Work required to move capacitor plates

AI Thread Summary
The discussion focuses on calculating the work required to move the plates of an ideal parallel plate capacitor from a distance D to a multiple of D. Participants explore various equations related to electric fields, potential energy, and force, emphasizing the need for an integral approach to find the work done. There is a debate on whether to use energy density or potential energy equations, with concerns about grading based on method diversity. Clarification is sought on whether the capacitor is charged and disconnected from the battery, as this affects the relationship between charge and voltage. The conversation highlights the complexity of the problem and the necessity for consistent results across different calculation methods.
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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
\Delta V = U/q
\DeltaU = -W
\DeltaV= 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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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.
 
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?
 
Why not use the appropriate equation for energy stored in a capacitor?
 
well since you know the force between the two plates

Work is

W=Integral (Fdr)
 
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?
 
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.
 
Also remember the potential energy of a system is (CV^2)/2
 
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