Electric potential of a configuration of charges

[fluidistic]

Homework Statement

The configuration is as follows: there's a charge q separated by another charge q (the distance is a) which is separated by a charge -q. And it continues to infinity. The distance of separation is always "a" and all the charges are linearly situated. I must show that for r>>a, the potential can be written as $$V(r)=k \left [ \frac{q}{r}+\frac{2qa}{r^2} \right ]$$

2. The attempt at a solution
I'm a bit confused about the distribution of charges. On the sketch, there's "..." after the 3 first charges. Hence I guess the configuration is q---q---(-q)---q---q---(-q)---q...
My other doubt is... what did they take as V(0)? It seems like the first charge, but I'm not sure. My intuition tells me what I'll have to depreciate terms of order greater than 2 at some moment.
The potential of a single charge q is $$\frac{kq}{r}$$. Of the 2 first charges of the problem: $$\frac{kq}{r}+\frac{kq}{r+a}$$. Of the 3 first: $$\frac{kq}{r}+\frac{kq}{r+a}-\frac{kq}{r+2a}$$. But I don't think I'm going into the right direction to solve the problem.

I'd appreciate a very little tip. Thanks.

 

[kuruman]

This "it continues to infinity" confuses me. When you use the form V = kq/r for the potential due to a point charge, you implicitly assume that the reference point (zero potential) is at infinity. If you put charges all the way out to infinity, then the potential cannot be taken to be zero there.

Another point that needs verification is whether this is a semi-infinite chain of charges with a beginning but no end. Also, where is point r? Is it off the line or on the line at some distance from the first charge on the chain?

Finally, something to consider is this. Suppose you only have three equally spaced charges, +q, +q and -q equally spaced on a straight line. Put the origin on the middle charge, the outer +q at x = +a and the outer -q at x = -a. Go distance x >>a on the positive x-axis and look at the distribution. You will see a pure point charge and a pure point dipole at the origin. What is the potential due to this distribution where you are?

 

[fluidistic]

Thanks once again kuruman!
Ok, I post the picture as it is in the exercise. I hope this answer your questions. (Notice however that the exercise doesn't say anything about the semi-infinite ocnfiguration of charges. It just talk about the "Configuration of charges seen in the figure".)

Finally, something to consider is this. Suppose you only have three equally spaced charges, +q, +q and -q equally spaced on a straight line. Put the origin on the middle charge, the outer +q at x = +a and the outer -q at x = -a. Go distance x >>a on the positive x-axis and look at the distribution. You will see a pure point charge and a pure point dipole at the origin. What is the potential due to this distribution where you are?

I appreciate very much your question, I hope it will train me.
Ok, so if I am in the middle charge, $$V (r)=kq \left ( \frac{1}{r-a} -\frac{1}{r+a} + \frac{1}{r} \right )$$. So if $$r>>a$$, $$V (r) \approx \frac{kq}{r}$$ which make sense since the potential of the dipole decreases with distance much faster than a point charge I believe. I don't really know how to prove it though, but I'd love to do it.

Hmm but maybe $$V(t)\approx k \left [ \frac{q}{r}+\frac{2qa}{r^2} \right ]$$, which is what I have to show. Now I guess I need to do some math. EDIT: Oh well, you rock! $$\frac{1}{r-a}-\frac{1}{r+a} \approx \frac{ 2a} {r^2}$$ so I've reached the result. I guess there wasn't an infinity of charges... strange.
I have a question. Why did you chose to put the 0 in the middle charge, and not, say the bottom one? It should reaches the same result right? Did it simplify the arithmetic?

 

[kuruman]

I think you misunderstood the dots in the figure. They don't mean "more charges"; they probably indicate the negative side of the axis, although strictly speaking r should not be thought of as negative.

Why did you chose to put the 0 in the middle charge, and not, say the bottom one? It should reaches the same result right? Did it simplify the arithmetic?

This is an extremely good question. When you have a distribution of N discrete charges, the general definition for the monopole term is

$$Q=\sum^{N}_{i=1}q_i$$

and the definition for the dipole term is

$$\vec{p}=\sum^{N}_{i=1}q_i\vec{r_i}$$

Note that (strange as it may seem) if you put a single point charge at distance a from the origin, you have both a monopole and a dipole term. In other words, you don't have a "pure" monopole. Putting the origin where I did, makes the charge distribution have a pure monopole term

Q = q and a pure dipole term p = 2aq.

 

[fluidistic]

Ok thanks a lot for the detailed explanation.
That was very interesting.

And what about if I don't have a distribution of discrete charges? Can you suggest a chapter of a book to explain it? (I own Purcell's one and I can check out Griffith's book).

 

[kuruman]

For continuous distributions

$$Q=\int\rho(\vec{r'})dV'$$

$$\vec{p}=\int \rho(\vec{r'})\vec{r'}dV'$$

where ρ(r') is the volume charge density and the primed coordinate indicates the location of the source. For discrete charges you may replace ρ(r') with an appropriate Dirac delta function.

 

[fluidistic]

Thanks.
After all I see it was almost obvious from the definitions for discrete charges.