Electrostatic speaker

In summary: V! So as d increases, the force on the charges diminishes.Even without the diaphragm, the two plates do experience a force. One nifty way to show this is via the principle of virtual work:
  • #1
Number2Pencil
208
1

Homework Statement



An electrostatic speaker is constructed using two conductive plates (stators) and an electrostatically charged diaphragm in the middle which vibrates. One of the conductors is grounded, and the other has an amplified voltage applied to it to drive the force.

A uniform charge density, ps (C/m^2) is maintained on the diaphragm. The stator conductors are separated by distance d.

Find the symbol equation for Pressure, P, as a function of Voltage, V(t). Use only the symbols Q, ps, A, E, V (where V=V(t)), d, and F.

Homework Equations



[tex]P=\frac{F}{A}[/tex]

[tex]F=Q(E+v \times B)[/tex]

[tex]Q=\rho_s A[/tex]

[tex]V = -\int{E \cdot dl}[/tex]



The Attempt at a Solution



I am having a hard time conceptualizing how to find voltage or electric field. Here is what I've done:

------------------
METHOD 1:
------------------

Since there are no magnetic fields in this problem...

[tex]F=QE[/tex]

Since all electric fields are parallel (angle = 0)

[tex]V = -\int{Edl}[/tex]
[tex]V = -Ed[/tex]
[tex]E = -\frac{V}{d}[/tex]

So now I have Q, and I have E, let's solve for F:

[tex]F = - \rho_s A \frac{V}{d}[/tex]

Solving for pressure:

[tex]P = \frac{F}{A} = -\rho_s \frac{V}{d}[/tex]

But alas, it says this is incorrect. I am guessing that the E-Field at the plate is more complicated than just -V/d

-----------------------------
--METHOD 2: Point Charges
-----------------------------

So next I thought, "If the charges are uniformly distributed, maybe I can look at just single point charges on the
conductors."

Coulombs Law:

[tex]F = \frac{k Q_{plate} Q}{r^2}[/tex]

We can leave Q as Q since that is a symbol in the answer, but we can use Gauss' Law to come up with a relationship
between electric fields and charge:

[tex]Q = \epsilon_0 E A[/tex]

I took the liberty of getting the results from someone doing a parallel plate capacitor example.

Similarily:

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

Giving us:

[tex]Q_{plate} = -\frac{\epsilon_0 V A}{d}[/tex]

and we know that the distance between the positive plate and the diaphragm is d/2, putting all this together:

[tex]F = \frac{4 k \epsilon_0 V A Q}{d^3}[/tex]

Change to pressure by dividing by Area:

[tex]P = \frac{F}{A} = \frac{4 k \epsilon_0 V Q}{d^3}[/tex]

Since our gap medium is air, we know that k is:

[tex]k = \frac{1}{4 \pi \epsilon_0}[/tex]

Plugging and canceling out common terms, I wind up with this answer:

[tex]P = -\frac{VQ}{\pi d^3}[/tex]


But the answer guide said it was incorrect. At this point I will just admit that I am unclear on how to approach this
problem...can someone do some nudging or hand-holding and explain the flaws in my shoddy physics?
 
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  • #2
I see nothing wrong with your method 1. Maybe the minus sign?

Anyone else?
 
  • #3
A quick negative sign swap did not appease the homework software...
 
  • #4
Turns out my method 1 answer was correct, and it was simply the homework answer analysis-algorithm that was faulty.
Yay...

I guess now that I know the answer, I can explain this in (hopefully) understanding terms for my own sanity. There was a constant electric field established by the voltage conductor and ground conductor throughout the gap between them, the same way there is an electric field inside a capacitor. It is true that the charged diaphragm also contained charge, and thus electric fields, but this didn't "counteract" the other electric fields...they just do what electric fields do. If the conductors were not rigidly attached to a frame, they would also experience a force.

The equation F=QE talks of a force on a charge which is produced when that charge is exposed to an external electric field. Q was the charge of the Diaphragm, and E was the field it was exposed to (created by the two conductors).

Sound good?
 
  • #5
Number2Pencil said:
Turns out my method 1 answer was correct, and it was simply the homework answer analysis-algorithm that was faulty.
Yay...

I guess now that I know the answer, I can explain this in (hopefully) understanding terms for my own sanity. There was a constant electric field established by the voltage conductor and ground conductor throughout the gap between them, the same way there is an electric field inside a capacitor. It is true that the charged diaphragm also contained charge, and thus electric fields, but this didn't "counteract" the other electric fields...they just do what electric fields do. If the conductors were not rigidly attached to a frame, they would also experience a force.

The equation F=QE talks of a force on a charge which is produced when that charge is exposed to an external electric field. Q was the charge of the Diaphragm, and E was the field it was exposed to (created by the two conductors).

Sound good?

Sure does.

Even without the diaphragm, the two plates do experience a force. One nifty way to show this is via the principle of virtual work:

Consider 1 m2 of plates separated by d. The E field is V/d and the energy of this part of the field is εE2d/2 = εV2/2d. Now if d were increased a small amount δd, the new energy per m2 would be as above but with d → d + δd. The difference in energy under 1 m2 of plates would be ≈ -εV2/2 (δd/d2) = F δd, so F = - εV2/2d2 per sq. meter or call it pressure P.
 

1. What is an electrostatic speaker?

An electrostatic speaker is a type of loudspeaker that uses a thin, electrically charged diaphragm to produce sound. It works by converting electrical energy into sound waves, without the use of a traditional cone or coil.

2. How does an electrostatic speaker work?

An electrostatic speaker works by applying an electric charge to a thin, lightweight diaphragm, usually made of a conductive material such as aluminum or Mylar. The diaphragm is placed between two static plates, one positively charged and one negatively charged. As the electrical signal from the audio source is applied to the plates, the diaphragm is attracted and repelled by the plates, causing it to vibrate and produce sound waves.

3. What are the advantages of using an electrostatic speaker?

Electrostatic speakers are known for their high fidelity and accuracy in reproducing sound, as they have a flat frequency response and low distortion. They also have a very low mass, which allows for quick response time and detailed sound reproduction. Additionally, electrostatic speakers have a wide dispersion pattern, meaning they can fill a room with sound without the need for multiple speakers.

4. What are the limitations of electrostatic speakers?

One of the main limitations of electrostatic speakers is their size and cost. They are typically larger and more expensive than traditional cone speakers, making them less accessible to the average consumer. Additionally, electrostatic speakers require a high voltage power supply, which can be a safety concern for some. They also have a limited bass response compared to traditional speakers, although this can be improved with the use of a subwoofer.

5. How do I maintain and care for my electrostatic speaker?

To maintain your electrostatic speaker, it is important to keep it in a clean and dry environment. Dust and moisture can negatively impact the performance of the speaker. It is also recommended to periodically check the connections and wires to ensure they are secure. If your speaker has a removable cover, you can gently wipe the diaphragm with a dry, soft cloth to remove any dust or debris. If you notice any buzzing or distortion in the sound, it may be a sign that the diaphragm needs to be cleaned. It is also important to follow the manufacturer's instructions for powering on and off the speaker to prevent any damage to the delicate components.

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