For context, I am a high school physics teacher.

I am teaching students about the basics of electromagnetic force between two point charges. The equation we use is $F=\frac{kq_1q_2}{r^2}$.

This gives us the instantaneous force and also gives us the instantaneous acceleration. I have a student asking me why we do not analyze the changing forces and acceleration as the two bodies move towards or away from each other. For example a proton and electron starting a distance apart and accelerating towards each other.

The best answer I can give is that we cannot analyze this without the use of calculus to allow for a changing acceleration. This particular student is fairly advanced, but I am looking for an answer that may be used as I move forward in my career.

Overall my question is if there is there a better explanation I can give them that will give a more in-depth understanding? Or does any understanding of changing acceleration require the use of calculus in all or most circumstances? I am not sure if I just need a more in depth understanding of where the equation is derived from or how these interactions are summarized into the simple equation.

This is also of course for my own knowledge base.

  • 1
    $\begingroup$ How do you find the work done by a position-dependent force? In the case of a linear force $F=kx$ I was taught in high-school that it could be done by finding the area in the $F$-$x$ plane. That also explained quite intuitively why the work done by a more complicated force couldn't be evaluated as easily. Maybe you could do reduce the problem of finding the trajectory to the computation of an area also for the present case.. $\endgroup$
    – pp.ch.te
    Commented May 26, 2021 at 20:26
  • 1
    $\begingroup$ I will say the student really asked a brilliant question but it requires special relativity and magnetism to answer. I will attempt to write something about this question $\endgroup$ Commented May 26, 2021 at 20:56
  • 1
    $\begingroup$ Possibly related question about why solving this question by energy is easier than with calculus: physics.stackexchange.com/questions/431151/… $\endgroup$
    – lamplamp
    Commented May 26, 2021 at 22:27
  • 7
    $\begingroup$ @pp.ch.te that's essentially using calculus, while pretending you are not using it $\endgroup$
    – fqq
    Commented May 26, 2021 at 23:43
  • 3
    $\begingroup$ @DakotaRuschkowski don't forget to "tick" (accept) the answer you find most useful: this not only gives both you & the writer additional rep points, it also helps the site operate as intended. [However, it's best to allow a couple of days before accepting an answer, to give others a chance to post something that might be even better or more useful.] Also, you don't have to accept the answer with the most votes: your choice should reflect what you found most helpful. Lastly, if you're planning to ask more questions, there's useful guidance at How to Ask, and don't forget the Physics.SE Tour! :-) $\endgroup$ Commented May 27, 2021 at 23:14

14 Answers 14


You do, in fact, have to take into account the change in separation distance between charged objects when analyzing the dynamics of the system. This is done mathematically through the use of differential equations, which are "self-updating" equations that tell you how quantities are expected to change in the next (infinitesimal) time step, given the current configuration of positions and velocities.

So to answer your students' question, you would have to write out a differential equation and solve it using calculus methods, as you've mentioned. For this particular problem, the relevant differential equation would be $F=ma$, or $$m\frac{\text{d}^2\mathbf{r}}{\text{d}t^2} = \frac{kqq' \hat{\mathbf{r}}}{|\mathbf{r}|^2},$$ and finding the dynamics $\mathbf{r}(t)$ as a function of time is accomplished by solving this second-order differential equation, and plugging in initial conditions such at $\mathbf{r}(t=0) = \mathbf{r}_0$, $\mathbf{v} = \partial_t\mathbf{r} = 0$, and $\mathbf{a} = \partial_t^2\mathbf{r} = 0$.

Pedagogically, you could use this exact scenario as a motivation for the need of differential equations. You can construct discrete time steps $t_1, t_2, \dots$ and compute at each time, what the positions, velocities, and accelerations are (you could make a neat table). The solution provided by calculus is then the limit as the separation between time steps approaches zero.

