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Three Plate Capacitor and Force

  • Thread starter schaefera
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Homework Statement


This is basically question 3.18 in Purcell. You have a plate being suspended into the middle of two other plates, and then you need to determine the force downward on the hanging plate given that the separation between it and either side of the other conductor is s; the plates have length b; the amount of the inside dangling plate between the other two is height y.


Homework Equations


F= (1/2)Q^2 * d/dx(1/C)


The Attempt at a Solution


I'm trying to say that this thing acts as two capacitors side by side, but only for the distance y by which it is dangling down. Since U= (1/2)CV^2, the total energy of the system is 2*(1/2)*C*V^2, and C=(by)/(4*pi*s). So I get the total energy as a function of capacitance; I differentiate the energy to get the force on it, and in the end I find that:

F= -bV^2/(4*pi*s).

Is this right?
Can the field between the plates be determined as 4*pi*sigma, or is it 2*pi*sigma since the center plate really has 2*pi*sigma amount of charge per area on each side?
 

Answers and Replies

  • #2
TSny
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F= -bV^2/(4*pi*s).

Is this right?
That's what I get also. Doesn't guarantee it's correct of course.
Can the field between the plates be determined as 4*pi*sigma, or is it 2*pi*sigma since the center plate really has 2*pi*sigma amount of charge per area on each side?
I'm not following the statement "...the center plate really has 2*pi*sigma amount of charge per area on each side". Suppose sigma is the charge density on each of the outer plates. What would be the charge density on each face of the inner (center) plate?
 
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  • #3
rude man
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You haven't given the potential difference between the two outer plates. You also nebulously describe the geometrical setup. If s is the disatance between the cente and either outer plate, what is y again??

It would be much preferable if you quoted the problem verbatim.
 
  • #4
TSny
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Here's the setup. As you can see, the two outer plates are actually part of one bent sheet. The problem refers to Eq. 27 of the text which is [itex]F = \frac{Q^2}{2}\frac{d}{dx}(\frac{1}{C})[/itex].
 

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  • #5
rude man
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Oh, how a picture is worth a thousand words!

Hint: invoke virtual work! i.e. F*Δy = Δenergy in the E fields.
 
  • #6
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I did exactly that. It would appear we already found the solution, but thanks!
 

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