Confusion regarding conductors and their behavior Consider two thin plates and one thick plate(all conducting). A and C are the thin plates, B is the thick one. Now B is inserted between the parallel plates A and C. A is given a charge 2q and C is given -q. What charges will be induced on eavh side of B? Although this may seem simple, one can obviously realise that it wont be -2q and q on the sides of B, as the plate must be neutral. I tried hard, but found no explanation leading to the answer. Does this have to be determined experimentally only?
Note: the plates have infinite dimensions.
 A: You state that the plates are infinite in extent.  
One problem this creates is that a finite charge placed on them creates infinitessimal surface charge density and therefore no (or infinitessimal) electric field.  So instead I shall interpret your question to mean that the plates are finite in extent but the distances between them are much smaller than their lateral dimensions.  Then fringing effects can be neglected. Alternatively you can consider plates which extend to infinity but the charges given result in finite surface charge densities.

Possibly the easiest way to arrive at the answer is by using the Superposition Principle, which states that the interaction between two charges is not affected by the presence of other charges.
So we can imagine that the 3 conductors are initially uncharged, in the order A, B, C from left to right, and then consider the effect of placing charges on each in turn.
The charge of $+2q$ placed on A distributes itself equally over both faces, $+q$ on each. The charge of $+q$ on the RH face of A induces charges of $-q$ and $+q$ on the LH and RH faces of B [1].  The latter induces charges of $-q$ and $+q$ on the LH and RH faces of C.
B is not charged so we do no need to consider this.
The charge of $-q$ placed on C distributes itself equally over both faces, $-\frac12q$ on each. The charge of $-\frac12q$ on the LH face of C induces charges of $+\frac12q$ and $-\frac12q$ on the RH and LH faces of B.  The latter induces charges of $+\frac12q$ and $-\frac12q$ on the RH and LH faces of A.
Finally we add the charges on each face :  
charge from A :  $.+q.| A | .+q ..... .-q. | B | .+q ..... .-q. | C | .+q$
charge from C :  $-\frac12q | A | +\frac12q ..... -\frac12q | B | +\frac12q ..... -\frac12q | C | -\frac12q$  
total charges :  $+\frac12q | A | +\frac32q ..... -\frac32q | B | +\frac32q ..... -\frac32q | C | +\frac12q$

Note [1] :  I am not entirely happy with this argument.  If the presence of charge on A causes a polarisation of charge on B, doesn't the polarisation of B cause a re-distribution of charge on A, as it has done with C?  Of course we can make B infinitessimally thin, as we can with A and C as gatsu suggests in the comment below. But if we do so, the concepts of induced polarisation and the charge on each face become meaningless.
Those answering the linked question  Why isn't the electrical field between two parallel conducting plates quadrupled?  also had some difficulty with the situation. 

For reference, the solution suggested by Previous is demonstrated in the following videos.
https://www.youtube.com/watch?v=wo83ZpOY5xw
https://www.youtube.com/watch?v=UTYQk110Xhg
