Work required to move capacitor plates

In summary: V is the voltage across the capacitor, not the battery.In summary, the problem involves determining the work required to move the plates of an ideal parallel plate capacitor from a separation distance of D to an integer multiple of D, assuming only one plate moves. The equation F = q^2 / (2 e_0 A) for the field strength between the plates can be rearranged to U = Qdq / e_0 A, where U is the energy stored in the capacitor. The integral approach can be used to solve this problem, but it may be simpler to use the equation for energy stored in a capacitor, (CV^2)/2. It is important to note that the potential energy of the system is (CV^2)/
  • #1
Frostfire
50
0

Homework Statement



Determine the work required to move the plates of an ideal parallel plate capacitor from a separation distance of D to an integer multiple of D, (assume only one moves for simplicity). The each plate has an area A.

Homework Equations


F = q^2 /( 2 e_0 A) field strength between the plates
[tex]\Delta[/tex] V = U/q
[tex]\Delta[/tex]U = -W
[tex]\Delta[/tex]V= Qd/e_0 A



The Attempt at a Solution



moving the terms above around I came up with U = Qdq/e_0A
I know I need an integral here, probably evaluated between D and multiple of D(its a general solution that's needed)
Am I on the right track?
 
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  • #2
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Though it's a little cheap and probably not what your professor had in mind, you can use the equation for energy density(J/m^2) stored in an electric field to solve this problem. Do final energy stored minus initial energy stored to calculate what energy was added (or done) to the system.
 
  • #3
I appreciate the reply. I thought about that, but I use that in another part of the problem, not that I can't use it again since its different variables, but I am fairly certain I'll be graded on the diversity of methods used.

Anyone know if that integral approach would work or am I grasping at fog?
 
  • #4
Why not use the appropriate equation for energy stored in a capacitor?
 
  • #5
well since you know the force between the two plates

Work is

W=Integral (Fdr)
 
  • #6
Thats what I thought but I encounter a problem, when I compare the integral result to one obtained by use of potential energy, I am getting a different result depending on different terms. Considering that the work should be the same, I am stuck. Is there anyway Q^2= Vbat?
 
  • #7
In the question has the capacitor been charged and disconnected from the battery?If so Q remains constant and V changes if the plate separation is changed.If the capacitor remains connected to the battery V remains constant and Q changes.
 
  • #8
Also remember the potential energy of a system is (CV^2)/2
 

1. What is the definition of work required to move capacitor plates?

The work required to move capacitor plates is the amount of energy needed to change the position of the plates in a capacitor. It is a measure of the force and distance involved in moving the plates.

2. How is the work required to move capacitor plates calculated?

The work required to move capacitor plates is calculated using the formula W = 0.5 * C * V^2, where W is the work, C is the capacitance of the capacitor, and V is the potential difference between the plates.

3. What factors affect the work required to move capacitor plates?

The work required to move capacitor plates is affected by the capacitance of the capacitor, the potential difference between the plates, and the distance the plates are moved.

4. Why is the work required to move capacitor plates important?

The work required to move capacitor plates is important because it is a measure of the energy stored in a capacitor. It also determines the amount of energy needed to charge and discharge a capacitor.

5. How does the work required to move capacitor plates relate to the overall energy of a circuit?

The work required to move capacitor plates is a component of the overall energy of a circuit. It contributes to the total energy stored in the capacitor and can be converted into other forms of energy, such as electrical or thermal, when the capacitor is discharged.

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