Breaking the chain of charges

In summary: Fo(r+s) / (rd + sd)^2 or Fo(r+s) / (rd - sd)^2For alternating signs, I can write (-1)^(r+s).In summary, the chain breaks when stable equilibrium is no longer possible due to the net force of attraction between two pieces of the chain becoming greater than the chain's ability to withstand external force. The sign of the product of the charges between two beads on the chain depends on the parity of their positions, and the force between them can be expressed as Fo(r+s) / (rd +/- sd)^2, where r and s are integers representing the positions of the beads, d is the distance between the beads, and Fo is a constant.
  • #36
Pranav-Arora said:
To the right of 2nd charge, there are infinite charges so the force due to them is ##F_0\ln(2)## but maybe I went wrong here.

If I consider that there are ##\infty-1## charges to the right, do I subtract the force due to first charge on left of it?
The ln(2) only arises when both sides of the link under study are infinite.
With 1 on the left of that link and infinity on the right (1:∞) we have π2/12.
When we introduce a second atom on the left, the force it feels from the right side is the same as for its neighbour, but missing the first term of the series.
 
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  • #37
TSny said:
Now wait a minute. The first charge feels the force of all of the charges to it's right (essentially an infinite number). And you said that the force is 8.22 N. Good.

The second charge has essentially an infinite number of charges to it's right. How do you get that the net force of these charges on the second charge is ##F_0\ln(2)##?

Besides the force from all the charges to the right of 2, there is also the force on 2 due to particle 1 on the left. How much is that force? Is it in the same direction or the opposite direction as the force due to the particles on the right?

haruspex said:
The ln(2) only arises when both sides of the link under study are infinite.
With 1 on the left of that link and infinity on the right (1:∞) we have π2/12.
When we introduce a second atom on the left, the force it feels from the right side is the same as for its neighbour, but missing the first term of the series.

I did the following. I calculated the net force on the second particle due to the charges on the right and it comes out to be ##F_0 \pi^2/12##. This force is towards right. The force due to first charge on the second charge is towards left and has the magnitude ##F_0##. Hence, the net force on the second particle is ##F_0(1-\pi^2/12)=1..7\,N## towards left.

Similarly, the net force on the third particle is: ##F_0(\pi^2/12+1/4-1)=0.724\,N## towards right.
 
  • #38
Looks good.
 
  • #39
TSny said:
Looks good.

Erm...:redface:

What to do with the net forces I have calculated? :confused:

Btw, looking at my post #35, I wanted to write ##F_0\pi^2/12##, I don't know why I wrote ##F_0\ln 2##. :tongue:
 
  • #40
Pranav-Arora said:
What to do with the net forces I have calculated?

You're looking for the "weakest link". The first link is the force that is felt by particle 1 alone due to all of the other particles. The second link is the force felt by particles 1 and 2 together due to all of the other particles (3, 4, 5, ...). One way to obtain this is to add together the force felt by 1 alone and the force felt by 2 alone.

Btw, looking at my post #35, I wanted to write ##F_0\pi^2/12##, I don't know why I wrote ##F_0\ln 2##. :tongue:

Ah, I feel better now.
 

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