# Homework Help: Need help in applying Binomial Aproximation when y L?

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1. Mar 21, 2015

### Aristotle

1. The problem statement, all variables and given/known data
So I am working on an Electric Potential problem. There is a point P that is located on top of this rod ( this rod is aligned horizontally & is length L). I've solved this problem and got an answer.

I want to find when y>>L using Binomial Approximation except I am quite lost on how to apply it? If someone can please help.

The answer that I got while doing this from finding the electric potential is:

V = 2kα [(L^2 / 16) + y^2 )^(1/2) - y ]

How would I use Binomial Approximation to get y>>L? Thanks in advance!

2. Relevant equations

Last edited: Mar 21, 2015
2. Mar 21, 2015

### Matternot

The (L2/16)1/2 can be simplified to L/4. The equation doesn't look like that which should be binomially expanded. If y>>L, the L/4 can be neglected, resulting in -y2kα.

Binomial expansion can be used on something of the form you have given in the "Relevant equation". I would suggest the 1/2 is maybe around the whole bracket, but then the equation isn't dimensionally correct.

3. Mar 21, 2015

### Aristotle

Oh my mistake I wrote the answer incorrectly:

V = 2kα [(L^2 / 16) + y^2 )^(1/2) - y ]

4. Mar 21, 2015

### Matternot

How sure are you with that equation? It doesn't seem right. You would expect it to reduce to the point particle potential for y<<L. It doesn't really seem dimensionally consistent with a potential.

I would expect it to be divided by something like (sqrt(ay2+bL2)

5. Mar 21, 2015

### Simon Bridge

We need to see the original problem statement - how do you know your answer is correct?

6. Mar 21, 2015

### Aristotle

The problem states a rod -- nonuniformly. A point P located above center-rod (y distance away). Find e-potential at that point.

7. Mar 21, 2015

### Matternot

Aah, that can be expanded binomially:

Fisrtly, factor out y to reduce to (L2/16y+1)1/2

Then use (1+x)n = 1 + nx/1! + n(n-1)(x2)/2! +...

When you see L2/y2 or higher terms, they can be ignored. This will reduce the formula down

8. Mar 21, 2015

### Matternot

Given the problem you suggested, I would really consider looking at the first part again. The solution should reduce to the typical point particle potential if y>>L... I don't see this happening with your current solution

9. Mar 21, 2015

### Aristotle

How did you factor the y out of

V = 2kα [(L^2 / 16) + y^2 )^(1/2) - y ]

Especially with the y^2 being in the square root still?

10. Mar 21, 2015

### Matternot

[L2+y2]1/2 = y[L2/y2+1]1/2

11. Mar 21, 2015

### Matternot

But again.... I still expect something of the form

E=kQ/y[L2+y2]1/2

with maybe a few more constants flying around

Try looking at this website

12. Mar 21, 2015

### Aristotle

I do agree with you that it indeed should look like a point charge when y>>L.

Also for the answer you provided:
[L2+y2]1/2 = y[L2/y2+1]1/2

Where did you L^2/4 (the 4 part in denominator go?)

13. Mar 21, 2015

### Matternot

I just ignored the constant... If I must:

[(L2/16) + y2 )1/2 - y ] = y[[L2/16*y2+1]1/2-1]

14. Mar 21, 2015

### Aristotle

Okay I got an answer of 4kQ/y using the binomial expansion.

15. Mar 21, 2015

### Matternot

Are you sure about the 4? I expect kQ/y

16. Mar 21, 2015

### Aristotle

Yeah sorry for being so brief about the problem, but I substituted in my alpha--which I found during the process of the problem within the first step. I really appreciate your help.

As for the problem so far, when y approaches zero, I get the same answer as when y>>L but with the denominator being L.

For when y>>L: kQ/y.

17. Mar 21, 2015

### Matternot

18. Mar 21, 2015

### Aristotle

If i were to graph both of those as E-potential vs y distance.

Wouldn't it just be a 1/r graph on the positive side with it being a nonuniform rod? --Correct?

19. Mar 21, 2015

### Aristotle

And yeah my bad I forgot to cancel out the 4s-- got kq/y :)

20. Mar 21, 2015

### Matternot

21. Mar 21, 2015

### Aristotle

Oh I know, I didnt use binomial expansion for when y approaches 0. I figured that one already.

But my question was if i were to graph both of those as E-potential vs y distance.

Wouldn't it just be a 1/r graph on the positive side with it being a nonuniform rod? --Correct?
I mean the rod itself was lamda equaling to alpha |x|.

22. Mar 21, 2015

### Matternot

It would look similar to a 1/y, but not exactly. The potential depends on the physical distance the charge is away. At large values of y, the charge is almost all focused on a point. At small values of y, this charge's distance varies.

23. Mar 21, 2015

### Matternot

It will look like a |1/y| graph in the sense that when y→0 V→∞ when y→∞ V→0 and it has no turning point etc

24. Mar 21, 2015

### Aristotle

Ah I see. Just making sure, something like this?

25. Mar 21, 2015

### Matternot

yes, but not as extreme as 1/y2