What would be the charge distribution of a conducting sphere in front of a positive point charge? What would be the charge distribution of a conducting sphere in front of a positive point charge? I mean if it's a positive charge then it should induce negative charge in the near side and positive on the other side. But as it's conducting then it should distribute the charge all over the sphere. So it should make the sphere nutral. Or something extra-ordinary might happen. Assume the sphere is isolated. 
 A: This electrostatic problem of a point charge $q$ in vacuum at a distance $L$ from the center of an isolated conducting sphere with radius $R$ can be easily solved by the method of images, which introduces a virtual image charge $$q'= -q\frac {R}{L}$$ at a distance $$l_1=\frac {R^2}{L}$$ from the center in the sphere's interior on the connecting line between the center and the outer charge. Using the superposition of the Coulomb potentials (or electric fields) of the charges $q$ and $q'$ one obtains the total potential and electric field outside the sphere and thus also the normal electric field $E_n$ on the surface of the sphere. 
From this follows the surface charge distribution on the sphere $$\sigma=\epsilon_0 E_n$$   
A: 
if it's a positive charge then it should induce negative charge in the
  near side and positive on the other side.

That's correct.

But as it's conducting then it should distribute the charge all over
  the sphere. So it should make the sphere neutral.

Since the sphere is isolated, it remains neutral at all times. The electrons moving toward the external positive point charge will each leave behind one positive ion.
As the electrons are moving closer to the side where the point charge is, they will start experiencing the increasing repulsion from each other and the attraction from the ions left behind and, at some point, when these forces will equalize the attraction force from the point charge, the electrons will stop moving.
At that point, there wont'be any field or force inside the sphere or along the surface of the sphere - otherwise the electrons would continue moving. When there is no field, there is no potential difference, so we say that the sphere has reached the equipotential state. 

This does not mean that the potential of the sphere will be zero: it will be positive due to the presence of the positive point charge. There will be electric field around the sphere, but all the field lines will be normal to the surface of the sphere, so they would not be causing electrons to move along the surface, but rather try to pull electrons or ions away from the sphere, which won't happen unless the field is very strong.
The charges would be evenly distributed along the surface of the sphere, if (a) there was a net charge on the sphere and (b) there was no external field to bias it.  
A: The answer using the method of images is for a grounded sphere, but to correct for this you simply need to add a second image charge
$$
q ^{\prime \prime} = -q^{\prime}
$$
at the center of the sphere. This gives a total "induced" charge of
$$
q_{induced} = q^{\prime}+q^{\prime \prime} = 0
$$
so you have two image charges, one at the center of the sphere with charge $q^{\prime\prime} = q \frac{R}{L} $ and one at $ l = \frac {R^2}{L} $ from the center toward the point charge with charge. $q^{\prime} = -q\frac{R}{L} $.
