manipulating inverse square graph into straight line graph
1. The problem statement, all variables and given/known data
In the course of performing several trials I determined that if you have 2 charged metal spheres (with the same amount of charge on them) and they are both positively charged, then the force of repulsion between them is changed by 1/4 each time you double the distance between them. This experiment no doubt sounds very familiar as it has been done many times before. I collected my data (16N@1.0cm, 4N@2.0cm, 1N@4.0cm, and 0.25N@8.0cm) and graphed it and of course I end up with a classic looking descending, concave looking graph which is the inverse square graph, I think. 2. Relevant equations Now I have been puzzling over this for quite a while and even asked my boss and some coworkers to help and even together we can not figure it out  the question is: "manipulate the data so that the graph is a straight line graph" 3. The attempt at a solution I have tried doubling the charge, halving the distance, square rooting the distance, etc. It always seems to end up with the same shape. I think I might be missing the fundamental lesson here. Hopefully someone here can offer some insight. 
Re: manipulating inverse square graph into straight line graph
You have a relationship that is force = some_factor/distance^2
And a straight line is y=mx + c So you need to put it into this form. y (ie. force) = m x (ie. 1/r^2 ) 
Re: manipulating inverse square graph into straight line graph
Quote:
I have been out of school for several years and am taking a course to upgrade and the formula for the graph of a line is something I have completely forgotten! So I need to manipulate my F=kq/r^2 into a form that matches y=mx + c? I will make an attempt to do so. 
Re: manipulating inverse square graph into straight line graph
Quote:

Re: manipulating inverse square graph into straight line graph
Quote:
If i used y=mx or in this case F=m(1/r^2) I am going to get the same answers am I not? I can't see the difference between F=m/r^2 and F=m(1/r^2) where m is equal to kq. 
All times are GMT 5. The time now is 04:29 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums