity when two balls collide elastically. Don’t worry why I need this—just trust me.

After working on this for a short bit and making an error, I realized what I need to do. I need to blog about it. A blog is the perfect place to work things out.

So, here is the situation. A ball of mass 10 kg (ball A) is moving with a speed of 0.1 m/s in the positive x-direction. This collides with a 1 kg ball (ball B) moving at 0.1 m/s in the negative x-direction. What is the final velocity of the two balls if the collision is perfectly elastic.

For a perfectly elastic collision, the following two things are true:

  • Momentum is conserved. The total momentum before the collision is equal to the total momentum after the collision.
  • Kinetic energy is conserved. The total kinetic energy is the same before and after the collision.

In one dimension, I can write this as the following two equations. I’m going to drop the “x” notation since you already know it’s in the x-direction. Also, I am going to use A1 for the velocity of A before the collision and A2 for after. Same for ball B.

m_Av_{A1}+m_B v_{B1} = m_Av_{A2}+m_B v_{B2}

\frac{1}{2}m_A v_{A1}^2+\frac{1}{2}m_B v_{B1}^2 = \frac{1}{2}m_A v_{A2}^2+\frac{1}{2}m_B v_{B2}^2

That’s two equations and two unknowns (the two final velocities). Before solving this, I want to find the answer with a numerical calculation.

Numerical Solution

Here’s the basic plan (I’m not going over all the deets).

  • Model the two masses as points with springs on them (not really going to show the springs).
  • When the two masses “overlap” there is a spring force pushing them apart. The strength of this force depends on the amount they overlap.
  • Calculate the position and force on each ball (the force would be the zero-vector in cases where they aren’t “touching”).
  • Update the momentum of the balls.
  • Update the position of the balls.
  • Repeat until you get bored.

Oh, make sure you set your fake spring constant high enough. If it’s too low, the two masses can just pass through each other (which would still be an elastic collision).

Here is what it looks like.

Here is the code (you should take a look). Oh, the final velocities are 0.0636 m/s for ball A and 0.2639 m/s for ball B. Also, here is a plot of the momentum so you can see momentum is conserved.

What about the kinetic energy? Here you go.

Actually, notice that KE is NOT conserved. During the collision there is a decrease in the total KE because of the elastic potential energy. I just thought that was cool.

Analytical Solution

Now let’s get to solving this sucker. I’m going to start with a trick—a trick that I’m pretty sure will work (but not positive). Instead of having the two balls moving towards each other at a speed of 0.1 m/s each, I am going to use the reference frame that has ball B with an initial speed of 0 m/s and ball A with a speed of 0.2 m/s.

Since I am switching reference frames, I am going to rename the velocities. I am going to call ball A velocity C1 and C2 and then ball B will be D1 and D2 (for final and initial). Technically, I should use prime notation – but I think it will just get messy.

So, here is how it looks in the new reference frame.

In general, the initial C velocity would be:

v_{C1}=v_{A1}-v_{B1}

Now I get the following for the momentum and kinetic energy conservation equations.

m_A v_{C1}=m_A v_{C2}+m_B v_{D2}

\frac{1}{2}m_A v_{C1}^2=\frac{1}{2}m_A v_{C2}^2 +\frac{1}{2}m_B v_{D2}^2

Now we have two equations two unknowns. I’m going to cheat. I worked this out on paper and I’m just going to take a picture of it.

Here is the final solution (in case you can’t read it).

v_{C2} = v_{C1}-\frac{m_B}{m_A}v_{D2}

v_{D2} = \frac{2v_{C1}}{\frac{m_B}{m_A}+1}

So, you can get a value for vD2 and then plug that into vC2. After that, you can convert them back to the stationary reference frame to get vA2 and vB2.

Boom. It works. Here is my calculation. Just to be clear, it looks like this:

The output looks like this:

Winning. That agrees with my numerical model.

Projectile in Polar Coordinates

Why?  There is no why.

Suppose a ball is shot at an angle θ with an initial velocity v0.  What will the motion be like in polar coordinates?

First, let me start with Newton’s 2nd Law in polar coordinates (I derived this in class).

F_r = m(\ddot{r}-r\dot{\theta}^2)
F_\theta = m(r\ddot{\theta}+2\dot{r}\dot{\theta})

Once the ball is in the air (and ignoring air resistance) the only force on the ball is the gravitational force. Yes, this would be -mg in the y-direction, but we don’t have a y-direction. Instead, we have polar coordinates. Maybe this picture will help.

Photo - Google Photos 2018-02-18 08-55-07.png

The r and θ components of the gravitational force will change as:

F_r = -mg\sin\theta
F_\theta = -mg\cos\theta

If I use these forces with Newton’s law in polar coordinates, I get:

F_r = -mg\sin\theta=m(\ddot{r}-r\dot{\theta}^2)
F_\theta = -mg\cos\theta = m(r\ddot{\theta}+2\dot{r}\dot{\theta})

Of course the mass cancels – but now I can solve the first equation for \ddot{r} and the second equation for \ddot{\theta}.

