Voltage caused by change in capacitor plate seperation

In summary, the peak to peak input voltage of the amplifier will be equal to the capacitor's maximum potential difference multiplied by the capacitor's capacitance.
  • #1
satellitelove
1
0

Homework Statement



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.


Homework Equations



ε = 8.854 x 10^-12

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 separation cause a change in charge across the plates. But I'm looking for a change in voltage. Any ideas are appreciated.
 
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  • #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
 
  • #3
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:
 

1. What is the relationship between voltage and capacitor plate separation?

The voltage across a capacitor is directly proportional to the change in separation between its plates. This means that as the plate separation increases, the voltage also increases, and vice versa.

2. How does changing the capacitor plate separation affect the voltage?

Changing the plate separation of a capacitor changes the distance between the plates, which alters the electric field between them. This change in electric field results in a change in voltage across the capacitor.

3. Can changing the capacitor plate separation affect the capacitance?

Yes, changing the plate separation of a capacitor can affect its capacitance. The capacitance of a capacitor is directly proportional to the plate area and inversely proportional to the plate separation. So, as the plate separation increases, the capacitance decreases.

4. How does the voltage change if the capacitor plate separation is decreased?

If the capacitor plate separation is decreased, the electric field between the plates increases, resulting in an increase in voltage across the capacitor. This is because the electric field is directly proportional to the voltage.

5. Why does changing the capacitor plate separation cause a change in voltage?

Changing the capacitor plate separation alters the distance between the plates, which affects the amount of charge that can be stored on the plates. This change in charge results in a change in electric potential energy, which is measured as voltage.

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