Need help in applying Binomial Aproximation when y L?

[Aristotle]

Homework Statement

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!

Homework Equations

 

[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.

 

[Aristotle]

Oh my mistake I wrote the answer incorrectly:
What I meant was this:

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

 

[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)

 

[Simon Bridge]

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

 

[Aristotle]

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

 

[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

 

[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

 

[Aristotle]

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

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

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?

 

[Matternot]

[L²+y²]^1/2 = y[L²/y²+1]^1/2

 

[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

 

[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?)

 

[Matternot]

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

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

 

[Aristotle]

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

 

[Matternot]

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

 

[Aristotle]

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

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

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.

 

[Matternot]

Nice! Glad I could help

 

[Aristotle]

Nice! Glad I could help

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?

 

[Aristotle]

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

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

 

[Matternot]

Try this link: http://www.slideshare.net/adcosmology/electrycity-2-p-c-r-o-1

The binomial expansion can't be used when y approaches 0. We ignored terms in it assuming y>>L

 

[Aristotle]

Try this link: http://www.slideshare.net/adcosmology/electrycity-2-p-c-r-o-1

The binomial expansion can't be used when y approaches 0. We ignored terms in it assuming y>>L

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|.

 

[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.

 

[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

 

[Aristotle]

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

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

 

[Matternot]

yes, but not as extreme as 1/y²

 

[Aristotle]

yes, but not as extreme as 1/y²

yes, but not as extreme as 1/y²

Ah okay thanks! -- only wanted to make sure that we had to include one coming down on the negative side of the graph, but I guess that makes sense because the (q) density is alpha|x|.

 

[Matternot]

Yes, it clearly has to be an even function by symmetry

 

[Aristotle]

Yes, it clearly has to be an even function by symmetry

Awesome! -- Thanks for being a great help!

 

[Matternot]

No problem