  • $\begingroup$ Your differential equation has a vector on one side and a scalar on the other. I'd recommend focusing on one-dimensional movement in order to avoid this issue. Also, maybe it should be mentioned that it makes a difference whether the location of the second object is kept fixed or not (in the question, there is two bodies moving towards each other). $\endgroup$
    – Noiralef
    Commented May 28, 2021 at 9:56
  • $\begingroup$ @Noiralef Thanks for catching that. $\endgroup$
    – jsborne
    Commented May 28, 2021 at 12:22

First, give them an example of "trying to do it without calculus". It'd look like this:

The acceleration of a particle at time $t$ seconds is, say, $a=t^2$. What's the change in velocity between $t$=$0$ and $t$=$2$s?

1.Acceleration in the first $0.5$ seconds varies from $0^2$ to $ 0.5^2$, or from $0$ to $0.25$. Since the change is relatively small, let's take the acceleration of the entire interval to be a constant of $0.25$. Change in velocity is acceleration multiplied by time-interval=$0.25 *0.5=0.125$

  1. Acceleration in the next $0.5$ seconds varies from $0.5^2$ to $1^2$, or from $0.25$ to $1$. Again, let's just do a crude approximation and take the acceleration of this entire interval to be $1$. So, the change in velocity is $1*0.5=0.5$

  2. Similarly, in the next $0.5s$, the change is $1.5^2 \cdot 0.5=1.125$.

  3. In the last $0.5s$, the change is $2^2\cdot 0.5=2$

Now just add up the changes to get the total change in velocity: $0.125+0.25+1.125+2=3.75$ $ms^{-1}$

With calculus:

First, tell them why the above answer of $3.75 ms^{-1}$ is wrong. It's because it's an overestimation. For example, the acceleration in the first half second increased smoothly from $0$ to $0.25$. But our calculation pretended as if the acceleration was $0.25$ for that entire interval. If we had been more precise, we'd have gotten a better estimate.

Can we ever be as precise as possible? Probably not if we don't use calculus. Here, the acceleration has a different value for EACH point of time. At $1s$, it's $1$. Just $0.01$ seconds later, it's $1.01^2=1.0201$. Accounting for the infinite number of acceleration values would be like doing an infinite number of multiplications and additions.

Let's look at the question differently. The acceleration function $t^2$ is responsible for instantaneous changes in the velocity function $v(t)$. It's, by definition, the derivative of $v(t)$. Our job is to guess the answer to "What function has the derivative $t^2$?"

Depending on the derivative, it's not always possible to make this guess. A closed-form solution doesn't always exist. But the case of $t^2$ is simple enough to have an answer. We look up our derivative tables and find that $t^3/3+C$ has the derivative $t^2$. So $t^3/3+C$ is the velocity function. Now we can precisely calculate the change in velocity by simply subtracting the velocities at $t=0s$ and $t=2s$. The unknown constant $C$ cancels out.

Some example problems which are do-able without calculus:

  1. When the acceleration changes are abrupt- Say, a particle has an acceleration of $1$ $ms^{-2}$ for $2$ seconds and then of $3 ms^{-2}$ for $3$ seconds. As long as the acceleration values stay constant for a finite time, we can solve the problem with basic arithmetic.

  2. $a=5t$

This acceleration function is continuously changing but is linear in time. When you graph this, the problem of finding the change in velocity here is equivalent to finding the area of a triangle.

We obviously have a well-known formula for the area of a triangle, so we don't NEED calculus here. The integration tables are basically just "area-formula" tables for more general curves.

  • $\begingroup$ Approximating as average=final rather than average=midpoint for each time period isn't the best you can do, though, so it's almost a straw-man argument. It will obviously only be exact in the limit as the interval approaches zero. If you were doing a numerical simulation with tiny but finite time steps, you might well do a linear approximation over the interval. (Then the error is how non-linear it is, rather than scaling with the slope and simulation step-size.) $\endgroup$ Commented May 30, 2021 at 3:57
  • $\begingroup$ Numerical simulation using small time steps is a real thing (usually using computers), and is especially useful for systems that are hard to solve analytically even with calculus, so perhaps worth mentioning to a curious student. Making the time steps small enough allows using approximations and still getting close to the right answer. $\endgroup$ Commented May 30, 2021 at 3:59
  • $\begingroup$ @PeterCordes We can also integrate the function's Taylor series in the cases where the function can't be integrated in terms of elementary functions. Many integrals are defined in terms of infinite series. Even our elementary functions are basically just names given to infinite series. $\endgroup$
    – Ryder Rude
    Commented May 30, 2021 at 8:43

