Electromagnetic force on particles forming a square

[FS98]

Homework Statement

[/B]
(a) At each corner of a square is a particle with charge q. Fixed at
the center of the square is a point charge of opposite sign, of magnitude Q. What value must Q have to make the total force on each of the four particles zero?
(b) With Q taking on the value you just found, show that the potential energy of the system is zero, consistent with the result from Problem 1.6.

Homework Equations

[/B]
F = kq1q2/r^2

The Attempt at a Solution

For part a, I set up the square with 4 particles at the corners and 1 particle at the center. I then set the x-axis to be along the line connecting three of the particles. After that I went on to find all of the forces acting on one of th corner particles by all other corner particles. The y compenents of all forces cancel out. For the x component, I found that for the two closer particles there was a force of kq^2/r^2 and for the farther particle there was a force of 1/4 of that because the value for r is twice as great. Setting the magnitude this force equal to the magnitude of the force applied by the particle in the center I got the answer Q = 9/4q.

Now for the second part I’m not quite sure what to do. I found the equation U = kqQ/r for the potential energy of two particles, but I’m not quite sure what to do with there being 5 particles.

 

[kuruman]

Now for the second part I’m not quite sure what to do. I found the equation U = kqQ/r for the potential energy of two particles, but I’m not quite sure what to do with there being 5 particles.

You find an expression for the sum of all possible pairings kq_iq_j/r_ij. How many pairs are there?

 

[tnich]

Here is another thought about the potential energy. What would be the potential energy of the system if the size of the square were infinite (if all four particles of the square were an infinite distance from the central particle)? How would the force acting on each particle change as you brought them closer together (in the square configuration)? Remember that change in potential energy is force integrated over distance.

 

[FS98]

You find an expression for the sum of all possible pairings kq_iq_j/r_ij. How many pairs are there?

I believe 10.

 

[kuruman]

I believe 10.

That is correct. However, there is duplication in that there are three distinct pairings of the same energy, so you only have three terms to calculate. The suggestion by @tnich is a good shortcut, however it is instructive to figure out how to answer in the general case when you have more than two charges in some arbitrary configuration.

 

[tnich]

Homework Statement

I then set the x-axis to be along the line connecting three of the particles. . . For the x component, I found that for the two closer particles there was a force of kq^2/r^2 and for the farther particle there was a force of 1/4 of that because the value for r is twice as great. Setting the magnitude this force equal to the magnitude of the force applied by the particle in the center I got the answer Q = 9/4q.

I think your calculations of the x components of the forces are incorrect. If the distance between the central particle and a corner particle is L, then the magnitude of the force between opposite corner particles is \frac{kq^2}{4L^2} and the force between neighbor corner particles is \frac{kq^2}{2L^2}The x-component of the force between neighbor corner particles is \frac{L}{\sqrt{2}L}\frac{kq^2}{2L^2}