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The Capacitor Lab Report

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Submitted By cheerup
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Phys261-203, the Capacitor
Author:.
09/24/2015
Instructor: Swabir Silayi
Abstract:
In this experiment, we are going to investigate the relationship between the distance between two parallel plates and the capacitance, and the how the material of the plates impact the capacitance of the capacitor. In the first part, we are going to set vary the distance between two plates at interval of 0.5cm from 0.25cm to 4 cm to get sufficient data. In the second part of this experiment, we are going to change the material between two parallel plates in order to change the dielectric constant.
Introduction:
1. We use a large parallel-plate capacitor in this experiment. It has two conductors and separated by an insulator. Because it stores capacitance, we call it capacitor. We connected one side of the capacitor with positive pole and the other side with negative pole. The amount of the charge on the capacitor is about the difference V and the magnitude of the capacitance on this capacitor: Q=CV 2. A simple parallel plate capacitor consists of two parallel conductors and is split by a distance. We also have a formula to descript the relationship between capacitance and some constants and variables.
C= κεA/d
Where κ is the dielectric constant, ε is the permittivity of free space (8.85*10^-12 C2/Nm2), and A is the cross sectional area of the parallel plates. Meanwhile, different sorts of material have different dielectric constant.
With the increasing of distance between two plates, the capacitance is going to decrease. In addition, different material of the plates will have different capacitances.
Material and Methods: 1. In the first part of the experiment, we will change the distance of the capacitor to investigate the relationship between capacitance and the distance. The capacitor is connected to a meter, one side with positive pole and the other side with negative pole.

We should check the all the equipment to ensure that the plates are parallel and the meter is zero at first, and then put the scale at pF. Next, it’s necessary to find the Co before the experiment so that we can get the more precise data. We need to separate the two parallel plates as far as we can but still on the set to determine the Co of the capacitor. Then, we need to measure the capacitance by varying the distance between two parallel plates from 0.25cm to 4 cm at interval of 0.5cm so that we can achieve sufficient data. Repeat the step at least 5 times. Moreover, we will use excel to make the graph of C vs. d. From the graph we can clearly see, it’s not linear. So we add the power trendline on the graph. Finally, use the meter stick to measure the diameter of the parallel plates so that we can calculate the cross sectional area of it. We do have the dielectric constant of air that is one, the permittivity of free space and the cross sectional area, so we can calculate the different capacitances with various distances. Compare the measured value with theoretical value and discuss.

2. In the second part of the experiment, we will vary the material of the plates into Plexiglas and rubber to find how materials influence the capacitance.
Change the material between the plates into Plexiglas or rubber respectively. We still need to change the distance from 0.5cm to 4cm at interval of 0.5cm in this part to get enough data. Finally, keep the distance in constant, and compare the data we got in this part to previous part. .

Results
In part one: Distance | STDDEV | AVG | TDC | DIFF | how large(%) | 0.25 | 2.086145 | 102.22 | 84.252 | 17.968 | 2.040838 | 0.75 | 0.965401 | 34.82 | 28.084 | 6.736 | 2.772549 | 1.25 | 0.905539 | 20.2 | 16.8504 | 3.3496 | 4.482864 | 1.75 | 0.544977 | 14.32 | 12.036 | 2.284 | 3.805706 | 2.25 | 0.2 | 10.8 | 9.36 | 1.44 | 1.851852 | 2.75 | 0.244949 | 8.7 | 7.66 | 1.04 | 2.815505 | 3.25 | 0.45607 | 6.74 | 6.48 | 0.26 | 6.76662 | 3.75 | 0.270185 | 5.84 | 5.6168 | 0.2232 | 4.626458 | 4 | 0.089443 | 5.36 | 5.26575 | 0.09425 | 1.668707 |

In part two: distance(cm) | c(pF)glass | C(air) pF | C glass/C air | c(pF)rubber | C rubber/ C air | 0.25 | 271 | 102.22 | 2.65114459 | 205 | 2.00547838 | 0.75 | 68.6 | 34.82 | 1.970132108 | 77.2 | 2.2171166 | 1.25 | 41.8 | 20.2 | 2.069306931 | 39.4 | 1.95049505 | 1.75 | 31.8 | 14.32 | 2.220670391 | 30.5 | 2.129888268 | 2.25 | 27.5 | 10.8 | 2.546296296 | 26.3 | 2.435185185 | 2.75 | 25.2 | 8.7 | 2.896551724 | 23.5 | 2.701149425 | 3.25 | 23.7 | 6.74 | 3.516320475 | 21.5 | 3.189910979 | 3.75 | 22.7 | 5.84 | 3.886986301 | 20.5 | 3.510273973 | 4 | 22.3 | 5.36 | 4.160447761 | 19.7 | 3.675373134 |

