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Measuring the electric field strength of a capacitor with and without a dielectric?

  1. Oct 22, 2005 #1
    I wanted to find the dielectric constant of different materials placed in a parallel plate capacitor using the equation:

    [tex]K = \frac{E_{dielectric}}{E_{original}}[/tex]

    Where [tex]E = \frac{V}{d}[/tex]

    I would use sensors connected to a computer to measure the voltage and i could easily measure the plate separation distance.

    Would the original electric field strength be measured using the maximum voltage reached when the capacitor is attached to the battery? Or would i have to remove the battery before measuring this voltage?

    When the dielectric is inserted I think that i have to remove the battery to measure the voltage difference, otherwise will the voltage will just be the same? If this is true will i get accurate readings? I mean the voltage won't decay unless i am actually discharging the capacitor right?

    I'm just not sure if this will work especially since i was thinking of making the capacitor myself :confused:
  2. jcsd
  3. Oct 22, 2005 #2


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    Most standard voltmeters have a 10 megahom impedance. This will discharge your capacitor with a time constant of (10 megahom * C), i.e if c is 1 microfard, you'll have a 10 second time constant, but if C is .001 microfarads, you'd only have a .1 second time constant.

    Constructing a much higher impedance voltemeter, often called an "electrometer" is a doable task, the difficulty of which depends on the degree of performance that is wanted.

    The wikipedia entry on electrometers has a couple of links to schematics of various degrees of sophistication - the "unity gain" buffer circut from


    is a simple op-amp design which could be used to drive a conventional voltmeter, and is rated as having over 10^13 ohm imput impedance (not sure of the leakage current spec on the unit).
    This design probably isn't optimized for your purpose, the low-pass filter section in particular may be unwanted, but it gives you a place to start.
  4. Oct 23, 2005 #3
    Thanks for the information pervect but that sounds too difficult, perhaps i could use the other equation:

    [tex]K = \frac{C_{dielectric}}{C_{original}}[/tex]

    Where [tex]C = \frac{Q}{V}[/tex]

    By using this equation i wouldn't have to dissconnect the battery which would prevent any problems with decay, i would just have to work out the the maximum charge the plates can carry without and with different dielectrics inserted in between the plates.

    My equipment would be set up like this: a battery connected to a current sensor in series, connected to a parallel plate capacitor that has a voltage sensor connected to it in parallel.

    The only problem would be that i wouldn't be able to measure the charge on the plates directly. However i know that [tex]Q = I \times t[/tex].

    So my first question is, if i recorded the current against time from the moment the battery was switched on to the moment the current reached zero (or close enough to it) could i then integrate the area beneath this exponential decay curve in order to find the charge?

    Could i then discharge the capacitor and repeat this method with the dielectric inserted to find the different maximum charge and thus the different capacitances?

    Also, how could i make a simple parallel plate capacitor? I've looked at some pictures of some and they look very precise so much so that they might be too difficult to make? What material should i use for the plates? Is there a specific conductor which is good to use? Would i minimise the "edge effect" by making the plates larger?

    Thanks in advance for more responses.

    <EDIT> I should really change the title of my thread now except i don't know how :P
    Last edited: Oct 23, 2005
  5. Oct 23, 2005 #4


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    Lots of approaches are possible. What you need to do is take stock of the equipment that you own, can buy, or can build, and evaluate if that equipment has the capability to do the experiments you want with the size of capacitance you want to measure.
  6. Oct 24, 2005 #5
    Point taken Pervect, all the equipment i need to try the method in my second post is available except...the parallel plate capacitor (i don't think my college has one in stock, although this is expecting the worst). I've e-mailed the university (just up the road from my college) asking if i could borrow their one (if they have one) because it'll save me building/buying one.

    It is the simplest type of capacitor apparently and looking at internet diagrams such as:


    It makes it look easy, but then when i look again at different images like:


    It makes it look a lot harder to build :uhh:

    Any answers to the questions in my second post? I need to know if my method is possible with my standard battery and computer sensors.
  7. Oct 24, 2005 #6


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    It seems a bit elaborate, but it could work, I = C dv/dt. To answer your question you need to know

    1) what sort of currents can your current sensor measure (magnitude - does it measure milliamps? microamps?)

    2) (if it's a computer) what sort of sampling rate does it have (i.e. what sort of time periods can you integrate over and how many samples per unit time do you get? If your capacitor charges in one cycle, you won't get a good integration

    3) About how big is your capacitor? This is a key issue.

    Actually one of the cheapest ideas that comes to my mind for a quick measurment would go something like this:

    you take a 555 timer chip, whose frequency depends on a RC time constant. You then measure the frequency of oscillation with the timer with the unknown capacitor replacing the timing capacitor.

