Calculating force between capactor plates using virtual work

[Pushoam]

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

W_{by 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,
W_{by me} = (Q²/2Aε₀)z
dW_{by me} = (Q²/2Aε₀)dz, (1)

dW= F.dz, (2)
From (1) and (2),
F_{by me}=(Q²/2Aε₀) along +ve z-axis
F_attractive= - F_{by me}=(Q²/2Aε₀) along -ve z-axis

where dW_{by me} is work done by me in moving negative plate along +ve z-axis by an infinitesimal amount dz,
F_{by me} is forced applied by me on the system
F_attractive is the attractive force between conductors in the parallel plate capacitor
(b)
keeping a fixed the voltage between the plates
W_{by me}= (1/2)(Aε₀/z)V²
dW_{by me} = (-1/2)(Aε₀)(V/z)²dz, (3)
dW= F.dz, (4)
From (1) and (2),
F_{by me}= (1/2)(Aε₀)(V/z)² along negative z-axis
F_attractive= - F_{by me}=
(1/2)(Aε₀)(V/z)² along +ve z-axis

While I think that should be F_{by me} in positive z-axis ( as I am pulling the negative plate upward , so both F_{by me} and the displacement dz is in the upward direction and hence dW_{by me} in (3) should be +ve).
I can't understand physically how dW_{by me} in (3) can be negative?
Where am I wrong?

 

[rude man]

I can't understand physically how dW_{by me} in (3) can be negative?
Where am I wrong?

The case of fixed V is complicated by the fact that some of the energy is supplied by or given to the source of emf (the battery for example).
So the energy balance is work done by you + energy supplied by battery = change in E field energy

 

[Pushoam]

The case of fixed V is complicated by the fact that some of the energy is supplied by or given to the source of emf (the battery for example).
So the energy balance is work done by you + energy supplied by battery = change in E field energy

Thank you for the reply.
By definition of work,
dW_{by me}= F_{by me}.dz
Here, both F_{by me} and dz are in the +ve z direction, so dW_{by me} is supposed to be +ve.
What is wrong with this argument?

 

[rude man]

Thank you for the reply.
By definition of work,
dW_{by me}= F_{by me}.dz
Here, both F_{by me} and dz are in the +ve z direction, so dW_{by me} is supposed to be +ve.
What is wrong with this argument?

Nothing! If you do it right and take the battery into account, you will find that you did do positive work. The field energy went negative but the battery got charged by twice that amount in magnitude, so bottom line you charged up the battery by the work you did, with the (loss in) field energy contributing an equal amount.

 

[Pushoam]

Nothing! If you do it right and take the battery into account, you will find that you did do positive work. The field energy went negative but the battery got charged by twice that amount in magnitude, so bottom line you charged up the battery by the work you did, with the (loss in) field energy contributing an equal amount.

After moving negative plate along +ve z-axis by an infinitesimal amount dz,keeping V constant,
the charge on the capacitor has got decreased to Q' = (Aε₀/(z+dz))V i.e. Q'<Q,
So, moving negative plate along +ve z-axis by an infinitesimal amount dz, keeping V constant, discharges the capacitor.
Hence, the potential energy of the capacitor gets decreased.
Now,according to you this loss in P.E. and work done by me goes to the battery
and as a consequence, the battery gets charged up.
What is meant by charging up the battery , here, the battery is still supplying the same potential difference V?
Experimentally, how will I verify that the battery has got charged or not?
Does it mean that lifetime of the battery has got increased?
How do you know that the battery got charged by twice (how twice?) that amount (which amount?)in magnitude?If I am right, then, W_{by me} is not equal to loss in P.E..
W_{by me} = (total energy given to the battery - loss in P.E)
How to calculate total energy given to the battery?
How to solve the second part of the question?

 

[rude man]

Experimentally, how will I verify that the battery has got charged or not?

Why "experimentally"? You are supposed to be analyzing on paper.

Does it mean that lifetime of the battery has got increased?

the battery lifetime is not an issue. And it doesn't have to be a battery; it could be a rectified ac generator output or many other kinds of dc emf sources.

How do you know that the battery got charged by twice (how twice?) that amount (which amount?)in magnitude?

