Work required to move capacitor plates

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 5K views
Frostfire
Messages
50
Reaction score
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?
 
Physics news on Phys.org
e6909617aaece890a9d11784ece37423.png


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