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

Question

please answer thes question fast............................

Answer 1

Based on Phil's comment below, here is a solution to the problem that does not assume the charges on the corners are permanently fixed.

 

In this case, let's take the side of the square to have length "a."

 

With this in mind, let's look at the forces one corner charge -Q gets from all the other corner charges, also -Q, as well as from the center charge q.  We will assume the system is in equilibrium, and then compute the value of q that makes this true.

 

We also assume the system is held somehow until all five charges are in place and in equilibrium.

 

With these assumptions, let's look at the forces on the bottom right corner charge -Q.  It is a distance "a" from the upper right and lower left electric charges.  The diagonal of a square equals its side length times √2, so the bottom right charge is a distance of (a√2) from the upper left charge.

 

What magnitude and direction of force does each other corner charge exert on the bottom right?  Let's look at them individually.  Drawing a quick sketch of the situation may help keep all this straight:

 

FROM UPPER RIGHT

 

magnitude:  (kQQ)/a² = (kQ²)/a²

 

direction:  since the charges have the same sign, they repel along a line between the charges, so this drives the bottom right charge AWAY from the upper right charge, which means the direction of the force on the bottom right charge is DOWN

 

 

FROM LOWER LEFT

 

 

magnitude:  (kQQ)/a² = (kQ²)/a²

 

direction:  the charges repel again, so this drives the bottom right charge directly away from the bottom left charge, so the direction of the force on the bottom right charge is to the RIGHT.

 

 

FROM UPPER LEFT

 

magnitude: (kQQ)/(a√2)² = (kQ²)/(2a²)

 

direction:  the two charges repel along a line between the two (the diagonal of the square), so the bottom right charge is driven away from the upper left along that diagonal line; thus, in a direction 45^o BELOW THE HORIZONTAL, DOWN AND TO THE RIGHT.

 

Now, let's combine these forces to get he net force on the bottom right charge.

 

RESULTANT FORCE ON BOTTOM RIGHT

 

 

The forces on the bottom right charge from the upper right charge and lower left charge are equal in magnitude and perpendicular in direction.  Thus, they form the legs of a 45-45-90 right triangle, and the resultant force of those two (which we can call F_UR+BL), is directed along the hypotenuse.

 

The magnitude of the resultant force is the value of the hypotenuse, which equals the value of the legs times √2.  So

 

magnitude of F_UR+BL = √2(kQ²/a²)

 

direction of F_UR+BL:  45^o below the horizontal, towards the lower right

 

This direction is exactly parallel to the force on the bottom right charge from the upper left charge.  So to get the overall resultant net force on the bottom right charge from the other corner charges, we just add F_UL+BR and the force from the upper left charge.  So:

 

F_net on bottom right charge from other corner charges = √2(kQ²/a²) + kQ²/2a²

 

Now, the force between the bottom right charge and the center charge is attractive, an thus pulls the bottom right charge along the diagonal toward the center.  And the distance between the charges equals half the length of the square diagonal (= (a√2)/2).  That distance squared is (a²/2).  So, for the force between the center charge and the bottom right charge, we have:

 

magnitude:  F_C-BR = kqQ/(a²/2) = 2(kqQ)/a²

 

direction:  at 45^o above the horizontal up and to the left

 

The direction is opposite the resultant force of the other corner charges.  So, to keep that bottom right corner charge stationary, in equilibrium, we need the magnitudes of F_C-BR and F_net to be the same.  So we need:

 

((2kqQ)/a²) = [(√2(kQ²/a²) + (kQ²/2a²)]  

 

You can solve this expression for the value of q in terms of Q.  Notice that the k constants and the a² values cancel out of all terms, so that the size of the square (and the value of the Coulomb constant) does not matter.  The value of q depends only on the value of Q.

 

You can make a symmetric argument for every corner of the square.  At every corner, net net force on that charge from the other three corner charges will have the same magnitude we calculated above, and point away from the center of the square along a 45^o diagonal line.  Hence, this value for q will hold all corner chargesl in equilibrium.  And each corner charge will exert an equal magnitude force on the center charge that will hold it in equilibrium, since each charge pulls on the center charge in a direction opposite the way the charge in the other corner pulls on it.

 

So the value of q solved for above in terms of q will hold the entire system of charges in equilibrium.

 

Without a diagram, this is a lot of steps to try to keep straight, and there is still some algebra for you to do.  So, if you have any more questions about this, just let me know, and I or another tutor should be able to help you with the details.

 

Hope this helps!