Answers

The Brainliest Answer!
2014-12-01T06:48:43+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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
wait , i am finishing answer please
click on thank you and select best answer
u r welcome