How Do You Determine When a Chain of Charges Breaks?

[Saitama]

Homework Statement

Homework Equations

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!

 

[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.

 

[Saitama]

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.

Should I write down the expression for potential energy?

 

[voko]

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

 

[haruspex]

Imagine the chain breaking at some point of an infinite chain.
What is the net force of attraction between the two pieces?
Hint:

Spoiler

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.)

 

[haruspex]

In my hint I wrote:

Btw, I don't find the given identity to be relevant, which is a worry.

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.

 

[Saitama]

Hi haruspex!

Take as origin one of the beads, considering that bead to be in the positive half.

I am not sure, do you mean that the bead at origin is of charge +q?

For a bead at position -r (r>0) and a bead at position +s (s>=0), what is the force between them?

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?

 

[haruspex]

Hi haruspex!


I am not sure, do you mean that the bead at origin is of charge +q?

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.

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?

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.

 

[Saitama]

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.

Please look at the attachment, did I interpret correctly?

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

 

[haruspex]

Please look at the attachment, did I interpret correctly?

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

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?

 

[Saitama]

In terms of that distance, what is the product of the charge signs?

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.

 

[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 F_o)

 

[haruspex]

I was taking the distances to be rd, sd, so r and s are just integers.

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

Is r+s even or odd here?

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

Is r+s even or odd here?

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

Is r+s even or odd here?

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

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?

 

[Saitama]

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 F_o)

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?

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:
$$F=\frac{(-1)^{r+s}kq^2}{(r+s)^2d^2}=\frac{(-1)^{r+s}F_o}{(r+s)^2}$$

 

[TSny]

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

 

[Saitama]

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

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

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

 

[TSny]

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

 

[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?

 

[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

 

[haruspex]

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?

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?

 

[Saitama]

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

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

When r+s=1, number of terms are 1.

When r+s=2, number of terms are 2.

When r+s=3, number of terms are 3.

Proceeding this way, I have the summation:
$$\sum_{n=1}^{\infty} \frac{n(-1)^n}{n^2}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n}=-\ln(2)$$

I doubt what I have done is correct because I don't seem to be able to use the formula given in the question.

If you saw it in integral form, you'd do a change of co-ordinates, no?

What do you mean by this?

 

[TSny]

When r+s=1, number of terms are 1.

When r+s=2, number of terms are 2.

When r+s=3, number of terms are 3.

Proceeding this way, I have the summation:
$$\sum_{n=1}^{\infty} \frac{n(-1)^n}{n^2}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$$

Very good. See if you can relate this to one of the common series here

[EDIT: Oops, I see you've already done that!]

 

[TSny]

I doubt what I have done is correct because I don't seem to be able to use the formula given in the question.

Recall that haruspex already indicated that you don't need the given formula except to maybe compare the result that you got to the force felt by an atom at the end of the molecule due to all the other atoms.

 

[Saitama]

Very good. See if you can relate this to one of the common series here

Umm..what about the series mentioned in the problem? Just to confuse the students? :P

So the magnitude of net force is $F_0\ln(2)$. I am not sure what to do now. Is this the maximum force which can be applied to pull the string of charges?

 

[Saitama]

Recall that haruspex already indicated that you don't need the given formula except to maybe compare the result that you got to the force felt by an atom at the end of the molecule due to all the other atoms.

Yes, I remember that I used that series to calculate force on one of the charges.

 

[haruspex]

Yes, I remember that I used that series to calculate force on one of the charges.

So which is greater, the force between two semi-infinite chains or that between a single atom on one end and the rest of the chain?
That asked, it leaves open the question of 2 versus infinity, 3 versus infinity, and so on. To prove ln(2) is the right answer, you need to show that adding a pair of atoms on one end, e.g. going from n:∞ to (n+2):∞, always reduces the net force. The provided formula might be useful for that, but if you can prove it without resort to the formula then you don't need the formula at all.

If you saw it in integral form, you'd do a change of co-ordinates, no?

I meant, presented with ∫_x>0∫_y>0f(x+y).dxdy you'd substitute u = x+y, v = x-y, or somesuch.

 

[Saitama]

So which is greater, the force between two semi-infinite chains or that between a single atom on one end and the rest of the chain?

The force acting on the charge at one end due to infinite chain is:
$$F_0\frac{\pi^2}{12}=8.22 \,N$$
(the force is attractive)

That asked, it leaves open the question of 2 versus infinity, 3 versus infinity, and so on.

Do you want me to evaluate the net force on 2nd and 3rd charge from one end?

To prove ln(2) is the right answer...

Unfortunately, its incorrect. $F_0\ln(2)=6.93\,N$,the correct answer is $6.45 \,N$. :(

, you need to show that adding a pair of atoms on one end, e.g. going from n:∞ to (n+2):∞, always reduces the net force. The provided formula might be useful for that, but if you can prove it without resort to the formula then you don't need the formula at all.

How does the net force decrease? I always get the same summation since the number of charges are infinite.

 

[TSny]

Unfortunately, its incorrect. $F_0\ln(2)=6.93\,N$,the correct answer is $6.45 \,N$. :(

Interesting. I don't see it. Maybe haruspex does.

I did note that by chance $F_0(2\frac{\pi^2}{12} - 1)$ is about 6.45 N. [EDIT: I think this is the net coulomb force on the first two particles in the chain.]

But I don't see an argument for that being the answer.

 

[TSny]

Well, maybe I do. Consider the net Coulomb force on the first particle, the first two particles, the first three particles, etc.

(I'm not sure about this problem because I can't see how the molecule can hang together with just the point-particle Coulomb forces.)

 

[haruspex]

Interesting. I don't see it. Maybe haruspex does.

Yes, I blundered. I thought I had proved to myself that adding one increased the force, but that's clearly wrong.
1:∞ gives $\Sigma_1 \frac {(-1)^n}{n^2} = -\frac{\pi^2}{12}$
2:∞ gives $-\frac{\pi^2}{12} + \Sigma_2 \frac {(-1)^n}{n^2} = 1 - 2\frac{\pi^2}{12}$
3:∞ gives $1 -2\frac{\pi^2}{12} + \Sigma_3 \frac {(-1)^n}{n^2} = 2 - 1/4 - 3\frac{\pi^2}{12}$
This gives an alternating sequence converging to -ln(2), but the smallest magnitude term is the second one.

 

[TSny]

This gives an alternating sequence converging to -ln(2), but the smallest magnitude term is the second one.

Here's a graph for fun.

 

[Saitama]

Well, maybe I do. Consider the net Coulomb force on the first particle, the first two particles, the first three particles, etc.

https://www.physicsforums.com/attachment.php?attachmentid=66869&d=1393004218

I calculate the net force on the first particle already which came out to be 8.22 N. The force is towards positive x-axis.

Won't the net force on the first two particles be simply the twice of 8.22 N?

 

[TSny]

https://www.physicsforums.com/attachment.php?attachmentid=66869&d=1393004218

Won't the net force on the first two particles be simply the twice of 8.22 N?

Can you explain how you got this? For the second particle, how many other particles are to the right of the 2nd particle and how many other particles are to the left of the 2nd particle?

 

[Saitama]

For the second particle, how many other particles are to the right of the 2nd particle and how many other particles are to the left of the 2nd particle?

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?

 

[TSny]

To the right of 2nd charge, there are infinite charges so the force due to them is $F_0\ln(2)$

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?