Mastering Classical Physics: Solving a Square Array of Charges Problem

[ggups]

HI,
I just have a simple question if you have a square array of charges of +1 where the charges lie on the vertices of the square(so the charges form a suare patten where 4 of them form a unit square). The way you can balance the charges is by placing a -1 charge in the center of each square or two -1/2 along the diagnols of each square thus making the system neutral.Now is it possible to use simple force balance to come up with a psoition where the -1/2 charges will lie.

Anybody need more clarification tell me.
thanks

 

[dannyboy]

Yeah, it is possible.

Step one is in focusing on a diagonal, and realizing that the forces due to the charges on the other diagonal cancel through symmetry (i.e., we can neglect them. So, the problem simpifies to this scheme:

q-----------(-q/2)-------(-q/2)----------q

For this system to be in equilibrium along the line shown, we require that a test charge q' placed at the centre of the line must experience no net force. This is only possible if the negative charges (-q/2) are equidistant from the centre of the line, say, at a distance (a/2) each side of the centre (so that the separation of each negative charge is a).

By applying Coulomb's law to the test charge, it is then a straightforward step to obtain a relationship between a and the distance r between the positive charges. This turns out to be a=r/\sqrt{2}; since, by geometry, r=\sqrt{2}, we have a=1.

In other words, the negative charges form a square with sides of length 1/4 at the centre of the square of positive charges.

 

[ggups]

thanks a further clarification

That was very good analysis however i still need some more fundamental(i know i may sound lame but i am just beginning with real physics).
How did you use coulombs law to find your relation. I got zero. Please elaborate.

thanks

 

[ggups]

your argument was erroneous

Yeah, it is possible.

Step one is in focusing on a diagonal, and realizing that the forces due to the charges on the other diagonal cancel through symmetry (i.e., we can neglect them.

no the argument that the two other chargejs produce forces would cancel out is erroneous. In fact the two charges have a components in the direction that you just drew. The force is oblique angle so it has one component that cancels but other one does contribute.

 

[turin]

... is it possible to use simple force balance to come up with a psoition where the -1/2 charges will lie.

What do you mean by "simple?" Do you know trig? As dannyboy pointed out, there is a high degree of symmetry. However, as you noticed, you cannot use this symmetry to ignore the oblique sources.

 

[ggups]

What do you mean by "simple?" Do you know trig?

Yes fortunately i do and i do find it simple. From what i did if i considered oblique sources i was getting a complex solution. This is what is bothering me by the way.

 

[turin]

ggups,
Do you mean "complex" as in not "real-valued", or "complicated?" I expect the solution to be real-valued and complicated; however, I can't immediately rule out the complex-valued possibility by inspection. If the solution does indeed turn out to be not real-valued, then this result indicates that there is no way to arrange the four -q/2 charges to achieve balance.

 

[ggups]

ya i mean mathematically a complex valued solution. Can you give it a try i just want to check my math. By the way there are not four -q/2 charges but only two. There are four +1 charges located around the vertices of square and two -q/2 charges lie along one of the diagnol. I know that the solution is possible i just don't know how to get there through my math.
thanks anyways

 

[turin]

Well, wait a second. Now I'm not so convinced that a nontrivial solution is possible, because the situation has now become non-symmetrical. Also, consider this: If you have only two -q/2 charges that lie along the diagonal, and you already know that a -q charge at the exact center is a solution, then this seems to suggest that both -q/2 charges must lie at the center, thus returning to the original situation. Of course, given the quadratic nature of the force, I suppose there is another solution out there, in principle.