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Four charges equal to -Q are placed at the four corners of a square and a charge -q is at the center. If the system is in equilibrium, the value of q is?

the charge at the centre is q not -q
okay i will rewrite answer



This Is a Certified Answer

See the diagram for the forces between the charges. I have shown forces on a charge at B in the square ABCD.  The system consists of all charges.  So we have to balance forces on all charges -Q and +q.

Forces on -q:

For any positive or negative value of q or Q, the charge q will be in equilibrium.  This is because if symmetry and cancellation of the forces of the four -Q on q.

Let the size of square be L.  The attractive force F_BO  exerted by one -Q on -q at the center is:
   F_BO = 1/4πε * (-Q)*q / (L/√2)²  = - 1/4πε * 2 Qq / L²

This attraction force is along the diagonal and towards -Q.  Now, The other -Q charge at the opposite vertex of the square also exerts the same attractive force F in the opposite direction. So they are cancelled.  Similarly for the other two charges -Q at other corners of the square. 

   Hence the charge +q at the center O is in equilibrium.

Forces on -Q at any corner

Now the for the charge - Q at the vertices.  Each of the charges is repelled by the other three charges at other vertices

   On the charge at B, the forces exerted on -Q are ,  F_AB, F_CB, F_DB and F_OB respectively.  F_AB = repelling force of charge -Q at A on charge at B.  Repelling force F_CB is due to the charge at C on charge at B.  F_DB is the repelling force on charge at B by charge at D.   Now the attractive force F_OB is due to the charge +q on charge at B.

    Magnitude of F_AB = F_CB = (1/4π∈ ) * (-Q)²/L²

Net Resulting force of these two forces is along the direction of diagonal DB and it is equal to  F_AB * √2  because the angle between F_AB and diagonal is 45°, AB and CB are at 90°.  Vector addition of F_AB and F_CB gives us that.
     Force  F_DB = (1/4π∈) * (-Q)² / (L√2)²    is along DB direction.

   Adding vector wise, we get net resultant force of F_AB, F_CB and F_DB as

   F_net  = Q²/(4π∈) [ √2 + 1/2) /L²  = [Q²/4π∈L² ] * (1+2√2)/2

  The attractive force of -q is along BO and is = F_OB = - (1/4π∈) * qQ/ (L/√2)²
         = (1/4π∈)* 2 qQ /L²

   equating these two forces F_net and F_OB, we get
         q = Q(1+2√2)/4

2 5 2
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