On a historical note, Isaac Newton studied the same problem for gravity, and presented the solution in Principia Mathematica without using calculus, although it turns out that he had privately developed and used calculus to develop the solution, but did not want to publish it. He therefore expressed his arguments entirely in terms of Euclidean geometry, familiar to other mathematicians and physicists.

Thus, it can be done without calculus, but the main reason science adopted calculus so enthusiastically was that it made it far easier to understand and calculate with than Newton's purely geometric approach.

Newton's Principia is very long and pretty impenetrable to a modern lay reader, but there is an excellent and much shorter book on a lecture by Richard Feynman where he redevelops a piece of Newton's argument. Feynman himself was an excellent and insightful teacher, and the authors expand and explain Feynman's argument step by step in even greater detail. They also go into the historical background, both of Newton's time, and a bit about Feynman. I can't say whether it would all be within range for a smart high school student, but I think a lot of it would be, the background about Newton and Feynman makes an inspiring story all on its own, and to the extent that it's hard, goes a long way towards answering the question of why we don't teach it pre-calculus. I'd buy a dozen copies and lend them out to the most curious students.

  • 1
    $\begingroup$ "I'd buy a dozen copies and lend them out to the most curious students." Or talk to the school library and ask them if they'd be able to find the budget to buy a few copies. $\endgroup$
    – nick012000
    Commented May 28, 2021 at 3:37

There are different levels at which you can explain this stuff.

At the first level you explain that between two point charges there is an inverse square law. This is hard enough to grasp in high school but this relatively simple law can already tell a lot about the universe. The inverse square law not only occurs for static charges but also for gravity and the intensity of sunlight as a function of distance from the sun. Historically this was one of the first laws discovered concerning electricity if not the first.

At the second level you could imagine the inverse square law also works for moving charges. Then you get the following laws \begin{align} m_1\frac {d^2\mathbf r_1}{dt^2}&=k\frac{q_1q_2}{|\mathbf r_2-\mathbf r_1|^2} \mathbf{\hat r}_{12}\\ m_2\frac {d^2\mathbf r_2}{dt^2}&=k\frac{q_1q_2}{|\mathbf r_2-\mathbf r_1|^2}\mathbf{\hat r}_{21} \end{align} Here bold face means the variables are vectors. Also $\mathbf{\hat r}_{12}$ means the vector with length 1 pointing from charge 2 to charge 1. These laws are already very complicated but if you have two opposite charges (one positive, one negative) this is identical to gravity. For two bodies the result is known and the paths of the two charges are ellipses

enter image description here

Reality is even more complicated. The two laws I mentioned above are not true for moving charges, they are only approximations. In reality the two charges produce both electric and magnetic field and when they orbit they lose energy in the form of electromagnetic radiation. This means eventually they would spiral into each other. At the third level the force on a moving charge is given by the Lorentz law $$\mathbf F=q(\mathbf E+\mathbf v\times \mathbf B)$$ where $\mathbf E$ is the electric field, $\mathbf B$ the magnetic field and $\mathbf v$ the velocity of the charge. You would have to solve for the electric and magnetic field by solving Maxwell's equations.

This still isn't the complete picture so the take home message is that things get complicated very quickly and each level is only an approximation/special case of the more deeper level.

Source of image: https://en.wikipedia.org/wiki/Two-body_problem


You could tell the students about the potential energy equation $P.E.=\frac{kq_1q_2}{r}$

It comes from integrating the force equation, but they don't have to know the details...

You could give them an example where they calculate the gain in kinetic energy and hence velocity of a charged particle that is repelled, from rest from a fixed charge and moves from one point to another. The gain in kinetic energy would be from the difference of the potential energy of the particle at the two points.