If I force charge Q into a source of emf, by how much have I increased its energy? Use this answer and the energy balance equation I gave you before. That is all you need.

If I am right, then, W_{by me} is not equal to loss in P.E..
W_{by me} = (total energy given to the battery - loss in P.E)

this is correct.

How to calculate total energy given to the battery?
How to solve the second part of the question?

see above

 

[Pushoam]

Thank you, Thanks a lot.

For calculating energy given to the battery :
The charge (Q-Q') goes to the positive terminal of the battery at constant potential V and -(Q-Q') goes to the negative terminal of the battery at constant potential 0V. Let's assume that inside the battery the negative charge remains at the negative terminal and the positive charge moves to the negative terminal to make it neutral and thus to maintain the constant potential difference between the two terminals. In doing so, the work done by me in this case is (-V)(Q-Q') = the energy given to the battery = E_batt. I am getting E_batt negative here. Where am I wrong?

But battery's mechanism ,in general , is such that it collects positive charge at negative terminal and send it to the positive terminal hence work done on transporting a positive charge Q by the battery is positive i.e.QV.
I think ,here in this case, it is work done by the battery which is negative, i.e. (-V)(Q-Q') is work done by the battery and work done by me = - (work done by the battery ) = V(Q-Q') = E_batt
So, assuming that total energy given to the battery is positive,
total energy given to the battery = V(Q-Q') =dW_{by me} + loss in P.E. of the capacitor,
where loss in P.E. of the capacitor = -(P.E._final - P.E._initial) = (V/2)(Q-Q')
dW_{by me} = total energy given to the battery - (loss in P.E. of the capacitor)
= V(Q-Q') - (V/2)(Q-Q')
= (V/2) [ (Aε₀/(z))V -(Aε₀/(z+dz))V]
=(V²/2)Aε₀[(1/z)-(1/z+dz)]
=(V²/2)Aε₀[(1/z²)dz]
So, F_{by me} = (V²/2)Aε₀[(1/z²) along positive z-axisIs this o.k.?

Now, work done by attractive electrostatic force is = dW_elec = F_elecdz,
where F_elec <0, dz>0
(P.E._final - P.E._initial) = - dW_elec
(V/2)(Q-Q') = - dW_elec = - F_elecdz,
F_elec = (V²/2)Aε₀[(1/z²) along negative z-axis on negative plate
Hence, F_elec = - F_{by me}

 

[rude man]

You are getting closer but not there yet.
I move a charge Q from a potential of zero (the battery negative terminal) to +V (the battery positive terminal). What is the work done by me?
You're worrying too much time about battery details instead of just sticking to fundamentals. What's important is potential difference, not battery "innards". The voltage source could just as easily have been a large capacitor, large in comparison to the capacitance of your plates, such that effectively for that capacitor, dV = 0 for adding or subtracting charge dQ.
Keep it up, you're making good progress.

 

[Pushoam]

I move a charge Q from a potential of zero (the battery negative terminal) to +V (the battery positive terminal). What is the work done by me?

The work done by you is QV.
But in this problem,(Q-Q') comes to the positive terminal from the positive plate and reaches to the negative plate via negative terminal and thus the work done by me is negative i.e.-(Q-Q') V.
Is this correct?

 

[rude man]

[QUOTE="Pushoam, post: 5771099, member: 619344[/quote]
OK, how about this: you know that battery energy W is reduced if I shunt the battery with a resistor R. Define positive current as current entering the battery at the + end and leaving at the - end, so we have negative battery current, and Q in the battery decreases. The battery energy change is ΔW = -Vt/R where t = time of current flow. Or, ΔW = -QV where Q = ∫i dt = -Vt/R.

Getting back to your problem, you know you lose charge dQ when you move the plates apart dz, right? Where does that charge go? Is there positive or negative battery current as consequence? Compare with the resistor case.

Or again, if you think of a large capacitor instead of a battery, and you know charge was lost on your plates, then did you gain or lose charge on your large capacitor? Did the large capacitor gain the charge lost by the plates? And if so, did the large capacitor not increase its stored energy?

 

[Pushoam]

know that battery energy W is reduced if I shunt the battery with a resistor R.

