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Homework Help: Describe how to show charge is proportional to pd on a capacitor

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Describe how you would show experimentally that the pd across a capacitor of fixed capacitance is proportional to the charge stored on the plates?

    2. Relevant equations

    3. The attempt at a solution
    So the most obvious thing would be to place a voltmeter in parallel to the capacitor

    To measure the current I could use an ammeter in series with the capacitor.

    I know that Q = It but surely the charge stored on a charged capacitor is fixed?
    Even If I had a variable resistor in the circuit the current in the circuit would change BUT SO WOULD THE potential difference.

    The MS states the following points but it confuses me slightly:
    - charge the capacitor through a large resistor at a constant rate
    - by using a variable resistor to gradually reduce the resistancd in the circuit
    - monitor the current with a micro-ammeter keeping the current constant
    - record pd with a voltemter across capacitor
    - pd should increase at a constant rate

    HOW CAN THE CURRENT STAY CONSTANT - V = IR BUT the power sources does not change so the only thing that changes with R is I :S

    please can someone explain to me where I am going wrong
  2. jcsd
  3. Mar 27, 2012 #2
    in fact would this be a sensible experiment:
    - charge a capacitor to a particular p.d and check with a voltmeter
    - using a coloumeter identify charge stored
    - repeat for many different p.d (use a range of power sources)
    - plot a graph
    (repeat with other capacitors)
  4. Mar 27, 2012 #3
    is the coloumeter is parallel or series with capacitor?
  5. Mar 27, 2012 #4
    I have not met a coulomb meter!!!!!
  6. Mar 27, 2012 #5
    well it is also called a voltameter
  7. Mar 27, 2012 #6
    the technique I know uses a vibrating (reed) switch run from a fixed frequency supply to charge and discharge a capacitor. The current measured can be used to find the charge on the capacitor.
    This is a standard textbook technique to determine capacitance/charge
  8. Mar 27, 2012 #7
    how does it measure the charge on the capacitor?
  9. Mar 27, 2012 #8
    This is the circuit arrangement. The voltage of the battery is known. The reed switch is operated by , maybe 100Hz AC. The reed switch alternately charges the capacitor from the battery and discharges the capacitor through a micro ammeter. The microammeter gives the average current flowing from the capacitor..i.e the charge per second from which the charge can be determined.
    For accuracy the time constant of the discharge circuit must be much less than the time period of the AC driving the reed switch.
    How does a vlotameter work?

    Attached Files:

    • cap.jpg
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  10. Mar 27, 2012 #9
    as I understand, a voltameter uses a capacitor of ENORMOUS capacitace (known) and is placed in parallel with the desired capacitor.
    The charge on the desired capacitor is shared equally according to the ratio of the capacitors so basically all charge goes to the known capacitor
    using Q = CV we can find Q as we can find pd across capacitor

    Although this does seem foolish as we are using the equation (Q=CV) for the equation we are trying to verify (Q=CV) :S

    Why is the reed switch NOT in the circuit?
  11. Mar 27, 2012 #10

    so would this be a viable expt: (i have never used reed switch - not on syllabus either)
    - monitor the discharge current with time simultaniously monitoring the discharge voltage (using a data logger for both)
    - find the area under the current curve to get charge
    - plot charge against voltage to give a straight line?
  12. Mar 27, 2012 #11
    I can see the logic behind your voltameter technique !!!
    The vibrating switch has its limitations because of time constant implications but then every measurement in physics has some limitations.
    The only other technique I know of for determining charge uses a "ballistic galvanometer"..... I leave you Js to find out more about this !!!!
    I do know that the vibrating switch is a good technique, having used it regularly.
    The purpose of the switch is only to connect the battery to the capacitor then the capacitor to the microammeter. It is not part of the "circuit"
  13. Mar 27, 2012 #12
    details of the technique:
    Voltage...this is no problem, it is the battery voltage. Lets say 9V
    As an example, if you have a reed switch vibrating at 100Hz and measure a steady current of 100microAmps then 1 micro coulomb is transferred each second.
    If you ignore time constant considerations this means that 9V charges a capacitor with 1 microCoulomb of charge
    So the capacitance must be C=Q/V = (1x10^-6)/9 = 1.1x10^-7F
    You can manipulate these figures to find any of the unknown quantities.
    These days reed switches are readily available and this is an easy experiment to do.
    If you need any more help with this technique let me know.

    PS a signal generator is needed to drive the reed switch at a known frequency
    Last edited: Mar 27, 2012
  14. Mar 27, 2012 #13
    it seems a ballistic galvonometer measures the charge DISCHARGED through it

    so if I know how voltage varies with time, I can find charge dishcarged with time and then plot a graph

    OR I could use the suggestion of monitoring change in current and integrating curve the nplotting resutls against voltage
  15. Mar 27, 2012 #14
    You are correct regarding the ballistic galvanometer. If you have access to one it is a possible technique.
    You need to obtain measurements of voltage applied (easy with a multimeter) and charge (several techniques now available to you!!!)
    Then plot a graph of volts against charge....should be a straight line
  16. Mar 27, 2012 #15
    yey..sorry to nag but was this method also correct:
    -monitor the discharge current with time simultaniously monitoring the discharge voltage (using a data logger for both)
    - find the area under the current curve to get charge
    - plot charge against voltage to give a straight line?
  17. Mar 27, 2012 #16
    That sounds OK but I think I would use the total area under the current against time graph to give the charge and use the steady value of the applied voltage.
    You have to be very careful analysing the information from graphs showing the voltage AND the current varying..... I would be inclined to keep one constant....the charging voltage each time
  18. Mar 27, 2012 #17
    I guess that makes sense
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