As I was studying about the electric field and its effect I came across the idea that to find the intensity of an electric field we consider a positive unit charge particle at that point.
If we have lots of charged particles in that particular electric field then the electric field will get disturbed by these charged particles and if it happen so then how we can consider the uniform electric field in the presence of these charge particles?
Don't define electric field strength, $\vec{E}$, at point P as force on a unit charge placed at P. For one thing this gives the wrong units for $\vec{E}$ (newton instead of $\text{newton coulomb}\ ^{-1}$). For another thing, a unit charge in SI is the coulomb, and a compact charge of a coulomb (if it could be prevented from flying apart) would have a huge electric field around it. This field would no doubt displace the very charges ('source' charges) whose resultant field strength at P you are supposed to be measuring! [It would probably destroy the laboratory and beyond!]
I recommend this definition…
The electric field strength, $\vec{E}$, at a point P is defined by $$\vec{E}=\frac{\vec{F}}{q}$$in which $\vec{F}$ is the force that acts on a small 'test' charge, q, placed at P.
The units are now right. The test charge can be as small a charge as you like, and doesn't have to be positive. [If the test charge is negative, you'll have a negative quantity on the bottom line of the fraction, making the direction of $\vec{E}$ in the opposite direction from that of $\vec{F}$, that is in the same direction as the force on a positive charge! Halving the charge of the test charge will halve the force it experiences, so, whatever the value of q, provided that it's not too large, the same value will be obtained for $\vec{E}$. This is just what you'd hope for, because $\vec{E}$ is supposed to be telling you something about the environment of point P, independently of your means of measurement.]
As long as the test charge doesn't displace source charges from their positions, it doesn't matter that the test charge creates its own field. The force experienced by the test charge is the resultant only of the forces due to the source charges, so $\vec{E}$ as measured using the test charge is the resultant field at P due to the source charges.
I think you have to solve an $N$-body problem in which you know the forces and the initial conditions (poisitions and velocities of the particles), and all you have to do is to solve $N$ equations for each time step and iterate them until you reach the time you want. When you do that you can find the position of the charge carriers at any given time and determine the electric field by summing all of the electric fields caused by the charge carriers plus the external electric field.
