Meaning of F = -dU/dx for capacitors

[phantomvommand]

I have noticed that F = -dU/dx in gravitation gives the attractive force experienced by both bodies.

For capacitors, does F = -dU/dx give the force experienced by each capacitor?

 

[vanhees71]

Can you define your quantities? What is "the force experienced by each capacitor"?

 

[phantomvommand]

Can you define your quantities? What is "the force experienced by each capacitor"?

Apologies for my lack of context :/
U = potential energy stored in the E Field of Capacitors.
X = separation between capacitor plates.
I am asking if F is the force experienced by each of the capacitor plates. Assume the capacitor is made up of 2 plates.

 

[alan123hk]

For capacitors, does F = -dU/dx give the force experienced by each capacitor?

How to find the value of $-\frac {dU} {dx}$ on the two plates of the capacitor?

 

[vanhees71]

So you want to calculate the force on the plate of a capacitor. What are your own thoughts of this problem? I think instead of giving directly the answer it's better to work it out as a "homework problem" as in the homework section.

Hints: There are at least three ways to get the result.

(a) One is to calculate the electric field generated by one of the plates and then get the force on the other plate by considering the force on the charges of this other plate in this electric field.

(b) Another method is to use Maxwell's stress tensor.

(c) You can also use your idea with the total energy in the electric field within the capacitor and interpret it as a potential for the interaction force between the plates.

 

[etotheipi]

Another thing to keep in mind is whether you're holding $V$ constant as you vary $x$, or if you're instead holding $\sigma$ constant. Using either of @vanhees71's suggestions you'll find that $F_x = {\sigma^2 A}/{2\varepsilon_0}$ (you should show it yourself, though!). The energy of the capacitor is $$U = \frac{\sigma^2 A x}{2\varepsilon_0} = \frac{\varepsilon_0 A V^2}{2x}$$Notice that$$\frac{dU}{dx} \big|_{\sigma} = \frac{\sigma^2 A}{2\varepsilon_0} = F_x$$whilst$$\frac{dU}{dx} \big|_{V} = -\frac{\varepsilon_0 A V^2}{2x^2} = - \frac{A \sigma^2}{2\varepsilon_0} = -F_x$$It's maybe a little tricky to understand the negative sign in the case at constant voltage [i.e. capacitor connected across a battery]. Always considering an equilibrium process, you can increase the separation by $\delta x$ by doing external work $F_x \delta x$ on the system (against the attractive forces between the plates). However, doing so results in the charge changing as$$Q = \frac{A\varepsilon_0 V}{x} \implies \delta Q = - \frac{A\varepsilon_0 V}{x^2} \delta x$$And this $\delta Q$ must pass through the battery $\mathcal{E} = V$, resulting in an energy change of $$\delta E = V \delta Q = - \frac{\varepsilon_0 AV^2}{x^2} \delta x = - \frac{\sigma^2 A}{\varepsilon_0} \delta x = -2F_x \delta x$$Overall then the energy changes like$$\delta U = F_x \delta x - 2F_x \delta x = -F_x \delta x$$Hence the negative sign!

 

[alan123hk]

Try to use drawing method to help understand. ☺