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Manipulating inverse square graph into straight line graph

  • Thread starter sync303
  • Start date
  • #1
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Homework Statement



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.


Homework Equations



Now I have been puzzling over this for quite a while and even asked my boss and some co-workers 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"


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.
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
12
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 )
 
  • #3
3
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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 )
Thank you very much - I knew it was something fundamental.

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.
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,774
12
So I need to manipulate my F=kq/r^2 into a form that matches y=mx + c?
Yes just plot force against 1/r^2
 
  • #5
3
0
Yes just plot force against 1/r^2
Ok, I am still struggling with this.

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.
 
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