# Breaking the chain of charges

1. Feb 19, 2014

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I am not sure how to proceed with the given problem. I can write down the net force on any of the charges but what should be the condition that the chain breaks?

Any help is appreciated. Thanks!

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• ###### chain of charges.png
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2. Feb 19, 2014

### voko

Just thinking aloud.

If the chain can withstand some external force F, it must be in a stable equilibrium. The chain breaks when stable equilibrium is no longer possible.

3. Feb 19, 2014

### Saitama

Should I write down the expression for potential energy?

4. Feb 19, 2014

### voko

I have not tried this, but I would give that a go.

5. Feb 19, 2014

### haruspex

Imagine the chain breaking at some point of an infinite chain.
What is the net force of attraction between the two pieces?
Hint:
Take as origin one of the beads, considering that bead to be in the positive half.
For a bead at position -r (r>0) and a bead at position +s (s>=0), what is the force between them?
Btw, I don't find the given identity to be relevant, which is a worry.)

Last edited: Feb 19, 2014
6. Feb 20, 2014

### haruspex

In my hint I wrote:
Now maybe I do see why they provide that formula, but it isn't really necessary. It's so that you can rule out one extreme case, but there's a relatively straightforward argument for ruling out all but the "two semi-infinite chains" case.

7. Feb 20, 2014

### Saitama

Hi haruspex!

I am not sure, do you mean that the bead at origin is of charge +q?
The distance between the beads is r+s but how do I comment on the sign of those beads i.e how do I know if its -q or +q at -r or +s?

8. Feb 20, 2014

### haruspex

No, I meant positive half of the axis on which the beads lie. I'm supposing there is a charge at position n for each integer n.
You don't care what the individual signs are - you only care whether they're the same or different. The distance r+s has enough information for that.

9. Feb 21, 2014

### Saitama

Please look at the attachment, did I interpret correctly?

The blue ones are positive charges and orange ones are negative.

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• ###### charges.png
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10. Feb 21, 2014

### haruspex

Yes, except that they also go off to infinity in the negative axis direction.
How far is it from the bead at -r (r>0) to a bead at +s (s>=0)?
In terms of that distance, what is the product of the charge signs?
What is the force between the beads?
What is the net force between all such pairs of beads?

11. Feb 22, 2014

### Saitama

I still don't see how can I comment on the product. I break this into cases.

Case i), if both r and s are even multiples of d, then the product is positive.

Case ii), if both r and s are odd multiples of d, then the product is again positive.

Case iii), if r is an even multiple of d and s is an odd multiple of d, the product is negative.

Case iv), if r is an off multiple of d and s is an odd multiple of d, the product is negative.

But I am not sure if you ask me this.

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12. Feb 22, 2014

### TSny

Following haruspex, let s and r be integers: s = 0, 1, 2, 3, ... and r = 1, 2, 3, .... which locate the particles on each side of the origin.

For a particle at location r and a particle at location s, how would you express the force between them in terms of r, s, d and q? (Or, better, in terms of r, s, and Fo)

#### Attached Files:

• ###### chain.png
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13. Feb 22, 2014

### haruspex

I was taking the distances to be rd, sd, so r and s are just integers.
Is r+s even or odd here?
Is r+s even or odd here?
Is r+s even or odd here?
I think you meant r odd, s even.
Is r+s even or odd here?
Looking through those, how does the parity of r+s relate to the sign of the product of the charges?
How do you write an alternating sign as an algebraic expression?

14. Feb 22, 2014

### Saitama

For case i) and ii), r+s is even and for case iii) and iv), r+s is odd.

Do I write the force between the two charges the following way:
$$F=\frac{(-1)^{r+s}kq^2}{(r+s)^2d^2}=\frac{(-1)^{r+s}F_o}{(r+s)^2}$$

Last edited: Feb 22, 2014
15. Feb 22, 2014

### TSny

Looks ok. Now set up an expression for the net force that all the s-beads exert on all the r-beads.

16. Feb 22, 2014

### Saitama

Before I proceed on doing the algebra, are my limits for summation correct?

$$F_{net}=F_0\sum_{r=1}^{\infty} \sum_{s=0}^{\infty} \frac{(-1)^{r+s}}{(r+s)^2}$$

17. Feb 23, 2014

### TSny

Yes, that looks good. Now take some time to think about a way to evaluate this.

18. Feb 23, 2014

### Saitama

Umm...I have been trying this for long but I can't see a way to proceed after this:
$$\sum_{r=1}^{\infty} \sum_{s=0}^{\infty} \frac{(-1){r+s}}{(r+s)^2}=\sum_{r=1}^{\infty} \frac{(-1)^r}{r^2}-\frac{(-1)^r}{(r+1)^2}+\frac{(-1)^r}{(r+2)^2}-\frac{(-1)^r}{(r+3)^2}\cdots$$
$$=\frac{-\pi^2}{12}-\left(\sum_{r=1}^{\infty} \frac{(-1)^r}{(r+1)^2}-\frac{(-1)^r}{(r+2)^2}+\frac{(-1)^r}{(r+3)^2}\cdots\right)$$
What to do after this?

19. Feb 23, 2014

### TSny

$$F_{net}=F_0\sum_{r=1}^{\infty} \sum_{s=0}^{\infty} \frac{(-1)^{r+s}}{(r+s)^2}$$

Suppose you regroup the terms, collecting all terms where r+s = 1, then all terms where r+s = 2, then r+s = 3, etc

20. Feb 23, 2014

### haruspex

Don't do that at all. Follow TSny's advice. If you saw it in integral form, you'd do a change of co-ordinates, no?