Force on a capacitor plate as the derivative of the energy -- is it a fluke?

[etotheipi]

Bit of a random question... capacitors can be a bit weird, in that if we connect one up to a source of EMF and do positive external work to separate the plates of the capacitor, the energy of the capacitor decreases (and instead the work you do plus the decrease in capacitor energy goes into the battery). So we get the strange property that if the electric field from one plate does positive work on the other plate, the potential energy of the plates actually increases (instead of decreasing, as we would expect from e.g. two point charges).

Specifically, we have that $U = \frac{A\epsilon_0 V^2}{2x}$ and that the force on the plate, obtained by multiplying the field from the other plate by the charge, is $F_x = -\frac{A\epsilon_0 V^2}{2x^2}$. We might notice that for this setup we have that the electric force on the plate relates to the potential energy of the plates via$$F_x = \frac{dU}{dx}$$However, this seems like big a red-herring since for a conservative force we would usually expect the relation$$F_x = -\frac{dU}{dx}$$with an important negative sign. That just leads me to believe that the fact that $F_x = \frac{dU}{dx}$ is a bit of a fluke. I wondered if maybe I'm being a bit too judgemental here... but would you agree?

 

[A.T.]

We might notice that for this setup we have that the electric force on the plate relates to the potential energy of the plates via $F_x = \frac{dU}{dx}$ However, this seems like big a red-herring since for a conservative force we would usually expect the relation $F_x = -\frac{dU}{dx}$ with an important negative sign.

Seems like you using the same symbol U for different things, that are not quite analogous. In the capacitor & battery case it should be the total potential energy of both.

 

[etotheipi]

Seems like you using the same symbol U for different things, that are not quite analogous. In the capacitor & battery case it should be the total potential energy of both.

Right yes, I guess that would work! That $-d_x U_{c,b} = -d_x (U_c + U_b)$, which also happens to be $d_x U_{c}$ would be the electric force on the plates.

This example is a lot harder for me to visualise (as compared to the potential functions of simple systems like point charges), but it seems to work out mathematically.

 

[hutchphd]

Again I ask my usual question: why worry about what did work on who? Energy is conserved.
You are charging the battery.

 

[etotheipi]

Again I ask my usual question: why worry about what did work on who? Energy is conserved.
You are charging the battery.

Well okay , I just thought it was a curious example that doesn't work (no pun intended...) exactly as we might first expect when we look more closely at the individual works done!

 

[hutchphd]

I understand, but increasing the complexity of the description does not necessarily imply better understanding. ...and my poor brain needs it simple...particularly in getting the signs correct!

 

[vanhees71]

Let's use two plates and let's use the usual approximation, neglecting the fringe fields, i.e., we make the distance of the plates, $d$ small compared to the extension of the plates. Let's put the positively charged plate in the $xy$ plane at $z=0$ and the negatively charged plate parallel to the $xy$ plane at $z=d$. Then the electric field is approximately
$$\vec{E}=\frac{Q}{\epsilon_0 A} \vec{e}_z$$
between the plates and 0 elsewhere.

However in your setup the potential difference is kept contant given by the voltage of the battery $V$, not $Q$. The potential obviously is
$$\Phi(z)=-\frac{Q z}{\epsilon_0 A}, \quad U=\Phi(0)-\Phi(d) \; \Rightarrow \; Q=\frac{\epsilon_0 A}{d} V.$$
Thus you have
$$\vec{E}=\frac{V}{d} \vec{e}_z.$$
Now let's calculate the electrostatic force on the lower plate in 2 ways:

(a) using the Maxwell stress tensor

$$T_{ij}=\epsilon_0 \left (E_i E_j-\frac{1}{2} \vec{E}^2 \delta_{ij} \right).$$
Obviously we only need
$$T_{zz} = \frac{\epsilon_0}{2} E_z^2=\frac{\epsilon_0 U^2}{2 d^2},$$
and the force is
$$F=\frac{\epsilon_0 A V^2}{2 d^2}.$$
This force must of course be compensated by mechanical stresses keeping the plates at the distance $d$.

Of course the same calculation holds for the upper plate, but the normal vector is in $-z$-direction and thus the upper plate has of course the opposite force than the lower plate (which is also intuitive, because the plates attract each other being oppositely charged).

(b) Using the mechanical potential

The energy density of the field between the plates is
$$u=\frac{\epsilon_0}{2} \vec{E}^2=\frac{\epsilon_0 V^2}{2 d^2},$$
i.e., the total em. energy
$$U=u A d= \frac{\epsilon_0 V^2}{2d}.$$
From energy conservation we have
$$U_{\text{mech}}+U=\text{const},$$
and thus
$$U_{\text{mech}}=\text{const}-U.$$
Now we have to interpret this as the interaction potential of the forces of the plates. So we have to write $d=|z_1-z_2|$, i.e.,
$$U_{\text{mech}}=-U+\text{const}=-\frac{\epsilon_0 V^2}{2d}$$
Make $z_2>z_1$, you get $d=z_2-z_1$, where $z_1$ is the position of the lower and $z_2$ the position of the upper plate. Thus
$$F_{\text{lower}}=-F_{\text{upper}}=-\partial_{z_1} U=+\frac{\epsilon_0 V^2}{2(z_2-z_1)}=\frac{\epsilon_0 V^2}{2d},$$
i.e., with both methods you get the same forces on the lower and upper plates.

 

[Meir Achuz]

Most EM textbooks show why $F_x = -\frac{dU}{dx}$ at constant Q, and $F_x = +\frac{dU}{dx}$ at constant V.
Just take the derivatives to see this.

 

[vanhees71]

But also at constant $Q$ the force on the lower plate is in $\vec{e}_z$ that of the upper plate is $-\vec{e}_z$ (setup of the coordinates as in #7) . The calculation is the same. With approach (b) the analysis is as follows. With constant $Q$ you have
$$\vec{E}=\frac{Q}{\epsilon_0 A}.$$
The em. energy density is
$$u=\frac{\epsilon_0}{2} \vec{E}^2=\frac{Q^2}{2 \epsilon_0 A^2},$$
and the total em. field energy is
$$U=u A d=\frac{Q^2}{2 \epsilon_0 A} d=\frac{Q^2}{2 \epsilon_0 A}(z_2-z_1).$$
But now this is a closed system, i.e., the plates are not connected to a battery, and thus in this case the mechanical potential is
$$U_{\text{mech}}=+U$$
and thus
$$F_{\text{lower}}=-F_{\text{upper}}=-\partial_{z_1} U=+\frac{Q^2}{2 \epsilon_0 A}.$$
In this case the force is constant (independent of $d$), because also $E$ stays constant.