Question: We all have elaborative discussion in physics about classical mechanics as well as interaction of particles through forces and certain laws which all particles obey.

I want to ask,Does a particle exert a force on itself?

**EDIT**

Thanks for the respectful answers and comments.I edited this question in order to make it more elaborated.

I just want to convey that I assumed the particle to be a standard model of point mass in classical mechanics. As I don't know why there is a minimum requirement of two particles to interact with fundamental forces of nature,in the similar manner I wanted to ask does a particle exerts a force on itself?

Answer: 
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).
This exact question is considered at the end of Jackson's (somewhat infamous) *Classical Electrodynamics*. I think it would be appropriate to simply quote the relevant passage:

> 
> In the preceding chapters the problems of electrodynamics have been
>  divided into two classes: one in which the sources of charge and
>  current are specified and the resulting electromagnetic fields are
>  calculated, and the other in which the external electromagnetic fields
>  are specified and the motions of charged particles or currents are
>  calculated...
> 
> 
> It is evident that this manner of handling problems in 
>  electrodynamics can be of only approximate validity. The motion of
>  charged particles in external force fields necessarily involves the
>  emission of radiation whenever the charges are accelerated. The
>  emitted radiation carries off energy, momentum, and angular momentum
>  and so must influence the subsequent motion of the charged particles.
>  Consequently the motion of the sources of radiation is determined, in
>  part, by the manner of emission of the radiation. A correct treatment
>  must include the reaction of the radiation on the motion of the
>  sources.
> 
> 
> Why is it that we have taken so long in our discussion of
>  electrodynamics to face this fact? Why is it that many answers
>  calculated in an apparently erroneous way agree so well with
>  experiment? A partial answer to the first question lies in the second.
>  There are very many problems in electrodynamics that can be put with
>  negligible error into one of the two categories described in the first
>  paragraph. Hence it is worthwhile discussing them without the
>  added and unnecessary complication of including reaction effects. The
>  remaining answer to the first question is that a completely
>  satisfactory classical treatment of the reactive effects of radiation
>  does not exist. The difficulties presented by this problem touch one
>  of the most fundamental aspects of physics, the nature of an
>  elementary particle. Although partial solutions, workable within
>  limited areas, can be given, the basic problem remains unsolved.
> 
> 
> 

There are ways to try to handle these self-interactions in the classical context which he discusses in this chapter, i.e. the Abraham-Lorentz force, but it is not fully satisfactory. 

However, a naive answer to the question is that really particles are excitations of fields, classical mechanics is simply a certain limit of quantum field theory, and therefore these self-interactions should be considered within that context. This is also not entirely satisfactory, as in quantum field theory it is assumed that the *fields* interact with themselves, and this interaction is treated only perturbatively. Ultimately there is no universally-accepted, non-perturbative description of what these interactions really are, though string theorists might disagree with me there.