Capacitor - Least squares fitting

[Woozah]

Homework Statement

We had a laboration for calculating ε_r in a parallel plate capacitor which we stuffed with plastic plates. All data we picked up was the area A, the distance d (and thus 1/d) and the capacitance C. We are now supposed to use the least squares-method to find ε_r, something we have never done before. She sent us a pdf regarding the least-squares but I am having it extremt hard understanding how to use it.

In an example (an example, so we can check our MATLAB code before using our own data) she has put d=2mm,4mm,6mm,8mm,10mm and C=353pF, 197pF, 141pF, 112pF, 97pF. She now states that using a linear regression you should get k=644pFmm and m=32pF.

Homework Equations

C=ε₀ε_rA/d

The Attempt at a Solution

What I don't understand is that if we want to try and find a linear fitting, what will i put as (k) (x) and m?
Since C=ε₀ε_rA/d, I am assuming that my x will be 1/d and k is ε₀ε_r, or am I wrong? No idea how to do this. They had forgotten that we were in the class and had not taken the course in which you learn this.

 

[TSny]

Welcome to PF!

Yes, your x values will be 1/d. Your y values will be C. So, if you plotted your data with x-axis horizontal and y-axis vertical, your data points would be of the form (x, y) = (1/d, C). The least-squares fit will produce the best straight line that fits the data. This line will have a slope k and a y-intercept m. (Apparently your class is using "k" for the slope and "m" for the y-intercept. In algebra classes, you usually use "m" for the slope of a line. Oh, well.)

According to your equation C = ε₀ε_rA/d, what should be the "theoretical" values for k and m? Hint: Replace C by y and 1/d by x.