Assume that the battery is nothing but a big capacitor with charge Q and potential difference V.
Now as I shunt the battery with a resistor R for time Δt, a +ve charge ΔQ goes from the positive plate to the negative plate of this capacitor and hence the charge of the capacitor gets reduced to Q - ΔQ. Consequently, the potential energy of the capacitor gets reduced to V(Q - ΔQ)/2 from V(Q /2).
According to conservation of energy ,
loss in the potential energy i.e. V(ΔQ /2) = work done on ΔQ by the capacitor = kinetic energy of the ΔQ + the thermal energy developed in the resistor
Hence, the work done on the charge ΔQ is ΔQ V= ΔQ V i.e.positive.
Looking it in another way,
+ve charge ΔQ goes from the positive plate at potential V to the negative plate at zero potential .
Hence, the work done on the charge ΔQ is ΔQ(0 - V)= -ΔQ V i.e. negative.

What is wrong here?

 

[vela]

For calculating energy given to the battery :
The charge (Q-Q') goes to the positive terminal of the battery at constant potential V and -(Q-Q') goes to the negative terminal of the battery at constant potential 0V. Let's assume that inside the battery the negative charge remains at the negative terminal and the positive charge moves to the negative terminal to make it neutral and thus to maintain the constant potential difference between the two terminals. In doing so, the work done by me in this case is (-V)(Q-Q') = the energy given to the battery = E_batt. I am getting E_batt negative here. Where am I wrong?

The work W = V(Q-Q') is the work needed to get the charge to the positive terminal of the battery from the positive plate of the capacitor.

In the scenario you describe, the electrical energy lost when the charge subsequently is moved down to the negative terminal is converted to chemical potential energy inside the battery. As rude man said, you don't really need to worry about that in this problem. All you care about is the energy required to get the charge to the positive terminal of the battery.

 

[rude man]

Assume that the battery is nothing but a big capacitor with charge Q and potential difference V.

OK.

Now as I shunt the battery with a resistor R for time Δt, a +ve charge ΔQ goes from the positive plate to the negative plate of this capacitor and hence the charge of the capacitor gets reduced to Q - ΔQ. Consequently, the potential energy of the capacitor gets reduced to V(Q - ΔQ)/2 from V(Q /2).

This is numerically incorrect. The large capacitor will also lose a small amount of voltage ΔV. The result is that the large capacitor will lose energy = V ΔQ, not V ΔQ/2. I show that calculation below.

Anyway, I'm not sure of the relevancy of what you're discussing to the problem at hand. We are trying to convince ourselves that (1) when you move the plates away from each other a distance dz, charge dQ is moved from the plates and deposited on the large capacitor, and (2) we need to determine the amount dQ. Hooking up a resistor to the large capacitor and trying to figure out what the work was done on the charge ΔQ seems irrelevant, and I don't know how to answer your question. I only invoked the resistor-large capacitor picture to convince you that if charge flows into the large capacitor's + terminal, that large capacitor has gained energy, and vice-versa. Which you probably knew anyway.

Let's stick to the problem at hand:
When you displace the plates by dz, charge dQ flows out of the plates:
Q = CV (of the plates)
dQ = C dV + V dC = V dC since V is effectively constant for the plates;
with dC = -ε₀A/z² dz
And energy loss in the plate field is d(CV²/2) = V dQ/2.
so charge dQ = V dC has left the plates and, by charge conservation, moved to the large capacitor, which gained energy, as follows:
The large capacitor has gained energy = (Q + dQ)(V + dV)/2 - QV/2 = (V dQ + Q dV)/2
but dV = dQ/C and C = Q/V so Q dV = V dQ
so energy change of large capacitor = V dQ for a plate displacement of dz.
So of the large-capacitor energy gain V dQ, half came from the plate field and so the other half had to come from F dz.

 

[Pushoam]

The work W = V(Q-Q') is the work needed to get the charge to the positive terminal of the battery from the positive plate of the capacitor.

The work needed to get the charge (Q-Q') to the positive terminal of the battery from the positive plate of the capacitor is zero as the potential difference between positive terminal of the battery positive plate of the capacitor is almost zero(as the charge (Q-Q') is infinitesimally small). .

