- #1
Pushoam
- 962
- 52
Homework Statement
Use “virtual work” to calculate the attractive force between conductors in the parallel plate capacitor (area A, separation z). That is, use conservation of energy to determine how much work must be done to move one plate by an infinitesimal amount, and then use the value of the work to determine the force. Do your virtual work computations in two ways:
(a)
keeping fixed the charges on the plates, and,
(b)
keeping a fixed the voltage between the plates.
Homework Equations
Wby me=P.E.=(1/2)(Aε0/z)V2=(Q2/2Aε0)z
dW= F.dz
The Attempt at a Solution
Let's consider the case when the negative plate is up and the positive plate is down along the z- axis.
The electrostatic force acting on the negative plate is along the -ve z-axis.
To move negative plate along +ve z-axis by an infinitesimal amount,
(a)
keeping fixed the charges on the plates,
Wby me = (Q2/2Aε0)z
dWby me = (Q2/2Aε0)dz, (1)
dW= F.dz, (2)
From (1) and (2),
Fby me=(Q2/2Aε0) along +ve z-axis
Fattractive= - Fby me=(Q2/2Aε0) along -ve z-axis
where dWby me is work done by me in moving negative plate along +ve z-axis by an infinitesimal amount dz,
Fby me is forced applied by me on the system
Fattractive is the attractive force between conductors in the parallel plate capacitor
(b)
keeping a fixed the voltage between the plates
Wby me= (1/2)(Aε0/z)V2
dWby me = (-1/2)(Aε0)(V/z)2dz, (3)
dW= F.dz, (4)
From (1) and (2),
Fby me= (1/2)(Aε0)(V/z)2 along negative z-axis
Fattractive= - Fby me=
(1/2)(Aε0)(V/z)2 along +ve z-axis
While I think that should be Fby me in positive z-axis ( as I am pulling the negative plate upward , so both Fby me and the displacement dz is in the upward direction and hence dWby me in (3) should be +ve).
I can't understand physically how dWby me in (3) can be negative?
Where am I wrong?