\ddot{r} = r\dot{\theta}^2-g\sin\theta
\ddot{\theta} = \frac{-2\dot{r}\dot{\theta}}{r} - \frac{g\cos\theta}{r}

It doesn’t matter that these second derivatives depend on the other stuff – I can still calculate them. Once I have those, I can create a numerical calculation to update the velocities and positions. Suppose I have everything know at time(1), then the stuff at time(2) would be:

\dot{r}_2 = \dot{r}_1 +\ddot{r}\Delta t
\dot{\theta}_2 = \dot{\theta}_1 +\ddot{\theta}\Delta t
r_2 = r_1+\dot{r}\Delta t
\theta_2 = \theta_1 +\dot{\theta}\Delta t
t_2 = t_1 +\Delta t

Now I’m all set to do a numerical calculation. Well, I still need the initial conditions. I could use this:

\dot{r}_0 = v_0
r_0 =0
\dot{\theta}_0 = 0
\theta_0 = \theta_0

But wait! There’s a problem. The calculation for \ddot{\theta} has a 1/r term. If r is zero the universe will explode. I can fix this by having the initial r a little bit bigger than zero. Problem solved.

Here is the program in Glowscript.

For the first run, I am going to calculate the x- and y-coordinates in each step and plot x vs y.  I know this should look like a parabola.

newplot (2).png

Boom.  That works.  Now for a plot of both r and theta as a function of time for a high launch angle.

newplot (3).png

Double boom.

Electric Field due to a Uniformly Charged Ring

Hold on to your pants. Let’s do this.

Suppose I have an electrically charged ring. The radius of this ring is R and the total charge is Q. The axis of the ring is on the x-axis. What is the value of the electric field along this x-axis?

Analytical Calculation

You can’t directly find the electric field due to a charge distribution like this. Instead, you have to break the object into a bunch of tiny pieces and use the superposition principle. The super position principle says that the total electric field at some point is the vector sum of the electric field due to individual point charges.

Of course the electric field due to a single point change can be found as:

\vec{E}=\frac{1}{4\pi \epsilon_0} \frac{q\hat{r}}{r^2}

So, if I break this ring into a bunch of tiny points I can find the electric field due to each of these points and add them up. Yes, the smaller the points the better the answer. In fact, if I take the limit in which the point size goes to zero I will turn this into an integral. Calculus for the win (CFTW).

Let’s take a look at one of these pieces on the ring of charge. Here is a view of the ring from the side.

I can find the little bit of electric field from this little bit of charge at the top of the ring. This electric field would be:

d\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{dq\hat{r}}{r^2}

Of course the direction of r and thus the electric field will change directions as you go around this ring and add up tiny electric fields. But WAIT! We can cheat. OK, it’s not cheating–it’s just being smart. What if you also consider the tiny piece of charge at the bottom of the ring? This would also make a tiny electric field at this same location. It would have the same magnitude as the electric from the top piece since it’s the same charge and distance. However, the y-components of these two electric fields would cancel and leave only an x-component of the field.

In fact, for every tiny piece on the ring there is a corresponding piece on the opposite side of the ring. The only surviving components of electric field will be in the x-direction. So that’s all we need to calculate.

Looking at the diagram above, I can find the x-component of this tiny electric field by using the angle θ (which I don’t explicitly know). However, I can still write down this new x-component of the tiny electric field.

dE_x = \frac{1}{4 \pi \epsilon_0} \frac{dq}{r^2}\cos\theta

Yes, this is no longer a vector. It’s the x-component of the electric field such that it’s just a scalar. But there’s still a problem. I want to add up (thus integrate) all these tiny electric fields due to the tiny charges. However, I have these two variables that I don’t like. There is θ and r. I need to replace those.

For the r, it’s clear that I can use R and x and the pythagorean triangle to get:

r^2=R^2+x^2

I can also replace the θ by again using the right triangle to write:

\cos \theta = \frac{x}{\sqrt{R^2+x^2}}

Updating the tiny x-electric field:

dE_x = \frac{1}{4 \pi \epsilon_0}\frac{dq}{R^2+x^2}\frac{x}{\sqrt{R^2+x^2}}

This simplifies to:

dE_x = \frac{1}{4 \pi \epsilon_0}\frac{xdq}{(R^2+x^2)^{3/2}}

Now we need to get a better integration variable—we can’t integrate over dq. Or can we? Yes, we can. Take a look at the equation above. As you move around the circle, which of those variables change? The answer: none of them.

As I integrate over dq, I just get the sum of the dq’s—which would be Q. Boom. That’s it. Here is my final expression for the x-component of the electric field along the x-axis of this ring.