Discussion: 1. A. For the first part, with the increasing of distance between two parallel plates, the capacitance decreases. And we add a power trend line and formula on it. The index number of this formula should be around -1, and we got -1.068 that is close to -1. There exist some errors in this experiment. According to the standard deviation, we get least precise data at distance 0.25cm due to its hard to set the distance at 0.25cm precisely. Meanwhile, the significant random uncertainties is the data when the distance between two plates is 0.25cm. In addition, from the error bar on graph 1, we can see that the first data has the biggest error interval.
B. From the table 1, we can see that difference from measurement and theory at various distance. Apart from the data at 0.25cm, the others are almost accurate.

C. I think thank the equipment and procedures give us precise value but not really. For instance, the capacitor bar is not long enough, so we cannot get more exact Co. Moreover, it’s not precise to use the meter stick to measure the diameter of the plates

2. A. With the changing of material, the capacitance definitely changes which also verifies the hypothesis.
B. Air doesn’t significantly change the measured dielectric’ value due to its dielectric value is 1. While we change the material of the two parallel plates, we can directly use the ratio of the material and air to calculate the capacitance.
Reference: Young and Freedom, University Physics, 12th Edition, Chapter 24.
Appendices:
For part 1: Distance(cm) | C1(pF) | C2(pF) | C3(pF) | C4(pF) | C5(pF) | 0.25 | 100.2 | 104.1 | 102 | 100.2 | 104.6 | 0.75 | 35.4 | 33.8 | 33.8 | 35.2 | 35.9 | 1.25 | 20 | 19.6 | 19.8 | 21.8 | 19.8 | 1.75 | 13.6 | 14.1 | 14.2 | 14.7 | 15 | 2.25 | 10.7 | 10.7 | 10.6 | 10.9 | 11.1 | 2.75 | 8.6 | 8.6 | 8.4 | 8.9 | 9 | 3.25 | 6.7 | 7 | 6.8 | 7.2 | 6 | 3.75 | 5.7 | 5.6 | 5.8 | 6.3 | 5.8 | 4 | 5.3 | 5.4 | 5.3 | 5.5 | 5.3 |

These are the raw data we got in measurement. Distance | STDDEV | AVG | TDC | DIFF | how large | 0.25 | 2.086145 | 102.22 | 84.252 | 17.968 | 2.040838 | 0.75 | 0.965401 | 34.82 | 28.084 | 6.736 | 2.772549 | 1.25 | 0.905539 | 20.2 | 16.8504 | 3.3496 | 4.482864 | 1.75 | 0.544977 | 14.32 | 12.036 | 2.284 | 3.805706 | 2.25 | 0.2 | 10.8 | 9.36 | 1.44 | 1.851852 | 2.75 | 0.244949 | 8.7 | 7.66 | 1.04 | 2.815505 | 3.25 | 0.45607 | 6.74 | 6.48 | 0.26 | 6.76662 | 3.75 | 0.270185 | 5.84 | 5.6168 | 0.2232 | 4.626458 | 4 | 0.089443 | 5.36 | 5.26575 | 0.09425 | 1.668707 |
For Theoretical data (at d=0.25cm, air, A=0.0238m^2, ε=8.85*10^(-12)) C= κεA/d=1*8.85*10^(-12)* 0.0238/0.0025=84.252 pF
DIFF: (at d=0.25cm) DIFF=AVG-TDC=102.22-84.252=17.968 pF
How large (d=0.25) STDV/AVG= 2.086145/102.22
For part 2: distance(cm) | c(pF)glass | C(air) pF | C glass/C air | c(pF)rubber | C rubber/ C air | 0.25 | 271 | 102.22 | 2.65114459 | 205 | 2.00547838 | 0.75 | 68.6 | 34.82 | 1.970132108 | 77.2 | 2.2171166 | 1.25 | 41.8 | 20.2 | 2.069306931 | 39.4 | 1.95049505 | 1.75 | 31.8 | 14.32 | 2.220670391 | 30.5 | 2.129888268 | 2.25 | 27.5 | 10.8 | 2.546296296 | 26.3 | 2.435185185 | 2.75 | 25.2 | 8.7 | 2.896551724 | 23.5 | 2.701149425 | 3.25 | 23.7 | 6.74 | 3.516320475 | 21.5 | 3.189910979 | 3.75 | 22.7 | 5.84 | 3.886986301 | 20.5 | 3.510273973 | 4 | 22.3 | 5.36 | 4.160447761 | 19.7 | 3.675373134 |

Crubber/Cair (d=0.25 cm)
=205/102.22=2.00547838…...

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