'Simple' capacitor problem

Question

I posted this question because I have a problem in grasping the connection between force on a charge and voltage potential equilibrium. So the problem is the following: we have a charged capacitor disconnected from the battery. If a negative capacitor plate is divided into two pieces but still connected to each other with a wire and we do the same to the positive plate, the charge should be equally distributed. Now, if we decide to bring one half of the negative plate closer to the half part of the positive plate we changed potential but as far as the charge on the plate isn't moving the far negative plate-half and the closer negative plate-half should equally act on the positive pieces as the electric field is uniform and not dependent on distance between plates. There shouldn't be any difference in attraction between all plates. Now, a current should flow through the wire when one part of the capacitor is brought closer as the voltage is different but on the other hand it shouldn't if we take that the electric field on the charge between plates doesn't change with distance (in an ideal case)?

Answer 1

Because the plates are connected by wires, they remain an equipotential surface. We can therefore safely assign both left plates (-) with $V=0$ and both right plates (+) with $V=V_0$. This MUST be true in both pictures A and B.

Now in what follows let $u$ denote the upper plates separated by distance $d_u$ and $l$ denote the lower plates, separated by distance $d_l$ with $d_u<d_l$ in situation B.

In situation B, $E_u = {\sigma_u\over\epsilon_0}$, and similarly for $E_l = {\sigma_l\over\epsilon_0}$.

$V_0 = {\sigma_u\over \epsilon_0}d_u = {\sigma_l\over \epsilon_0}d_l$

${\sigma_u\over\sigma_l} = {d_l\over d_u}$

Since the areas of each plate is identical,

${Q_u \over Q_l} = {d_l \over d_u}$.

The charges are not equal, and the fields are not equal. Only the voltage difference between the plates remains equal since the left plates must be equipotential, and the right plates must be equipotential.

Answer 2

As a guide to dealing with this, use the following:

Capacitance $C=\frac{\epsilon_0 A}{d}$

where d is the distance between the plates and A is area.

The situation is like two capacitors in parallel and the total capacitance $C_T = C_1+C_2$

Also the electric field between the plates does depend on distance, $E=\frac{V}{d}$

The voltage will be the same for both halves if the system is connected to a battery, e.g with a wire going from the middle of blue wire to the middle of the red wire via a battery or power supply of a certain voltage.

Finally $C=\frac{Q}{V}$

Answer 3

First we can solve direct problem with fixed voltage $\pm 1$ on the capacitor, divided capacitor and divided capacitor with moved part. As result we can define electric charge on every part of capacitors (we use FEM to solve these problems). Distribution of potential, electric field and electric charge on the capacitor

Distribution of potential, electric field and electric charge on the divided capacitor

Distribution of potential, electric field and electric charge on the divided capacitor with moved part

Figure 3 shows that due to the broken symmetry there are asymmetric distribution of charge on the capacitor parts. Now we can come back to the initial problem with fixed electric charge. It is clear that there is electric current through the wire in that time when part move to support electric charge distribution as shown in Figure 3.

Answer 4

Now, if we decide to bring one half of the negative plate closer to the half part of the positive plate we changed potential

The potential does not change, because by connecting the plates with a wire (of theoretically zero resistance) the voltage remains the same. Both FIG A and B are capacitors in parallel.

but as far as the charge on the plate isn't moving the far negative plate-half and the closer negative plate-half should equally act on the positive peaces as the electric field is uniform and not dependent on distance between plates.

The electric field is dependent on the distance between the plates. For a parallel plate capacitor, the relationship between the electric field $E$, voltage $V$, and plate separation $d$ is

$$E=\frac{V}{d}$$

Since in FIG B the voltage between the upper plates is the same as the voltage between the lower plates, and since the distance between the upper plates is less than the lower plates, the electric field $E$ is greater for the upper plates than the lower plates.

There shouldn't be any difference in attraction between all plates.

Since $E$ is greater for the upper plates in FIG B, the attraction is greater between the upper plates than the lower plates.

Now, a current should flow through the wire when one part of the capacitor is brought closer as the voltage is different but on the other hand it shouldn't if we take that the electric field on the charge between plates doesn't change with distance (in an ideal case)?