High school students are often used to this kind of thing with gravity - when they calculate the gain in kinetic energy of a falling object, due to the loss of gravitational potential energy.

see also https://en.wikipedia.org/wiki/Electric_potential_energy#Definition


How about keeping it simple:

  • From the above equation, the charges create a force.
  • We know that the a force will cause the objects to move,
  • therefore probably changing their separation ($r$ above).
  • This will change the force on the objects.
  • Which will cause them to move differently,
  • probably changing their separation, etc, etc, ad infinitum...

As you can see just using the basic equations leads to a circular calculation that cannot be solved in this manner.

The only way to solve this type of problem is rather than using the mathematics of static (non-changing) values (like instantaneous velocity or force etc) to use the mathematics of changing values (like the concept of velocity being a change of position relative to time) - which is calculus, specifically differential calculus.


why we do not analyze the changing forces and acceleration as the two bodies move towards or away from each other. For example a proton and electron starting a distance apart and accelerating towards each other.

When the charged bodies are macroscoplowic, this would be valid but:

  1. the resulting equations are the same as the equations of motion for a two-body system in newtonian mechanics (double star, or the Earth-Moon system), so mathematically the model is understood - the particles' trajectories are conic sections, either ellipses (bound state) or hyperbolae (infinite motion, scattering). This is still a topic teachable without methods of formal calculus, and the basic results are taught (Kepler's laws). Full analysis requires either calculus or advanced kinematics and geometry (conic sections, hodographs) and depending on the country and the school, this is either the highest level of mathematics the high school students are exposed to, or more often, too advanced and not taught.

  2. in practice this situation (charged macroscopic bodies accelerating) almost never occurs, because macroscopic bodies do not carry substantial charge, they tend to be close to neutral and the electric forces are either more complicated (because of polarized bodies) or negligible to other forces (gravity, pressure, friction...). So the model is a bit on the obscure side, describing improbable situation with little benefit to students.

When the particles are microscopic, such as the proton-electron system, this kind of model is less physically valid. People in the past (Thomson, Larmor, Rutherford and other pioneers of nuclear and particle physics) tried to analyze interaction of electron and nucleus(and the proton in particular) this way (through Coulomb forces and Newtonian mechanics), but they found this model has limited validity: it corresponds to experience only when the particles are far from each other and have small speed (Rutherford scattering), and then the motion is like the above-mentioned two-body system in newtonian mechanics. But when the particles get close, or the particles are very fast (close to speed of light), this model is not that fruitful. It can't describe properties of atoms such as their size, or their energy levels, or generation of new particles on scattering, without more advanced assumptions which completely change the model.

So big part of the reason one does not see this kind of model analyzed in textbooks is because of its small usefulness in teaching the students the important parts of physics.

  • 5
    $\begingroup$ " macroscoplowic" hm... did you mean macroscopic? $\endgroup$ Commented May 27, 2021 at 0:15
  • 3
    $\begingroup$ macroscoplowic: (adj.) seeing the big picture when lying in a ditch. From macro- ("long, abnormally large, on a large scale,") + -scope ("an instrument for seeing") + low ("not high, below the usual level") + -ic word-forming element making adjectives. $\endgroup$ Commented May 27, 2021 at 23:39
  • $\begingroup$ @ChappoHasn'tForgottenMonica Pretty sure that's not a real word, based on a Google search. $\endgroup$
    – nick012000
    Commented May 28, 2021 at 3:39
  • 2
    $\begingroup$ @ChappoHasn'tForgottenMonica that's hilarious, now I think we should keep that typo word there. $\endgroup$ Commented May 28, 2021 at 13:48
  • $\begingroup$ @Buraian I did! $\endgroup$ Commented May 28, 2021 at 13:48

I think the best way to explain why not to your more curious students is to show them how to.

First, it's important to note that conservation of energy (namely work) allows us to analyze situations where forces are not constant. What we can't (usually) directly do with conservation of energy is get a sense of what happens as a function of time.

