Homework Help: Point charges in a square Help please

1. Aug 23, 2004

sbraman

A point charge of -.70microC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A point charge q is fixed to each of the remaining corners. The net force acting on either of the charges q is zero. Find the magnitude and algebraic sign of q.

The only equation I know for this type of proble is F=[k(q1)(q2)]/r^2

sbraman

I have tried to work this out by setting up the above equation so that the r^2 would cancel since I don't have that. I also have tried to figure it out using a right triangle but all of my answers are incorrect.

Last edited: Aug 23, 2004
2. Aug 23, 2004

chroot

Staff Emeritus
You can first conclude that the unknown charges must be positive, because an ensemble of four negative charges would have nothing but repulsive forces and would fly apart.

To be definite, let's make the top-left corner one of the unknown positive charges, along with the bottom-right. The top-right and bottom-left corners have the -0.70 uC charges.

Now consider just one of the unknown positive corner charges, say, the top-left.

The two known negative -0.70 uC charges each produce an attractive force on this positive charge. Furthermore, the two negative charges are equidistant, so they together produce a symmetric force that pulls the unknown positive corner charge directly toward the center of the square.

The other unknown positive charge, at the bottom-right corner of the square, produces a repulsive force that pushes the top-left charge directly away from the center of the square.

We thus have two competing forces: one force (due to the negative charges acting together) pulls the top-left charge toward the center of square, while the other force (due to the other positive charge) pushes the top-left charge away from the center of the square.

The magnitude of the force due to the negative charges is just the sum of the magnitudes of the forces due to each individually. Use Coulomb's law to find the magnitude of the force due to one, and double it.

The positive charge at bottom-right must exert the same magnitude of force, just in the opposite direction. You can use Coulomb's law here to find the charge both positive particles must carry.

- Warren

3. Aug 23, 2004

HallsofIvy

One problem you have is that "F=[k(q1)(q2)]/r^2" is not sufficient. Magnetic force is a VECTOR quantity and you need vector equations. The F vector is ([k(q1)(q2)]/r^3)r where r is the vector from the first charge to the second. The point of the r^3 rather than r^2 is to cancel the length of the r vector. Calculate the x, y, z components of each force and set them equal to 0.

4. Aug 23, 2004

chroot

Staff Emeritus
Halls,

What magnetism? This is an electrostatic problem. And you really don't need the full power of vectors in this situation, because both forces involved are along the same line. Negative and positive will do just fine.

- Warren