Tiny-tim's proof: force on a capacitor plate

[tiny-tim]

is this a valid proof? it seems very simple, and gives the correct result, but i can't find it in a book, so perhaps there's something fundamentally wrong with it?

Given an infinite parallel plate capacitor, what is the force per charge on either plate in terms of the electric field, E?

The electric field E is defined so that the force on a charge q is the Lorentz force, qE, ie the Lorentz force per charge is simply E.

However, this applies to a charge in the field, not at the edge of the field, and the field E of a capacitor stops at the plate.

So imagine that there is an identically charged plate on the other side … the E field from this plate will have the same magnitude but the opposite direction (because it is from an equally charged plate, not from an oppositely charged plate), so the total force (from both outside plates) on the (now) middle plate is twice the original force.

But that total force now is the Lorentz force, since the field is now continuous.

Therefore the force per charge from one plate is E/2, ie half the "usual" Lorentz force per charge. ​

 

[Gordianus]

Check Feynman's book. He gives the same expalantion

 

[Borek]

Gmta

 

[codelieb]

Check Feynman's book. He gives the same expalantion

Feynman certainly does NOT use this hand-waving explanation in his book. To find the force between the plates of a capacitor, he applies the principle of virtual work to its electrostatic energy. See FLP Vol II Section 8-2 "The energy of a condenser. Forces on charged conductors."

Mike Gottlieb
Editor, The Feynman Lectures on Physics
---
www.feynmanlectures.info

 

[Gordianus]

My mistake. I was almost sure Feynman, in addition to resorting to the virtual work principle, had pointed out the average field is half the field between the plates because the field suddenly drops from a given value yo zero inside the conductor.

 

[codelieb]

I was almost sure Feynman, in addition to resorting to the virtual work principle, had pointed out the average field is half the field between the plates because the field suddenly drops from a given value to zero inside the conductor.

FLP does say something like that, but I can not see how it is related to tiny-tim's "proof" involving the introduction of a 3rd charged surface.

Specifically, FLP says "One would immediately guess that the force acting on one plate is the charge Q on the plate times the field acting on the charge. But we have a surprising factor of one-half. The reason is that E0 is not the field at the charges. If we imagine that the charge at the surface of the plate occupies a thin layer, as indicated in Fig. 8-4, the field will vary from zero at the inner boundary of the layer to E0 in the space outside of the plate. The average field acting on the surface charges is E0/2. That is why the factor one-half is in Eq. (8.18)."

Note that nothing is "proven" in this statement. It is a qualitative description of the phenomenon only. You can't get the factor "1/2" out of it. (To do that, you would first have to find the field inside the conductor.)

 

[tiny-tim]

Feynman certainly does NOT use this hand-waving explanation in his book. To find the force between the plates of a capacitor, he applies the principle of virtual work to its electrostatic energy. See FLP Vol II Section 8-2 "The energy of a condenser. Forces on charged conductors."

mmm … i thought it was unlikely that i'd missed it in feymman!

on reflection, even if my argument is valid, my premise seems unsustainable …

i don't see how i can contrive to have the same electric field on both sides of a charged plate

(which is what the "proof" relied on)

thanks mike! ​

 

[shoestring]

Given an infinite parallel plate capacitor, what is the force per charge on either plate in terms of the electric field, E?

How about this:

Because of symmetry, charges on the same plate won't exert a net force on the charge we're looking at. Only the field from charges on the opposite plate, E/2, matters.

 

[kcdodd]

I am having some issue with this. If we look at the difference in Electric field stress on either side of the plate, one should get the force per area on the plate, right?

$$\delta P = \frac{\epsilon_0}{2}(E_0 + \frac{\sigma}{2\epsilon_0})^2 - \frac{\epsilon_0}{2}(E_0 - \frac{\sigma}{2\epsilon_0})^2 = \sigma E_0$$

How does this 1/2 you all are talking about get in there? Unless I made a mistake.

edit: never mind. I realized you are talking about the field between the parallel plates, so the field pressure difference is:

$$\delta P = \frac{\epsilon_0}{2}E^2 - 0 = \frac{\epsilon_0}{2}E^2 = \frac{1}{2}\sigma E$$

 

[vanhees71]

I think, the correct way goes over the energy stored in the capacitor. Suppose you have two very large plates so that we can treat the electric field between the plates as homogeneous. The energy density of the electric field at each point is

$$\epsilon=\frac{1}{2} \vec{E}^2.$$

So we have to calculate the electric field between the plates, given that one is charged with charge, $$q$$, the other with charge, $$-q$$. The charges are attracting each other, so that the whole charges sit on the inner side of each plate.

Now we use Gauss's Law for a volume consisting of a cylinder parallel to the plates such that its "bottom" is inside the left plate, where the electric field is zero, and the "top" between the plates. Since the electric field is perpendicular to the plates and can be considered as homogeneous, we get

$$\int_{\partial Z} \mathrm{d}^2 \vec{S} \cdot \vec{E}=A E=\int_Z \mathrm{d}^3 \vec{x} \rho=q,$$

where $$A$$ is the area of the capacitor and $$E$$ the component of the electrical field. This gives

$$E=\frac{q}{A},$$

and thus the energy density is

$$\epsilon=\frac{1}{2} E^2=\frac{q^2}{2A^2}$$

The total energy is the volume integral over the whole interior of the capacitor. Since everyting is constant, we find

$$U=\epsilon V=\epsilon A d=\frac{q^2 d}{2 A}$$,

and thus the force component perpendicular to the plates is given by the derivative with respect to $$d$$, the distance of the plates, leading to

$$F=-\frac{\partial U}{\partial d}=-\frac{q^2}{2A}=-\frac{q E}{2}$$.

The sign is from the fact that the plates are attracting each other. It's best you make a little drawing of the whole situation to get the meaning of these signs right.

Note: I used Heaviside-Lorentz units for this calculation. If you want to use SI units, you have to put appropriate factors of $$\epsilon_0$$ everywhere.

 

[tiny-tim]

i think that the simplest proof is the energy of any capacitor = QV/2,

so the force per charge locally on either plate is:

(1/Q)∇(energy) = ∇V/2 = E/2

(applies to capacitors of any shape)