As already pointed out, the voltage between the upper and lower plates is the same. But since

$$C=\frac{\epsilon A}{d}$$

The capacitance of the upper plates is now greater than the lower plates. Then, from

$$C=\frac{q}{V}$$

If $C$ is greater and $V$ is the same, the charge on the upper plates is now greater than the charge on the lower plates. Since total charge is conserved, some charge from the lower plates has to migrate to the upper plates. To determine what the final charge is on the upper and lower plates, we need to know the relationship between their capacitances. If $d_1$ is the separation of the upper plates and $d_2$ the separation of the lower plates, then

$$C_{1}=\frac{\epsilon A}{d_{1}}$$ $$C_{2}=\frac{\epsilon A}{d_{2}}$$

Where $d_{1}<d_2$ and $A$ and $\epsilon$ is the same for both capacitors, we have

$$\frac{C_1}{C_2}=\frac {d_2}{d_1}=\frac{q_1}{q_2}$$

Where

$$q_{1}+q_{2}=2q$$

Is the force inequilibrium that is the cause of charge displacement?How does this force show up if the field lines of a plate capacitor are paralel implying the attractive force on electrons does not change changing the distance....there should be a force causing the electrons to start accumulating when the plate comes closer...or > not?

The strength of the electric field, $E$, is greater for the upper capacitor in FIG B for the reason explained, and that results in the charge redistribution. But you are misunderstanding what the "field lines" in the capacitor mean.

The fact that the lines are parallel only means the field is uniform between the plates, but it has no meaning regarding the strength of the field. In field line drawings, the relative strength of the field is represented by the relative density of the field lines (lines per unit area) in different areas of the same drawing.

So if we were to draw field lines for the upper and lower capacitors, the density of the lines would be shown to be greater for the upper capacitor. Even then, the lines would not provide any quantitative information on strength of the field, only the relative strength at different locations.

Hope this helps.

Answer 5

If I understand correctly, the argument you're making is this. You know that in situations A and B, the fields between the top and bottom capacitors aren't going to be the same: that's what is meant by "current must flow". The charges will redistribute, because you know what the steady state in situation A is, and you know what the steady state in situation B is, and so you know that there must be some kind of redistribution. This is what quantumwave's answer is saying.

However, the paradox is that if you could freeze the charges in place, and move the top capacitor plates closer together, then as we go from situation A to B, the electric field acting on the unequal charges in the top capacitor plates hasn't changed, and neither has it changed on the charges in the bottom plates, and so there is no force to pull extra charges from the bottom to the top plates.

I can think of three possible arguments to explain what's happening here.

• In order for the two plates to be brought closer together, you need to move them. This means they will have a non-zero velocity as we move from situations A to B. Therefore, there will be magnetic effects, and you can't use the results for the electrostatic steady state distribution of the charges on the capacitor plates. Although you can make this velocity $v$ as small as you like (quasi-static), the non-static field effects will be of the same order of magnitude as $v$, and so these could cause redistribution of the charges at a rate of order $v$, and if you integrate the non-static charge redistribution over the whole path, it will sum up to the required change of charge.
  I find the above argument not so convincing, as in theory, we could fix the electrical charges in place during the transition from A to B. For example, we could cut the wire, then bring the top plates closer together, and then reconnect the wire afterwards. In other words, there must still be a path from the non-equilibrium situation B to the steady state.

• The more convincing argument is that because of the introduction of the conducting wire between the bottom plates and the top plates, the charges in the wire and at the connection between the wire and the plates are not at equilibrium, so they will start moving, which will "kick-start" the motion of the charges towards equilibrium.

• Finally, another line of argument might be that it's caused by the edge effects of the capacitor.

I would like to be give more details, by explaining exactly how the wires modify the electric field, but it's probably going to be quite complicated. I thought it might be interesting to study a toy model, where the plates are all parallel: you put down a capacitor at $z = 0$, then another at $z = z_0$, connect the plates by a vertical wire at $x = y = 0$, and then to connect the topmost plate to the bottommost plate, you make the whole of space loop along the $z$ axis, either by repeating the pattern, or by modifying the topology of the space. Unfortunately I don't have time to work through it, but if somebody else would like to give it some thought, that would be great.