 

[Pushoam]

The large capacitor has gained energy = (Q + dQ)(V + dV)/2 - QV/2 = (V dQ + Q dV)/2

But,we have already assumed that the battery i.e.large capacitor maintains constant potential difference i.e.V and so, the large capacitor has gained energy = (Q + dQ)(V )/2 - QV/2 = (V dQ )/2.
If I assume that potential difference of large capacitor has got increased to V + dV, then the potential difference of small capacitor has also got increased to V + dV, as the two capacitors are connected by wire.
Then,

dQ = C dV + V dC = V dC since V is effectively constant for the plates;

it seems to me that you are taking dV=0 for small capacitor and dV= non-zero for large capacitor.
This I didn't understand. Will you please explain it?

 

[rude man]

But,we have already assumed that the battery i.e.large capacitor maintains constant potential difference i.e.V and so, the large capacitor has gained energy = (Q + dQ)(V )/2 - QV/2 = (V dQ )/2.
If I assume that potential difference of large capacitor has got increased to V + dV, then the potential difference of small capacitor has also got increased to V + dV, as the two capacitors are connected by wire.
Then,

it seems to me that you are taking dV=0 for small capacitor and dV= non-zero for large capacitor.
This I didn't understand. Will you please explain it?

That is a very good question and I will answer it later today. If you want to work on it in the meantime, expand d(1/2 CV²) which is the differential change in energy of the plate capacitor. Hint: dQ/Q' << dC/2C and dV = V dQ/Q'. Here Q' = large capacitor charge and C and V refer to the plates.

 

[rude man]

But,we have already assumed that the battery i.e.large capacitor maintains constant potential difference i.e.V and so, the large capacitor has gained energy = (Q + dQ)(V )/2 - QV/2 = (V dQ )/2.

You can easily show that that's wrong.

 

[rude man]

The work needed to get the charge (Q-Q') to the positive terminal of the battery from the positive plate of the capacitor is zero as the potential difference between positive terminal of the battery positive plate of the capacitor is almost zero(as the charge (Q-Q') is infinitesimally small). .

You move a charge dQ from the plates to the large capacitor. You have lost plate energy of V dQ/2 but have gained capacitor energy of V dQ. And you have done this as you say with zero work, You have just created a perpetual-motion machine!

 

[rude man]

As promised in post 16:
plate capacitor energy U = 1/2 CV²
We change dz differentially which incurs V + dV, C + dC and U + dU:
So differential change in U is dU = 1/2 (2CV dV + V² dC) = CV dV + V² dC/2
but dV = V dQ/Q' where Q' is charge on large capacitor
so dU = CV² dQ/Q' + V² dC/2
divide both terms by C: V² dQ/Q' + V² dC/2C
Now, I can make Q' as large as I want, so the first term can be << second term.
But the 1st term is in dV so dV has negligible effect on dU.
So dU = V² dC/2.
I will leave it to you to show that V² dC/2 = V dQ/2.
Remember, V is constant since that's the voltage before the displacements dz, dQ, dV etc.
Note also that dV > 0 but dU, dC < 0. dQ < 0 for the plates but > 0 for the large capacitor C'.
QED

 

[Pushoam]

I was wrong here.

But,we have already assumed that the battery i.e.large capacitor maintains constant potential difference i.e.V and so, the large capacitor has gained energy = (Q + dQ)(V )/2 - QV/2 = (V dQ )/2.

Change in the potential energy of the bigger capacitor = 1/2[ (Q'+dQ') (V'+dV') - (Q'V')], (1)
where Q' is the charge of the bigger capacitor,
V' is the potential of the bigger capacitor,
dQ' is the increase in charge of the bigger capacitor,
dV' is the increase in potential of the bigger capacitor,

