Capacitance Lab Report


Capacitance is defined as the ratio of the Charge to the voltage. .  For this lab parallel plate devices are considered and the capacitance of such devices is calculated by the formula

Where k is the dielectric constant for the material between the plates, for a it’s 1, and is the Permittivity of the vacuum which is a constant, 8.854*10^-12 F/m.  The constants are accepted measure of the ability to accept an electric field of the material between the plates. It’s important to consider the material and the geometry of the device. In the formula A is the area of the plates and d is the distance between them. Traditionally your told d must be very small compared to the area and many calculations are done assuming the plates are infinite, but a lab is real life and we’re going to settle for a set of circular plates 18cm in diameter separated by distances varying from 1mm to 50mm. Another problem is our inability to measure the capacitance, the electrometer only measures voltage. We can substitute into and we have

Where, since for any given charge, everything is constant and V is proportional to d. There are some complicating factors when dealing with capacitors. While still considered passive, they have a transient response. When power is first applied to them they act as a wire and current flows, but of course a capacitor isn’t a wire and as charge flows into them they become charged at which point they behave as an open circuit. If the circuit only has a capacitor, this is almost instantaneous and the capacitor goes from a short circuit to an open circuit. If there is resistance in the circuit, this happens in a short time determined by the values of the resistance and capacitance of the system. The other complicating factor is how to measure it. To determine the capacitance you have to charge the capacitor, but the measuring device will affect the measurement by draining off charge. There are meters for such measurement and just as with resistors a network of capacitors can be combined into a single capacitor.

In a series circuit, the sum of the voltages must add up to the total voltage, by KVL and conservation of energy. The charge on each capacitor has to equal the total charge and you can use  to determine the formula for the total capacitance in a series circuit.

In a parallel circuit, the voltage across each capacitor has to be the same as the total voltage supplied and the sum of the individual charges must equal the total charge, and now you can use  to find that the total capacitance is the sum of the individual capacitances.

Objective of the experiment

  • To review the concept of Capacitance
  • To learn, to calculate the total / equivalent capacitance of series and parallel circuit

Experiment procedure

Procedure one

  • Watched the first part of the video.
  • Recorded the criteria used, charge, voltage, plate separation
  • Opened the capstone file and selected the page with the voltage vs separation data.
  • Measured as many points as possible, as described in the video, recording them in a table for d and voltage.
  • Plotted the data recorded with a scatter graph.
  • Added another column to the table and inserted the formula Vplates= (1 + 644 d) V Clicked in a cell, type ‘=’ and the formula, substituting the cell references for d and Vsystem which is the measured value.
  • With the adjusted value of V, plotted a scatter graph of V vs d, where d is the independent variable.

Procedure 2

  • Watched the video for procedure 2.
  • Opened the capstone file and selected page 2 displayed as kappa.
  • Measured the starting voltage and the voltage with the dielectric inserted.

Procedure 2.5

  • Watched the video for procedure 2.5
  • Built the circuit demonstrated in the lab. Used a 10V battery, a 3𝛀 resistor, a 0.2F capacitor.
  • Setup the voltage and current charts to record the changes in time and used the stopwatch to measure the time.
  • Took a screenshot for the report.
  • Replaced the battery with an AC current source with the default value.
  • Recorded the changes in the current and voltage over time.
  • Took another screenshot of the display and included it in the report.

Procedure 3

  • Watched the video for procedure 3 and recorded the measurements for the capacitors and the series circuit, as they’re made.
  • Prepared a table for the measurements of the capacitors in a series circuit.
  • Used C1, C2, C3, and C4
  • Used the formulas and the values measured for the individual capacitors calculated the total capacitance and compared it with the measured value using percent difference.

Procedure 4

  • Watched the video for procedure 4 and recorded the measurements for the parallel circuit, as they’re made.
  • Prepared a table for the measurements of the capacitors in parallel.
  • used C1, C2, C3, and C4
  • Used the formulas and the values measured for the individual capacitors and calculated the total capacitance and compared it with the measured value using percent difference.

Procedure 5

  • Watched the video for procedure 5 and recorded the measurements for the combination circuit, as they’re made.
  • Prepared a table for the measurements of the capacitors in the combination circuit, as shown. Omitted the power supply.
  • Inserted a copy of the circuit diagram into your report.
  • Used the formulas and the values measured for the individual capacitors and calculated the total capacitance and compared it with the measured value using percent difference.



Distance between plates (mm) Voltage in V Vplates=(1+644d)Vsystem
5 29.5 124.49
10 34.2 254.45
15 36.0 383.76
20 36.8 510.78
25 37.3 637.83
30 37.5 762
35 37.7 887.46
40 37.8 1011.53
45 37.9 1136.24
50 38.0 1261.60
55 38.0 1383.96
60 38.1 1510.28
65 38.2 1637.25
70 38.2 1760.26
80 38.2 2006.26


The graph given for the v plate is linear and the intercept is zero.

The calculated charge is 5.853*10^-12 coulombs.



Procedure 2

In the video the change in dielectric caused the voltage to drop. Is that correct? Yes it is Explain why? This is because of decrease in electrostatic field between the plates that causes the voltage to drop because of charge cancellation by the dielectric.

Precise measurements are difficult in these cases, but from the voltages recorded and the formulas in the introduction, calculate the dielectric constant for the paper used in the experiment. The plates are 18cm in diameter and the separation is 8cm.

Voltages recorded are 27.5 and 23.6



For alternating current;

Procedure 3

The capacitors values in series

C1 = 2.0 uf, C2= 3.1 uf, C­3=4.4 uf and C4 = 6.8 uf.


procedure 4

Capacitors in parallel



Procedure 5

Combination of both series and parallel




  • Look at the data collected in capstone for procedure 1. What did you calculate for charge? 3 *10^-12.
  • There is a clear trend as the plate separation increases. Usually d must be small compared to the area of the plates, but we didn’t do this in procedure 1. If the separation, d, were to grow very large compared to the plates, we would expect the charged plates to act like point charges YES.
  • Do you see this trend in the data? YES Explain your answer. Conservative coulombs forces are associated with configuration point charges and its influence by its relative position as well as the its own electric charge
  • In procedure 2 you calculated the dielectric constant. Is it a reasonable value as compared to other papers? NO
  • In procedure 2.5 In a DC circuit it acts like an open circuit, blocking the current, is that true in an AC circuit? NO
  • In the remaining procedures you calculate the equivalent capacitance of the various circuit types and compare them to measured values. Discuss how well they agreed or did not agree. The measure valued agreed with calculated value by a big percentage only differed on small scale percentage



  • We didn’t use power supplies in the circuits, but if we had included a battery,
  • In a series circuit,
    • Which capacitor would have the greatest charge? C1
    • Which capacitor would have the greatest potential difference? C4
    • As more capacitors are added does the total capacitance go up or down? Go down
  • In a parallel circuit?
    • Which capacitor would have the greatest charge? C4
    • Which capacitor would have the greatest potential difference? C1
    • As more capacitors are added does the total capacitance go up or down? Go up





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