Parallel Plate Capacitor experiment
Analysis portions by Dr. R. Fisher, ISM
October 2015
For your Data and Analysis section. I have written this in the form and with the detail that I expect when you write.
Obviously, some of this is more complicated than we’ve covered, but I want you to learn how to handle complicated
concepts in labs and projects. This is my data, and you are allowed to use it, or to use your own from your experiment.
This section does not include anything on the Leyden Jar portion.
Data from part 1, fixed V with C variable, is shown in Table 1. The potential across the capacitor was held constant, so of
course the voltage measured across the capacitor is constant.
Table 1. Measurement of voltage across the capacitor as capacitor separation is varied.
d mm
V volts
1
2
3
4
5
6
7
8
9
29.5
29.5
29.5
29.5
29.5
29.5
29.5
29.5
29.5
10
29.5
In part 2, the data were more interesting. See Table 2, where the raw data of capacitor voltage is show as a function of
plate separation.
To analyze this data, we need to recognize that the parallel plate capacitor is not the only capacitance in the system.
The power supply and the wires also can act with capacitance, such that the total capacitance is measured by the total
voltage across the system (same as voltage across the capacitor)
Q/Vtotal = Ctotal = Cpp + Cwire leads + Celectrometer = 0 A/d + Csystem
equation 1
This can be rearranged to
1/V = 0 A/Q (1/d) + Csystem/Q
equation 2
Thus a plot of 1/V vs 1/d should be linear with slope 0 A/Q and y-intercept of Csystem/Q. The appropriate columns are
shown in Table 2 and this is plotted in Figure 1.
Further, we know that
Q = Ctotal * V,
equation 3
where V is also measured across the whole system, not just the capacitor in isolation. So a plot of V vs. Q should give a
line of slope 1/Ctotal.
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Table 2: voltage data (d and V) parallel plate capacitor with fixed q and variable C for two replicate runs.
1/d
(1/mm)
d (mm)
1
0.5
0.333333
0.25
0.2
0.166667
0.142857
0.125
0.111111
0.1
0.083333
0.066667
0.05
0.033333
1
2
3
4
5
6
7
8
9
10
12
15
20
30
V (volts)
run 1
29.5
42
55
62
67
71
75
77
80
82
84
86
87
88
V (volts)
run 2
30
50
60
66
70
72
75
77
79
80
81
83
85
86
1/V
run 1
0.033898
0.02381
0.018182
0.016129
0.014925
0.014085
0.013333
0.012987
0.0125
0.012195
0.011905
0.011628
0.011494
0.011364
1/v
run 2
0.033333
0.02
0.016667
0.015152
0.014286
0.013889
0.013333
0.012987
0.012658
0.0125
0.012346
0.012048
0.011765
0.011628
Figure 1: plot of 1/V vs 1/d for fixed charge
0.04
0.035
y = 0.0244x + 0.0101
R² = 0.9936
1/V (1/volts)
0.03
0.025
1/V run 1
0.02
1/v run 2
0.015
Linear (1/V run 1)
y = 0.0221x + 0.0102
R² = 0.9886
0.01
Linear (1/v run 2)
0.005
0
0
0.2
0.4
0.6
0.8
1
1.2
1/d (1/mm)
From this plot we can see that the expected behavior is displayed, giving an average slope of 0.0233 mm/V and an
average intercept of 0.0101 volts-1.
From part 3, we collected the data in Table 3, where the aliquots have been converted to charge. (Note that in my data
the aliquot carries different Q from your data. )
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Table 3. Data of voltage for fixed separation of 15mm and variable charge
aliquots
0
1
charge (nC)
0.00
1.65
V volts
14
25
2
3.30
35
3
4
5
6
7
8
9
4.95
6.60
8.25
9.90
11.55
13.20
14.85
45
55
65
75
84
93
102
This data is plotted in Figure 2, where the data are clearly linear, and the slope is 7.916 V/nC
capacitor (and system) potential (volts)
figure 2: plot of V vs. Q for fixed d = 15mm
120
y = 7.9156x + 15.309
R² = 0.9992
100
80
60
40
20
0
0.00
2.00
4.00
6.00
8.00
10.00
12.00
charge on the capacitor (nC)
We know from equation 3 that the slope in figure 2 is 1/C, so from that slope I calculate
Ctotal = 1.263*10-10 F.
Further, from equation 2 and the line fit, I know in figure 1 that the slope = 0 A/Q = 0.0233 mm/V = 0.0000233 m/V. For
this capacitor, A = 0.0962m2, and using e0 = 8.85*10-12 C2/N*m2, we can calculate Q = 3.65x10-8 C.
Then from figure 1 and the intercept of 0.0101 volts-1 which equals Csystem/Q, and using the Q above, we can calculate
Csystem = 3.69x10-10 F
And by difference, then Cpp = Ctotal – Csystem = -2.42x10-10 F. This is not physically possible, being negative, so we have an
error in the data or calculations. I cannot find that error yet. I am confident the model I used is accurate but I may have
made some mistake.
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