    This may be good only for fairly large capacitances, I'm still not clear on what sort of range of capacitance you need to measure.

    If you have an AC voltmeter or your computer can be configured as one, and you have access to an AC signal source (if you have a transformer you'll always have access to a 60 hz signal), you could consider trying to build a bridge circuit.

    Basically you put your unknown capacitor in series with a known capacitor, and determine the voltage division ratio.

    The frequency will determine the impedance of the circuit, if you use a bridge circuit, that is another way of generating the equivalent of a very high impedance voltemeter.


    has some bridge circuits.

    Anyway, there are a whole bunch of ways that you could measure a capacitance, a lot depends on how big the capacitance is.

    IIRC bridge circuits were popular in the old days for precision measurments. Nowadays people usually just buy a capacitance meter :-)

    Last edited by a moderator: May 2, 2017
  8. Oct 24, 2005 #7
    Thanks for all that information Pervect, i've just been trying to gather the information for those questions here's what i've got:

    1) I'm fairly certain this is the sensor i'll be using:


    I looked at the manual for it and the accuracy is plus or minus 2 mili amps and the max sample rate is 1000 per second, the default is 10.

    2) I'm not sure about this, i'll probably have to gather more information on the computer but, i know it's fairly modern and the sensor works with a software package called DataStudio by PASCO, i downloaded a trial version of it last night and it has an inbuilt tool which finds the area beneath a curve.

    3) I haven't decided but i think a small capacitance would be better because i'd want to keep the voltage low so that the temperature is about constant. But would a larger capacitance mean i could measure the charge differences more easily? If so maybe i should make the capacitance larger by bringing the plates closer together. Well there's my thoughts, what do you think would be a good capacitance to use to find dielectric constants?

    (I just realised how i should of answered that question :P) I haven't decided what dielectrics i'm going to test which, decides the range of capacitances i'm going to measure, but they'll probably be materials i can easily get my hands on like plastic, wood, paper, glass, rubber...what would you suggest for testing? (that's cheap and easy to find prefferably :) )

    Also should the dielectrics i'm using be about the same dimensions? I think they should...but i'm not sure why...something to do with more volume or surafce area interfering more with the electric field?
    Last edited: Oct 24, 2005
  9. Oct 24, 2005 #8


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    Larger C is going to be better, especially if you get stuck with that particular measuring instrument.

    Here is an exercise: What is the largest capacitance you can experimentally study? Compute [itex]C = \epsilon_0 A / d[/itex] where A is the plate area (meter^2) and d is the dielectric thickness (meters) for the maximum feasable plate area and the thinnest feasibile dielectric.
  10. Oct 25, 2005 #9
    Paper is probably going to be the thinnest dielectric i'm going to test which i think has a thickness of roughly about 0.01cm = 1x10^-4m. As for plate area i probably won't be able to get plates bigger than about 20cm by 20cm (assuming they're square) so that's 400cm^2 = 4m^2. E0 is about 8.85x10^-12...so...

    Maximum C will be about...3.54x10^-7F , hmmm i think that's pretty small...
  11. Oct 25, 2005 #10


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    Smaller than that, .2m x .2m is only .04 m^2, so it's actually about 3*10-9 F, otherwise known as about 3000 pf (picofarads).

    The most senstive "integrate the current" arrangement I can think of with your probe would work something like this


    x is connected up to a switch to a known voltage source
    c is your capacitor under test
    P is your probe
    R is some resistor, or some current source.

    The goal of the arrangement is to charge the probe up with as little current as possible so as to get the longest integration time.

    Remember that your probe has 1 megohm imput impedance, so that it will draw 10 microamperes at 10 volts. Thus the current supplied by your resistor or current source must be at least 10 microamperes for the probe to reach 10 volts.

    Question: how long does it take 10 microamperes to charge up 3000 pf to 10 volts, if we ignore the current drain by the probe and the variation of the current with time if we use a resistor and not a current source

    I = C (dv/dt)

    If you work this out, I think you might start to see the appeal of using a very high impedance probe and a much lower current.

    You don't have to integrate the current vs time manually if you replace the resitor 'R' with a known test capacitor 'C' in the circuit above -but you do need a probe with a very high impedance to make this approach work, either a FET input op-amp or a commercial electrometer would do.

    You could also use a bridge circuit and an AC signal to get an effective high impedance probe out of a lower impedance one.
  12. Oct 30, 2005 #11
    Maybe you could wire the capacitor to an inductor of known inductance and measure the resonance frequency.

    Just a thought.
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