1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Voltage caused by change in capacitor plate seperation

  1. Jun 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A parallel plate capacitor having two square plates each of side length 10mm has a plate separation of 100μm.

    It is used to make a microphone where the sound waves cause the plate separation to change by +/- 10μm.

    The microphone capacitor is connected in series with a constant 1 volt d.c. source of zero internal impedance and an amplifier with an input that can be regarded as a 10pF capacitor, the three components forming a complete circuit.

    What is the peak to peak input voltage of the amplifier caused by the plate movement of +/-10μm ?

    Assume the capacitor microphone has its gap filled with air.


    2. Relevant equations

    ε = 8.854 x 10^-12

    3. The attempt at a solution

    Having a bit of trouble. I'm not sure where exactly to start.

    Q = CV

    where C = epsilon x A/d

    A change in plate seperation cause a change in charge across the plates. But I'm looking for a change in voltage. Any ideas are appreciated.
     
    Last edited: Jun 8, 2010
  2. jcsd
  3. Jun 8, 2010 #2
    Okay well no-one else has responded, so I'll take a crack, but work with me and recheck everything until someone comes with an absolute answer.

    Here's a circuit diagram just for reference

    Circuit.jpg

    So, are we sure that Q changes as plate separation changes? Think about what causes charges on capacitors in a circuit

    What does 'V' represent?

    Have you done the proof for the 'effective capacitance' of capacitors in series? (Its not really relevant to the question per se, but understanding it would be a big help) If not then just disregard this
     
  4. Jun 8, 2010 #3

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    I'll give another response.

    Hello Satellitelove,

    Welcome to Physics Forums!

    Your first course of action is to calculate the capacitance of the microphone itself, at its peak displacements. In other words, you need to find the capacitance of the microphone at two conditions: one where the separation is at 100 + 10 μm, and another where the separation is at 100 - 10 μm.

    Secondly, you need to calculate the voltages across each of the components in the circuit (actually, you only really need the voltage across the 10 pF capacitor, but since all the voltages are related you might as well find them all). There are two concepts that will help you find this. You'll have to do everything below twice; once for each of your microphone capacitance values.

    (1) Kirchhoff's circuit laws. The sum of all the voltages around a loop add up to zero.

    (2) This second concept is a little more subtle. But I'm guessing you should assume the following: Before the components are connected together, the charge on each capacitor is zero. Which means the same thing as saying: Before the components are connected together, the voltage across the respective terminals of each capacitor is zero. Charge only builds up in the capacitors once the circuit is complete (including the 1 V DC source). Now here is the subtle part. Any current entering the microphone capacitor must also enter the amplifier capacitor. Since current is a measure of charge per unit time, it means that both capacitors have the same charge. If that doesn't make sense right away, think about it for awhile.

    Using the above concepts, you should be able to generate two equations. So you have 2 equations and 2 unknowns. You should be able to solve for the voltages. Perform the above twice. Once for each microphone capacitance value. :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook