Physics Experiment Help - Capacitors

In summary, the conversation discusses a physics experiment about parallel plates capacitors, where two round parallel plates with a radius of 0.0999 m are charged and discharged at a frequency of 20Khz. The capacitance is calculated using the equation C = \frac{\epsilon A}{d} and a graph is created comparing the theoretical capacitance with the measured capacitance. The conversation also touches upon the effects of bias error, reactance, and inductive loading on the results, as well as the potential impact of using round plates on fringing. The conversation ends with further suggestions for improvement and clarification.
  • #1
erik006
5
0
Note: This is a question about a physics experiment. As such, it does not completely fit into the provided template, but I'll do my best to stick to it anyway.

Homework Statement


The problem is about an experiment about parallel plates capacitors. The experimental setup is the following:

  1. Two round parallel plates with a radius = 0.0999 m.
  2. a constant current provided with magnitude 30uA
  3. The plates are charged and discharged at a circuit frequency of 20Khz using a solid-state switch
  4. -The resultant charging waveform is observed on an oscilloscope, and the potential difference between the bottom of the waveform and the top is recorded

The full description of the lab can be found here: http://epic.mcmaster.ca/%7Eglen/2a3/2a4_exp3.pdf, if you're interested.

I have several questions about this:
firstly, there is a derivation in the lab manual of the capacitance with respect to the rate of change of voltage with respect to time:

[tex]C = \frac{Q}{V}[/tex]
[tex]Q = C[/tex]
[tex]\Delta Q = C\Delta V[/tex]
[tex]\frac{\Delta Q}{\Delta T} = C * \frac{\Delta V}{\Delta T}[/tex]

but,

[tex]\frac{\Delta Q}{\Delta T} = I [/tex]

so we know that the V/T is inversely proportional to the capacitance.

Question 1: re-arranging the equations leads me to believe this constant of proportionality is I, is that correct?

I can use the following equation to calculate the theoretical capacitance of the plates:
[tex] C = \frac{\epsilon A}{d} [/tex]
Where A is the area of one of the plates (same as the other), [tex]\delta[/tex] is the permittivity of free space (~air), and d is the distance between the plates.

So I calculated the theoretical capacitance & the capacitance from my measurements. I then proceeded to graph them:

attachment.php?attachmentid=16044&stc=1&d=1224807694.jpg


Obviously there's a bias error that is affecting my results.
It seems to me that there are several things affecting these results, some of which I'm not sure what the exact effect of them is:
  1. The oscilliscope introduces resistive, capacitive and inductive loading
  2. the rest of the circuit has a small capacitance
  3. the frequency of the waveform (due to the current switching) leads to reactance (it will resist the change in voltage)?

Question 2: It is unclear to me how exactly reactance & inductive loading tie into this DC current, I've only learned about reactance, impedance, etc in an AC context, could someone enlighten me?

Question 3: Is my approach correct? Any further hints?

Kind regards,

Erik
 

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  • #2
Question 1: re-arranging the equations leads me to believe this constant of proportionality is I, is that correct?

Yes.

Question 2: It is unclear to me how exactly reactance & inductive loading tie into this DC current, I've only learned about reactance, impedance, etc in an AC context, could someone enlighten me?

It may not be necessary to give a detailed analysis of all this in your lab report. Look at your graph, how (in simple terms) do the two curves differ?

Question 3: Is my approach correct? Any further hints?

Your approach looks good. I suggest, however, that your graph's capacitance axis use either scientific notation or better yet unit prefixes (pF or nF). There are a lot of zeroes there that people must count to see just what the capacitance values are!
 
  • #3
Well, the only thing I can see is the bias error: the graph seems to be exactly the same, just shifted up by about 2nF. The circuit accounts for about 30pF of that (given). I figure the oscilloscope could add some to that (not quite sure how much).

In addition, I'm curious what the effect of round plates is on fringing. 2*pi*r < 8*r, so I'd assume less fringing?

Erik
 
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  • #4
I'm seeing a 0.2 nF or 200 pF shift between the graph's 2 curves, so you may want to double-check that number.

The scope capacitance should be printed next to the input jacks, though I see you used a voltage probe and not a direct connection to the scope input jack. The probe would have a capacitance too.

One thing I'm not sure of is how the stray inductance would affect the measurement.

In addition, I'm curious what the effect of round plates is on fringing. 2*pi*r < 8*r, so I'd assume less fringing?

Not sure what you're comparing here. Circular vs. square plates? A good number to look at is the ratio of perimeters for the two shapes, assuming equal areas (same theoretical capacitance) for the circular and square plates.

Good luck!

Regards,

Mark
 
  • #5
0.2 nF, yes. I made a mistake there.

In addition, I'm curious what the effect of round plates is on fringing. 2*pi*r < 8*r, so I'd assume less fringing?

Not sure what you're comparing here. Circular vs. square plates? A good number to look at is the ratio of perimeters for the two shapes, assuming equal areas (same theoretical capacitance) for the circular and square plates.

That's exactly what I was trying to compare. Your suggestion definitely makes a lot more though. I'll look into it! Thanks for helping me out!

Erik
 

Related to Physics Experiment Help - Capacitors

What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric. When a voltage is applied to a capacitor, one plate accumulates positive charge while the other accumulates negative charge. The capacitor can then release this stored energy when needed.

How do I calculate the capacitance of a capacitor?

The capacitance of a capacitor can be calculated using the formula C=Q/V, where C is capacitance in farads, Q is charge in coulombs, and V is voltage in volts. Alternatively, it can be calculated by multiplying the permittivity of the dielectric material by the surface area of the plates and dividing by the distance between them.

What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by three main factors: the area of the plates, the distance between the plates, and the type of dielectric material used. Capacitance increases with larger plate area, smaller plate distance, and higher permittivity of the dielectric material.

How can I use capacitors in a circuit?

Capacitors are commonly used in circuits for various purposes, such as energy storage, filtering, and timing. They can be connected in series or parallel to achieve different levels of capacitance. They can also be used in conjunction with resistors and other components to create different types of filters or oscillators.

What are some common applications of capacitors?

Capacitors have a wide range of applications in various electronic devices, including power supplies, audio equipment, and computers. They are also used in energy storage systems, motors, and sensors. In addition, capacitors are used in many industrial and medical applications, such as lasers, X-ray machines, and defibrillators.

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