Does a particle exert force on itself?

This is one of those terribly simple questions which is also astonishingly insightful and surprisingly a big deal in physics. I'd like to commend you for the question!

The classical mechanics answer is "because we say it doesn't." One of the peculiarities about science is that it doesn't tell you the true answer, in the philosophical sense. Science provides you with models which have a historical track record of being very good at letting you predict future outcomes. Particles do not apply forces to themselves in classical mechanics because the classical models which were effective for predicting the state of systems did not have them apply forces.

Now one could provide a justification in classical mechanics. Newton's laws state that every action has an equal and opposite reaction. If I push on my table with 50N of force, it pushes back on me with 50N of force in the opposite direction. If you think about it, a particle which pushes on itself with some force is then pushed back by itself in the opposite direction with an equal force. This is like you pushing your hands together really hard. You apply a lot of force, but your hands don't move anywhere because you're just pushing on yourself. Every time you push, you push back.

Now it gets more interesting in quantum mechanics. Without getting into the details, in quantum mechanics, we find that particles do indeed interact with themselves. And they have to interact with their own interactions, and so on and so forth. So once we get down to more fundamental levels, we actually do see meaningful self-interactions of particles. We just don't see them in classical mechanics.

Why? Well, going back to the idea of science creating models of the universe, self-interactions are messy. QM has to do all sorts of clever integration and normalization tricks to make them sane. In classical mechanics, we didn't need self-interactions to properly model how systems evolve over time, so we didn't include any of that complexity. In QM, we found that the models without self-interaction simply weren't effective at predicting what we see. We were forced to bring in self-interaction terms to explain what we saw.

In fact, these self-interactions turn out to be a real bugger. You may have heard of "quantum gravity." One of the things quantum mechanics does not explain very well is gravity. Gravity on these scales is typically too small to measure directly, so we can only infer what it should do. On the other end of the spectrum, general relativity is substantially focused on modeling how gravity works on a universal scale (where objects are big enough that measuring gravitational effects is relatively easy). In general relativity, we see the concept of gravity as distortions in space time, creating all sorts of wonderful visual images of objects resting on rubber sheets, distorting the fabric it rests on.

Unfortunately, these distortions cause a huge problem for quantum mechanics. The normalization techniques they use to deal with all of those self-interaction terms don't work in the distorted spaces that general relativity predicts. The numbers balloon and explode off towards infinity. We predict infinite energy for all particles, and yet there's no reason to believe that is accurate. We simply cannot seem to combine the distortion of space time modeled by Einstein's relativity and the self-interactions of particles in quantum mechanics.

So you ask a very simple question. It's well phrased. In fact, it is so well phrased that I can conclude by saying the answer to your question is one of the great questions physics is searching for to this very day. Entire teams of scientists are trying to tease apart this question of self-interaction and they search for models of gravity which function correctly in the quantum realm!

This question is never addressed by teachers, altough students start asking it more and more every year (surprisingly). Here are two possible arguments.

1. A particle is meant to have 0 volume. Maybe you're used to exert a force on yourself, but you are an extended body. Particles are points in space. I find it quite hard to exert a force on the same point. Your stating that the sender is the same as the receiver. It's like saying that one point is gaining momentum from itself! Because forces are a gain in momentum, after all. So how can we expect that some point increases its momentum alone? That violates the conservation of momentum principle.

2. A visual example (because this question usually arises in Electromagnetism with Coulomb's law):

   $$\vec{F}=K \frac{Qq}{r^2} \hat{r}$$

If $r=0$, the force is not defined, what's more, the vector $\hat{r}$ doesn't even exist. How could such force "know" where to point to? A point is spherically symmetric. What "arrow" (vector) would the force follow? If all directions are equivalent...

Well a point particle is just an idealization that has spherical symmetry, and we can imagine that in reality we have some finite volume associated with the "point", in which the total charge is distributed. The argument, at least in electromagnetism, is that the spherical symmetry of the charge together with its own spherically symmetric field will lead to a cancellation when computing the total force of the field on the charge distribution.

So we relax the idealization of a point particle and think of it as a little ball with radius $a$ and some uniform charge distribution: $\rho= \rho_{o}$ for $r<{a}$, and $\rho=0$ otherwise.

We first consider the $r<a$ region and draw a nice little Gaussian sphere of radius $r$ inside of the ball. We have: $$\int_{} \vec{E}\cdot{d\vec{A}} =\dfrac{Q_{enc}}{\epsilon_{0}}$$ $$4\pi r^{2}E(r) = \frac{1}{\epsilon_{0}}\frac{4}{3}\pi r^{3}\rho_{0} \qquad , \qquad r<a$$

Now we say that the total charge in this ball is $q=\frac{4}{3}\pi r^{3}\rho_{0}$, then we can take the previous line and do $$4\pi r^{2}E(r) = \frac{1}{\epsilon_{0}}\frac{4}{3}\pi a^{3}*\frac{r^{3}}{a^3}\rho_{0}=\frac{q}{\epsilon_0}\frac{r^{3}}{a^{3}}\rho_0$$

or

$$\vec{E}(r)=\frac{q}{4\pi\epsilon_{0}}\frac{r}{a^{3}}\hat{r} \qquad,\qquad r<a$$

Outside the ball, we have the usual: $$\vec{E}(r)=\frac{q}{4\pi\epsilon_{0}}\frac{1}{r^{2}}\hat{r} \qquad,\qquad r>a$$

So we see that even if the ball has a finite volume, it still looks like a point generating a spherically symmetric field if we're looking from the outside. This justifies our treatment of a point charge as a spherical distribution of charge instead (the point limit is just when $a$ goes to $0$).

Now we've established that the field that this finite-sized ball generates is also spherically symmetric, with the origin taken to be the origin of the ball. Since we now have a spherically symmetric charge distribution, centered at the origin of a spherically symmetric field, then the force that charge distribution feels from its own field is now

$$\vec{F}=\int \vec{E} \, dq =\int_{sphere}\vec{E} \rho dV = \int_{sphere} E(r)\hat{r}\rho dV$$

which will cancel due to spherical symmetry. I think this argument works in most cases where we have a spherically symmetric interaction (Coulomb, gravitational, etc).