Three Plate Capacitor and Force

In summary, the conversation discusses a problem involving a plate suspended between two other plates. The goal is to determine the force acting on the hanging plate given the distance between the plates, length of the plates, and the height of the inner dangling plate. The solution involves treating the setup as two capacitors in parallel and using the equation U=(1/2)CV^2 to find the total energy. By differentiating the energy, the force can be calculated to be -bV^2/(4*pi*s). The conversation also touches on determining the field between the plates using the charge density and potential difference. A picture of the setup is provided and a hint is given to use virtual work to solve the problem.
  • #1
schaefera
208
0

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?
 
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  • #2
schaefera said:
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?
 
Last edited:
  • #3
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
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].
 

Attachments

  • Prob 3-18 Purcell.jpg
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  • #5
Oh, how a picture is worth a thousand words!

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

Related to Three Plate Capacitor and Force

1. What is a three plate capacitor and how does it work?

A three plate capacitor is a type of capacitor that consists of three parallel plates, with the middle plate being connected to a voltage source. The two outer plates are connected to each other and create an electric field between them. The middle plate is able to store charges, and the electric field created by the two outer plates helps to increase the capacitance of the capacitor.

2. What is the capacitance of a three plate capacitor?

The capacitance of a three plate capacitor is determined by the distance between the plates, the surface area of the plates, and the dielectric constant of the material between the plates. It can be calculated using the formula C = (k * ε0 * A) / d, where k is the dielectric constant, ε0 is the permittivity of free space, A is the surface area of the plates, and d is the distance between the plates.

3. How does the force on a three plate capacitor change when the distance between the plates is altered?

The force on a three plate capacitor is directly proportional to the distance between the plates. This means that as the distance between the plates increases, the force decreases, and vice versa. This is because the electric field between the plates decreases as the distance increases, resulting in a weaker force.

4. Can a three plate capacitor store more charge than a two plate capacitor?

Yes, a three plate capacitor can store more charge than a two plate capacitor. This is because the additional middle plate in the three plate capacitor increases the surface area, which results in a higher capacitance. Therefore, the three plate capacitor is able to store more charge for a given voltage compared to a two plate capacitor.

5. What is the significance of the middle plate in a three plate capacitor?

The middle plate in a three plate capacitor is connected to a voltage source and is able to store charges. This helps to increase the capacitance of the capacitor and allows it to store more charge compared to a two plate capacitor. Additionally, the middle plate helps to create a more uniform electric field between the outer plates, resulting in a more stable and efficient capacitor.

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