E_x =  \frac{1}{4 \pi \epsilon_0}\frac{xQ}{(R^2+x^2)^{3/2}}

Now for a couple of checks on this result. I will let you make sure these are ok.

  • Does this expression have the correct units for electric field?
  • What about the limit that x>>R? Does this expression look like the field due to a point charge?
  • What is the electric field in the center of the ring?

One last thing to think about. What if you want to find the electric field at some location that is NOT on the x-axis? Then the above derivation wouldn’t work. Too bad.

Numerical Calculation

Let’s do this again. However, this time I am going to create a numerical calculation instead of an analytical calculation. What’s the difference?

  • The analytical calculation takes the sum of charge pieces in the limit as the size of the pieces goes to zero.
  • The numerical calculation uses numerical values for a finite number of pieces to calculate the electric field.

That’s really the only difference.

OK, let’s just get into this. I’m going to give you a link to the code and then I will go over every single important part of it. Honestly, you aren’t going to understand this code until you play with it and probably break it.

Here is the code on trinket.io. Sorry—I can’t embed the code in this wordpress blog (I wish I could).

This is where I start.

k=9e9
Q=10e-9
R=0.01

I like to start with some constants and stuff. Here I am using k for the constant. I also had to pick a value for the total charge and the radius of the ring. Remember, you can’t do a numerical calculation without numbers.

Now let’s make the ring.

cring=ring(pos=vector(0,0,0),axis=vector(1,0,0), radius=R,thickness=R/10, color=color.yellow)

Some notes.

  • This isn’t really needed for the calculation—but it allows us to make a visualization of the thingy.
  • “ring” is a built in object in VPython (glowscript).
  • The important attributes of this object are the position (pos), the axis—which is a vector in the direction of the axis of the ring. Radius and thickness and color should make sense.

For the next part, I need to break this ring into pieces. If I use N pieces, then I can find the location of each of the tiny charges by using an angle to indicate the location. I also need to find the charge on each tiny piece. Here is the important code.

N=20
theta=0
dtheta=2*pi/N
dq=Q/N

The “dtheta” is the angular shift between one piece and the next. It’s like a clock. It’s a clock that only ticks N times and doesn’t really tell time. But each “tick” is another tiny charge. Maybe that’s a terrible analogy.

But how do we deal with all these tiny charges? What if N is equal to 50? I don’t want to make 50 variables for the 50 charges. I’m too lazy. So instead, I am going to make a list. Lists in python are super awesome.

I won’t go into all the details of lists (maybe I will make tutorial later), instead I will just show the code and explain it. Here’s where I make the tiny charge pieces.

points=[]

while theta<2*pi:
  points=points+[sphere(pos=R*vector(0,cos(theta),sin(theta)),radius=R/8)]
  theta=theta+dtheta

Some notes.

  • First, I made an empty list called “points”.
  • Next I went through all the angular positions around the ring. That’s what the while loop does.
  • Now I create a sphere at that angular position and add it to the points list.
  • Update theta and repeat.
  • In the end I have a list of points—the first one at points[0].

Next part—make the observation location. This is the spot at which I will calculate the electric field.

obs=sphere(pos=vector(.03,0.03,0), radius=R/10)
E=vector(0,0,0)

The “E” is the base electric field, it starts at the zero vector.

Now for the physics.

for p in points:
  r=obs.pos-p.pos
  dE=k*dq*norm(r)/mag(r)**2
  E=E+dE

print("E = ",E," N/C")

Here you can see the full power of a list. Once I make the list of tiny charges, it is very simple to go through the list one tiny charge at a time—using the “for loop”.

Essentially, this loop does the following:

  • Take a tiny charge piece.
  • Find the vector from this piece to the observation location
  • Find the tiny component of the electric field using the equation for a point charge.
  • Add this tiny electric field to the total electric field and then move on to the next piece.

Boom. That’s it. Print that sucker out. Maybe you should compare this electric field to the analytical solution. Oh wait. There’s a bunch of homework questions. Actually, I was going to do some of these but this post is already longer than I anticipated.

Homework

  • Pick a value of N = 10 and an observation location of x = 0.1 meters. How well does the analytical and numerical calculations agree? What if you change to N = 50? What about N = 100?
  • Create a graph that shows the magnitude of the electric field as a function of x (along the ring axis). In this graph include the analytical solution and plots for N = 10, 30, 50, 100.

Actually, I wanted to make that last graph. It would be great.

Oh wait! I forgot about the most important thing. What if I want to calculate the electric field at a location that is NOT on the x-axis? Analytically, this is pretty much impossible. But it’s pretty easy with a numerical calculation. Here’s what that would look like.

Oh, if you like videos—here is the video version of this post.

