Electrostatics Equilibrium Problem.

$As\quad we\quad know,\\ \\ F=C\cfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \\ \\ let\quad length\quad of\quad side\quad is\quad a\quad \\ \\ Net\quad force\quad on\quad any\quad of\quad +q\quad will\quad be,\\ \\ F=2C\cfrac { { q }^{ 2 } }{ { a }^{ 2 } } cos{ 45 }^{ o }+C\cfrac { { q }^{ 2 } }{ { (a\sqrt { 2 } ) }^{ 2 } } +C\cfrac { Qq }{ { (\cfrac { a }{ \sqrt { 2 } } ) }^{ 2 } } \\ \\ \Rightarrow F=\cfrac { 2\sqrt { 2 } +1 }{ 2 } C\cfrac { { q }^{ 2 } }{ { a }^{ 2 } } +2C\cfrac { Qq }{ { a }^{ 2 } } (outword\quad along\quad diagonal)\\ \\ for\quad balance\quad F=0\quad \Rightarrow Q=-\cfrac { 2\sqrt { 2 } +1 }{ 4 } q\\ \\ \Rightarrow Q=-0.957q$

He has firstly not mentioned that we must take q=1C

now we want that any of the charges placed at the corner do not move simply we want hem to be in equilibrium equilibrium apply that NET FORCE ON THEM SHOULD BE ZERO NOW CONSIDER ANY ONE CHARGE FORCE ON PARTICLE IS GIVEN BY : KQ1Q2/R^2 Q1 = FIRST CHARGE , Q2 =SECOND CHARGE NOW FORCE ON ANY ONE CHARGE = kq^2/r^2+kq^2/r^2+kq^2/2r^2...........................(1) this last force is due to the one which is diagonally opposite . uptill now we have calculated the net force now the two forces which are equal get their one component cancelled and the other one added up with the force of diagonally opposite charge now total force will be like kq^2/2r^2 + 2^1/2kq^2/r^2 ( i mean one of the component of the perpendicular force cancelled up an the other which is of sin45 or cos45 adds up) now we want a charge at the centre let it name Q . now the force on the same charge due to th Q = 2kQq/r^2 .....................(2) equate 1 and 2 equation remember the polarity of charge should be negative so that it attract the corner charge