Download Finding the Dielectric Constant of Paper and more Study Guides, Projects, Research Engineering Physics in PDF only on Docsity!
Purpose:
In this lab, our goal is to find the dielectric constant of paper. We’re going to conduct an experiment that will help us analyze our data and reach our purpose of finding the dielectric constant of paper.
Procedure:
- First of all, we completed our setup by constructing a parallel plate capacitor using two
sheets of aluminum foil which we connected to the digital multimeter (DMM). We made sure that the two aluminum foils have the same area through out the experiment. Then we placed the two sheets of aluminum in a textbook at different pages. We made sure to use only 20 pages between the two sheets.
- After that, because we’re relating our data to C = ε(A/d) , we wanted to collect two different
sets of data following this criteria: one of the components of the formula must be constant and the capacitance would vary as function of the other component which we will be changing while collecting our data. We will be measuring the capacitance using the digital multimeter (DMM) , which we turned its knob from the off position to the parallel-capacitor setting.
- So for the first set of data we wanted to keep the area constant throughout the experiment
and then make the capacitance change as function of the distance, which is the distance between the two foil sheets. We started this set of data by first measuring the area of the two identical foil sheets using a ruler and this formula: Area= length x width= 0.134 x 0.13= 0.01742 m^2. After that we started measuring the capacitance while varying the distance. To make that possible, we first needed to know the distance of the thickness of each paper. What we did is that we measured the distance of the thickness of the 20 pages that we’re
starting off with. Using a caliper, we measures that distance which turned out to be: distance of 20 pages= 0.1cm= 0.001m. To get the distance of a page: 20 pages ———> 0.001 m 1 page ————>? We get (1)(0.001)/(20)= 5*10^-5 m for each page We made sure that one of the group members would sit on the textbook so that we would avoid any air pockets that would affect the value of the capacitance because the air between the capacitor can decrease the value of the dielectric constant. Starting off with 20 pages, we started measuring the capacitance. With every data we take We would decrease the pages by 2. So we had to calculate the distance every time and then measure the capacitance again. We did 8 trials, which I’ll show the data for in the data and graphs section of the lab report.
- Now that we’re done with the first set of data, we have to do the other set which is the one
where we left the distance constant and then varied the area and recorded the capacitance. So for the distance between the foil sheets we used 10 pages this time. That means d= 10 x 5 x 10^-5 = 5x10^-4 m. Now for the area, each time we would record the capacitance we would just fold the foil and measure the sides using a ruler to calculate the area which is A(m^2)=Length x width. So we started measuring the area and recording the capacitance until we finished 8 trials, I’ll include them in the data and graphs section of the lab report.
Lab 4 Finding the Dielectric Constant Lynn Said
Calculations:
To solve for the dielectric constant, first the capacitance was plotted as a function of area ( C vs. A). The plot revealed the following equation: y= (510^-8)x + 110^-10 that we can compare to the following formula: C = (εkA)/(d). I just wanted to also note that (110^-10) doesn’t actually belong in the comparison and it should be zero, but due to human error it turned out to be that way. Since the area is along the x-axis we can substitute it by x, so that will show us that (εk)/(d) is equal to the slope. Since they are equal, we can solve for k as the following: C = (εkA)/(d) C = (εkx)/(d) (because A=x) By comparison: (εk)/d = slope (εk)/d =510^- A(m^2) C(F) d(m)=5(10^-4) 0.017 9.51E- 0.016 8.45E- 0.014 8.38E- 0.012 7.44E- 0.0095 5.86E- 0.0072 4.85E- 0.0058 3.39E- 0.0036 2.83E- 0 1 2 3 4 5 6 7 8 9 1/d (m) y = 5E-08x + 1E- 10 R² = 0. 0 1E- 2E- 3E- 4E- 5E- 6E- 7E- 8E- 9E- 1E- 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0. C (F) A (m^2) C vs A
k= (510^-4)(510^-8)/8.85*10^- k=2. If we also take k= d(C)/εA and use it for the “C vs d” graph for each point we have, we get an average of k=3. And then if we do the same thing with the “C vs A” graph and use it for each point and then calculate the average, we get an average of k=3. So the average for both would be k=3.
Conclusion:
The dielectric constant we got using slope is k= 2.82 and the average dielectric constant using the average k for each graph is k=3.56 which is so close to the dielectric constant that I found by searching the internet, which is k= 3.6 and it was stated that the dielectric constant for paper varies between 2 and 4 which makes the dielectric constant we got valid. So I concluded that using the average k for all the points we got for each graph and then calculating the average for both of the dielectric constants gave a more accurate result and it is so close to k=3.6 that I got from the internet. We can say that we met our purpose of this lab, which is finding the dielectric constant of paper.