Modeling a falling slinky

I already posted some stuff about the MythBusters Jr. slinky defying gravity thing—here are those notes.

But how do you make a model of a falling slinky? Remember, you don’t fully understand something until you model it.

Also, with a model you can quickly test different situations. What happens if you put a car on one end of the slinky (or massive spring). What kind of spring constant do you need? What if the two masses are different?

All of these questions can be investigated with a model.

Let’s get to it. Of course, I am building my model with python—because I like python (and so should you). Here is my code. This is what most of it looks like (sorry, I can’t embed here).


Here is a gif of the output.

Some notes:

  • The balls wait a short time before dropping—just to make it dramatic.
  • I have calculated the position of the bottom mass so that it starts in equilibrium. If you don’t do that, the bottom mass will just oscillate up and down and ruin the whole thing.
  • I added two objects—a stick on the side and a free falling ball. That way you can see how the spring thingy falls.
  • Oh, you should absolutely try changing things up and running the model.

Here is how the model works.

  • There are two masses (the ball1 and ball2)—just ignore the other objects, they don’t matter.
  • Once the top mass is let go, there are two forces on the two balls. The downward gravitational force and then the spring force. Whatever the spring force on the bottom ball is, the top ball has the opposite.
  • The gravitational force is easy to calculate.
  • For the spring force, you need to know the natural length of the spring and the distance between the masses. The spring force depends on the difference between the distance and the natural length—then just multiply by the spring constant. Yes, I often mess up the sign on this force so that the two objects get pushed away in a weird motion.
  • After that, you are pretty much done. Use this force to update the momentum and then use the momentum to update the position.

Homework.

Here are some things for you to try.

  • What if the top mass is 0.1 times the bottom mass? Does this still work?
  • What if the bottom mass is 0.1 times the top mass?
  • See if you can calculate and plot the vertical motion of the center of mass of the two ball system.
  • What if the spring also has mass? There is a way to model this, but I’m going to make you think about it first.
  • Suppose I want to do this with a 2000 kg car. What spring constant would I need? What natural length of a spring should I use?

Analysis of a borked lab

It happens all the time. It even happens to you. There is a new lab you want to try out—or maybe you are just modifying a previous physics lab. You are trying to make things better. But when the class meets—things fall apart (sometimes literally).

Yes. This is what happened to me this week. And yes—it’s OK.

But let’s look at the lab and go over the problems so that I can make it even better for the future.

Finding the electric field due to a point charge

This is a lab for the algebra-based physics course. It’s always tough because many of the first things they cover in the lecture class don’t have lab activities with things you can measure. Oh sure—there is that electrically charged clear tape lab, but it will be a while before they get to circuits.

So, my idea was to have the students use python to calculate the electric field due to a point charge. This would give them a safe and friendly introduction to python so that we could use it later to get the electric field due to other things (line a dipole or a line charge). It would be great.

Here is the basic structure of the lab (based on this trinket.io stuff that I wrote – https://rhettallain_gmail_com.trinket.io/intro-to-electric-and-magnetic-fields#/introduction/vector-review

You can look at that stuff, but basically I give a workshop style presentation and have the students do the following:

  • Review vectors. Add two vectors on paper (not with python).
  • Find the displacement vector – given the vector for a point, find the vector from that point to another point (the vector r).
  • Find the unit vector and the magnitude of a vector (using python).
  • Next, find the electric field due to a point charge for the simple case with a charge at the origin and the observation point on the x-axis. Do this on paper.
  • Now do the same calculation with python.
  • Find the electric field at some location due to a charge not at the origin (in python).
  • Use python (or whatever) to make a graph of the electric field as a function of distance for a point charge. Graph paper is fine. If they wanted to, they could do the calculations by hand (or use python).
  • Finally, give a quick overview of the sphere() and arrow() object in glowscript.

So, that was the plan.

Lab problems

Here are the problems students had during this lab.

  • Computer problems. Yes—whenever using computers, someone is going to have a problem. In this case, it was partly my fault. There was one computer that was broken and some other ones weren’t updated. Honestly, the best option is for students to bring their own.
  • I can see that there are some students that just sort of “shut down” when they see computer code. They automatically assume it’s too complicated to grok.
  • Students working in big groups. I hate having 4 students use one computer. That’s just lame.
  • Too much lecture. The first time I did this, I spent too much time going over vectors with not enough breaks for students to practice. I partially fixed this for the second section of lab.
  • Some students were just lost on vectors.
  • Yes, the unit vector is a tough concept.
  • I’ve learned this before—but I guess I need to relearn. The visualization (sphere and arrow) are just too much for many students. That’s why I moved it to the end in my second section.

So, that’s it. I am going to rewrite the lab stuff on trinket.io. I am also going to change my material for the dipole stuff that they are doing next week. Hopefully it goes well. Let’s just see.