# Graphing Potential(V) as a function of distance (r) help fitting a power function

1. Jul 14, 2009

### dapavelko

I am graphing Potential(Volts) y axis vs distance (r) x axis. I am asked to fit a power function F(r)=Ar^n . Any ideas on how to do this? What is n supposed to represent?

2. Jul 14, 2009

### cepheid

Staff Emeritus
Re: Graphing Potential(V) as a function of distance (r) help fitting a power functio

Hi dapaveko, welcome to PF!

EDIT: 'n' is just some number (any number).

'n' is the power (the exponent). 'n' could be 1, or 2, or 3, or 4, or....

in which case the function would be linear, or quadratic, or cubic, or quartic, or ....

The point is that it's some sort of power law. What you get as a result for 'n' depends upon what exponent produces a curve that best fits your data. (I am assuming that you are trying to fit a power law function to some data. Otherwise your question doesn't make much sense).

Last edited: Jul 14, 2009
3. Jul 14, 2009

### dapavelko

Re: Graphing Potential(V) as a function of distance (r) help fitting a power functio

r V
1 8.7
2 4.5
3 2.9
4 2.2
5 1.8
6 1.5

that is my data I got, and I made an excel. I added a trendline curved and got y=8.7715x^-2.011. The data was curved not linear? I am trying to associate the power function F(r)=Ar^n and determine n. Wouldn't n be the -2.011? Sorry your answer confused me so I added the data.

4. Jul 14, 2009

### cepheid

Staff Emeritus
Re: Graphing Potential(V) as a function of distance (r) help fitting a power functio

Okay.

Are you asking me or telling me? Obviously if n is anything other than 1, then the data will not be linear. What were you expecting it to be like? Are the data curved? Does your fit look like a good fit?

Yes, of course. Excel has determined that n = -2.011 for you.

By the way, I edited my first post slightly.

What, specifically, about it confused you?

5. Jul 14, 2009

### dapavelko

Re: Graphing Potential(V) as a function of distance (r) help fitting a power functio

ok I reread your post and I thought you were telling me it should be linear.

The question I am asked is...What I would expect "n" to be. Well I would expect it to be other than 1 because it is not a linear relationship ,the potential difference decreases with the first power of the distance. such as V=kQ/r

but is there a way I can calculate an n value? as I need to compute percent error and I need to find an actual value being the experimental value is given by excel?

Thanks for the help I think I was trying to make more out of what I was being asked.

Last edited: Jul 14, 2009
6. Jul 14, 2009

### dapavelko

Re: Graphing Potential(V) as a function of distance (r) help fitting a power functio

also would the slope of the trend-line represent the constant value k? I plotted both V to r and and also the magnitude of E to R and I get a slope of aprox... 8.77 which seems close to the k value of 8.99*10^9 ?

Last edited: Jul 14, 2009