In plasma when we introduce a positive test charge then electron cloud shield it, why don't they stick to it and neutralise it? When a positive test charge is introduced into a plasma, then there is a formation of an electron cloud which shields it. Why don't these electron stick to the positive charge and just neutralise it? What prevents them from doing so?
 A: 
My question is when positive test charge is introduced in plasma then there is a formation of electron cloud which shield it but why don't these electron stick to the positive charge and just neutralise it? What prevent them?

All the other positive charges that were already part of the plasma prior to the new positive charge entering exert electric fields as well.  The whole concept of Debye shielding arises from the fact that electric fields do work to get rid of themselves.  You cannot ignore the electric fields of any particle in the system, all must be included when the new charge is introduced.  Once the system reorganizes due to the newly introduced particle's field, then particles outside of a Debye sphere effectively don't matter because their fields are balanced by other charges in other Debye spheres.
Update
Suppose we ask why Debye spheres are not points in the limit of zero temperature?
In principle, this is not far from the truth.  The Debye length is defined as:
$$
\lambda_{De} = \sqrt{ \frac{ \varepsilon_{o} \ k_{B} \ T_{e} }{ n_{e} \ e^{2} } } \tag{0}
$$
where $\varepsilon_{o}$ is the permittivity of free space, $k_{B}$ is the Boltzmann constant, $e$ is the fundamental charge, $T_{e}$ is the electron temperature, and $n_{e}$ is the electron number density.
We can approximate this numerically as:
$$
\lambda_{De}\left[ m \right] \approx 7.43394 \ \sqrt{ \frac{ T_{e}\left[ eV \right] }{ n_{e}\left[ cm^{-3} \right] } }
$$
where the characters within the []'s indicate the units of the parameters.  If we look at a typical lab plasma device, the parameters for the electrons are roughly $T_{e}$ ~ 5 eV and $n_{e}$ ~ 1015 cm-3.  This gives a Debye length of ~5 x 10-7 meters or roughly half a micron.  The solar wind, in contrast, has $T_{e}$ ~ 10 eV and $n_{e}$ ~ 5 cm-3 so the Debye length is ~10 meters.
In the limit that $T_{e} \rightarrow 0$, the Debye length should go to zero but this is completely unphysical unless this limit is treated as the limit where the number of free charges goes to zero (i.e., everything undergoes recombination).  It is unphysical because the creation of a plasma typically requires high energies.  It could be possible to cool the plasma a great deal such that $T_{e}$ becomes very small but then you are challenged with preventing recombination, i.e., if the mean electron kinetic is too low, it becomes easier and easier for the positive ions to capture them and neutralize.  If all constituent particles are neutral, then the only significant electric fields roughly exist within the radius of the electron clouds of the atoms (assuming monatomic particles).
Note that even electron-proton plasmas can exist and be stable well below the first ionization energy of hydrogen of ~13.6 eV.  For instance, Larson et al. [2000] doi:10.1029/1999GL003632 observed electrons in a coronal mass ejection (CME) that had $T_{e}$ < 1 eV.
A: You can not really shield an excess charge Q in an otherwise neutral plasma (equal number of positive and negative charges). From the outside you will always see the net charge Q. This follows from charge conservation.
Anyway, note that the Debye theory quoted in the first answer assumes thermodynamic equilibrium i.e. dominance of collisions over all other physical processes. In particular would it require the collision frequency to be much higher than the plasma frequency (which determines the time scale on which a plasma responds collectively to changes in applied electric fields). This condition is far from being valid in most practical applications. For the solar wind the electron collision frequency would be about $\nu_c=10$ -7/sec but the plasma frequency is $\nu_p=10$ 5/sec, so the former is 12 orders of magnitude too small for the Debye theory to be applicable. Even for the lab case we have $\nu_c=3*10$ 7/sec whilst $\nu_p=2*10$ 12/sec, so the collision frequency is still 5 orders of magnitude too small. Unless one has an extremely small electron energy/temperature (the Coulomb collision cross section increases strongly with smaller energies) one would need plasma densities of at least $10$ 24/cm3 i.e. fluids, solids or the interior of the sun (note that the Debye theory was actually developed for electrolytes i.e. fluids and solids).
So as far as gaseous plasmas are concerned, re-organization of charges due to applied fields will always happen by collective charge displacements. The force due to the applied field is then cancelled by the thus generated plasma polarization field, rather then by the pressure gradient as assumed in the Debye theory.
In either case though, as indicated above, the excess charge is not totally neutralized. It is only that it appears now at the surface of the volume (thus not contributing to the field inside).
