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Electrostatic speaker

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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]



    3. 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?
     
  2. jcsd
  3. Mar 20, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    I see nothing wrong with your method 1. Maybe the minus sign?

    Anyone else?
     
  4. Mar 20, 2013 #3
    A quick negative sign swap did not appease the homework software...
     
  5. Mar 21, 2013 #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?
     
  6. Mar 21, 2013 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

    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.
     
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