Perhaps a good way to start your explanation is with the simple kinematic equations. Your students should know that for 1D motion, the area under a $F$ vs. $x$ graph will be the work done by the force $F$. You can start with the kinematic equations, showing that work will simply be $F\Delta x$. Then, you can move on to a spring, to show that the work is $\dfrac12k\Delta\left(x^2\right)$, both of which correspond to the area under a graph. From there, you can proceed to the electric (or gravity) force -- the area under the force vs. position graph will correspond to the potential energy.


Several answers talk about using a step-like explanation: calculate the force, calculate the acceleration, move the particle a little bit, repeat. Then leave out the really fun step. Actually do the problem with a computer. Show the student what research and teaching physicists do.

Use this as a great motivation to teach coding and numerical methods. NASA does calculations this way! Solving a tractable calculus problem with the computer is the first step to solving near-impossible integrals.

Use Jupyter notebooks/python with the graphing (pyplot subset of the matplotlib package) to show that numerical approximation is a great physics tool.

Do a search on the phrase let's code physics and you will be amazed at the videos available to you (and your students).


There are two issues the students face in a problem where acceleration is a function of position $a = \mathrm{a}(x)$.

  1. The time domain problem. As the system evolves, and positions change (over time) so is the acceleration changing over time. But how is acceleration varying as a function of time is unknown at this point. Under constant acceleration, like with projectile motion, this problem is trivial to overcome. For more complex systems, it is usually the last step in the solution to introduce the time domain. Depending on what you know already you use one of the following calculus expressions

$$\begin{aligned} \Delta t & = \int \frac{1}{\mathrm{v}(x)} \,{\rm d}x \\ \Delta t & = \int \frac{1}{\mathrm{a}(v)} \,{\rm d}v \end{aligned}$$

  1. The area under the curve problem. Even before time domain is introduced and the problem is looked at from an energy conservation standpoint, the total work done by a position varying force involves the area under curve. $$ W =\int \mathrm{F}(x)\,{\rm d}x $$ For simpler systems like springs, the area is easy to estimate without calculus, but I do not expect the students to tackle the area under of $1/x^2$ without calculus, regardless of how advanced they are. This is the part where a potential function $\mathrm{V}(x)$ is needed which is the result of the above integration, $ W = \mathrm{V}(x_2) - \mathrm{V}(x_1)$

As I understand it, the question you want an answer to is not "How do I answer the student's question?", but "How do I explain to the student that I cannot answer their question?" In other words: "Why do we need calculus here?"

To calculate motion resulting from forces as a function of time, we use Newton's second law, $F=ma$. This is a differential equation, because $a$ is a rate of change (of velocity, which is itself the rate of change of position), and so to solve it, we need to use calculus.

Your student may reasonably wonder why you are able to solve other problems involving Newton's second law, like projectile motion. The answer is that, in certain cases, such as constant acceleration, the differential equations are simple enough that we can just look up the solutions in a book. (For example, for constant acceleration, the solutions are the kinematic equations, like $s=ut+\frac{1}{2}at^2$. Your students might also have seen solutions to the differential equations for other cases, like circular motion and simple harmonic motion.)

But in most cases, including for freely-moving point charges, we need to solve the differential equations ourselves, and for that we need to know how to use calculus.


It's high school and the student is taking physics. This means that they will also be likely to be taking at least math and possibly applied math also. European students taking math in their last but one year (equivalent to junior year in USA) in school cover calculus, differential and integral. So no trouble with those students taking math.

With students taking physics without math (the phenomena involved like optics, sound, electrostatics, current, magnetism, etc are quite interesting to teenagers) you would do well to use the approach of Ryder Rude and perhaps try using graph methods too. Graphs of Force and Acceleration vs Separation Distance might be useful here. They show students to see the non-linearity of the acceleration/force and how in this case we cannot apply any simple equations of motion under fixed acceleration.


A little history of calculus and why we use it is much simpler than actually teaching them calculus. And I find it is sufficient to explain why we need it here, so it should be accessible, even to an average student.