dV' = (V'/Q')dQ' (2)Now, here V' = V(potential of the smaller capacitor) (3)
dQ' = dQ (charge lost by the smaller capacitor) (4)
So, dV' = (V/Q')dQ, is very small as Q' could be as large as we want (5)
So, change in the potential difference across the bigger capacitor is effectively negligible.
Consequently , change in the potential difference across the smaller capacitor is also negligible.
But, this does not imply,
1/2[ (Q'+dQ') (V'+dV') - (Q'V')] = 1/2[ (Q'+dQ') (V'+0) - (Q'V')] =1/2[ (Q'+dQ') V' - (Q'V')] = 1/2 (dQ') V'
We can not substitute dV' = 0 here as substituting dV' =0 makes Q'dV' =0, which is wrong.
Hence , even though dV' is negligible, its contribution in the increase of potential energy of the bigger capacitor is not negligible because of Q' being large.
Hence,
1/2[ (Q'+dQ') (V'+dV') - (Q'V')] = 1/2[(dQ') V' + (dV') Q' ]
Using eqns. 3,4,5
= 1/2[(dQ) V + (dQ) V ]
=V dQ
( Learning : Before substituting dV' = 0 , I must check whether it is proper to do so.
I should not substitute blindly.)Thank you ,thanks a lot.

 

[Pushoam]

I will leave it to you to show that V² dC/2 = V dQ/2.

dQ = d(CV) =V dC + C dV
V dQ = V ² dC +Q dV

Q dV = Q (V/Q')dQ,
V ² dC = V ² (-ε₀A/z² dz)
because of Q' being very large , Q dV becomes negligible in comparison with V ² dC.
I understand this intuitively, but can't show mathematically.

V dQ = V ² dC +Q dV = V ² dC
V² dC/2 = V dQ/2, note that here both dQ and dC is negative as both charge and capacitance of the small capacitor has got decreased.

 

[rude man]

So, what's your final answer for F dz for part (b)?

 

[Pushoam]

So, what's your final answer for F dz for part (b)?

Work don by me = energy gain by the battery - energy lost by the capacitor
F dz = V dQ' - [-(V/2)dQ] = (V/2)dQ' = (V/2)V (-dC) [as here dQ = (V dC <0) = -dQ']

(dQ = d(CV) =V dC + C dV = V dC , as dV →0V
dC = - (ε₀A/z²) dz)F dz = ( V²/2) (ε₀A/z²) dz
F = ( V²/2) (ε₀A/z²)
F =( V²/2) (ε₀A/z²) along positive z-axis

Finally, done.
Thanks for being with me patiently and leading me towards the answer, Thanks a lot.

 

[rude man]

Work don by me = energy gain by the battery - energy lost by the capacitor
F dz = V dQ' - [-(V/2)dQ] = (V/2)dQ' = (V/2)V (-dC) [as here dQ = (V dC <0) = -dQ']

(dQ = d(CV) =V dC + C dV = V dC , as dV →0V
dC = - (ε₀A/z²) dz)F dz = ( V²/2) (ε₀A/z²) dz
F = ( V²/2) (ε₀A/z²)
F =( V²/2) (ε₀A/z²) along positive z-axis

Finally, done.
Thanks for being with me patiently and leading me towards the answer, Thanks a lot.

100% correct, and congrats for sticking it out. That's how we learn!

 

[rude man]

dQ = d(CV) =V dC + C dV
V dQ = V ² dC +Q dV
Q dV = Q (V/Q')dQ,
V ² dC = V ² (-ε₀A/z² dz)
because of Q' being very large , Q dV becomes negligible in comparison with V ² dC.
I understand this intuitively, but can't show mathematically.

Just follow post #19.

 

[Buffu]

“virtual work”

What is virtual work ? I am interested in this type of questions.

 

[Pushoam]

Just follow post #19.

V2 dQ/Q' + V2 dC/2C
Now, I can make Q' as large as I want, so the first term can be << second term.

I got it.
I can make Q' as large as I want, but I can't do anything with the second term.
So , by making Q' sufficiently large , I can make Q dV negligible in comparison with V ² dC.
Thank you.

 

[Pushoam]

What is virtual work ? I am interested in this type of questions.

This is what is written from where I got the question.
The method of virtual work is simply the use of conservation of energy to determine a force—in this case,the force on the surface of a conductor. The work is virtual in the sense that the entire system is static, so no work is actually done. Instead one asks how much the energy of a system would change if the surface were moved an infinitesimal amount ,and this change in energy is then used to compute the force.

But I don't understand how does it affect the calculation where the system is actually moved or it is moved in imagination?
I think a good way to understand the concept of virtual work is to understand the difference between work and virtual work using examples, and this I myself don't know.