You don't need calculus to solve for motion with changing accelerations. You need calculus to solve for motion in the general case. There are plenty of specialized systems with changing accelerations where you don't need calculus to arrive at an answer. For example, the motion of a satellite around a planet is easily done with algebra using Keplerian laws. In fact, Kepler had solved the motion of the planets about 30 years before Newton was even born!

The issue is that you always have to find some geometric trick to replace the integrals found in calculus based physics. In the case where velocity is not changing, that is easy. You just look at the area of triangles and average accelerations. In the case where velocity is changing but acceleration is not, it is trickier. It's really hard to explain why that $\frac{1}{2}$ appears in $\frac{1}{2}at^2$ without calculus, but it does indeed work. You can prove it empirically.

But what if accelerating is changing, and not in some convenient way with a nice simple answer. Well, if you don't have calculus, the best you can do is talk of "average acceleration," the usual $\frac{v_f-v_i}{\Delta t}$ equation. And then you can point out some obvious examples where average acceleration is insufficient. You can show that a sudden acceleration at the start followed by a constant velocity afterward yields a very different path from a constant velocity followed by a sudden acceleration at the end, even though the average accelerations and final velocities are the same!

At this point, the best thing to do is start pointing at the problems that lead to the desire to invent calculus. Observe that, if you break it up into two intervals, you get a more accurate measure of what happens because the inaccuracies involving assuming constant acceleration over a period (the average acceleration) holds, when we know it doesn't. You can start slicing this problem smaller and smaller. In calculus we call this breaking down of this problem a Riemann sum.

So you can get smaller and smaller slices, getting better and better approximations, but can you ever get it "right?" Can you ever come up with an actual closed-form solution for the motion? Note that the $\Delta t$ in the bottom keeps getting closer to 0... can you just set it equal to 0 and find the result? Well, no. It doesn't work. But can we talk about what happens as we make it smaller and smaller, getting infinitesimally small?

It turned out we couldn't for a long time. Zeno's paradoxes about how a man cannot move to the end of a football field because he must first get half way, but must first get half way to half way, and so on, ran rampant for thousands of years. Its hard to do math with infinitesimals. Lots of things that we think are just easy turn out to be hard.

The calculus Newton and Leibniz invented was originally called "the calculus of infinitesimals." A calculus is just a calculation method, so "calculus" is actually a poor name, but this calculus was such a big deal after thousands of years, that it eventually just got called "the calculus," and later took over the word entirely and became "calculus."

The thing that was special about their calculus is that it handled infinitesimals rigorously in a way which matched our intuition about how the real world worked. We knew Zeno was wrong -- you can walk to the other side of a field (he was actually going after a different metaphysical question, but that's fine). We just couldn't explain why we could do it! With the calculus of infinitesimals in hand, we could finally predict motion using simple equations: Newton's Laws of Motion. It could sidestep the ugly details of Zeno's supertasks and show exactly how things should move!

What's nice about this is that you can get arbitrarily close to the right motion without calculus, even in the general case, by just Riemann sums. Indeed, when computers calculate motion, we tend to do something almost exactly like that. So you don't need calculus to get close to the correct motion for general acceleration. However, you do need it to go from "close" to "the right answer."

And, something perhaps frustrating for a non-calculus based physics student: physics is so much simpler when it is calculus based! That ugly $x(t)=\frac{1}{2}at^2+v_0t+x_0$ equation that they have to memorize is just $F=ma$, integrated twice, in the very special case of $a$ being a constant. And when you get to rocketry, where mass changes and it becomes $F=ma+\frac{dm}{dt}v$, you find out that both of them are actually just $F=\frac{d}{dt}(mv)$, under different guises. Many of the things your student is having to memorize now are just a bunch of special cases of the simpler calculus based version.

And maybe (just maybe) that will encourage them to go learn calculus!


One historical reason is that Coulomb discovered this force law using a torsion balance, which directly measures the force between two objects. We do not need calculus, nor do we need to watch some particles moving and observe their locations as a function of time. Instead, the apparatus balances a known spring force with an unknown electromagnetic force, such that we can learn the unknown force by measuring how much the spring has been condensed (in the experiment, how much the apparatus has been twisted).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.