There is a little bit of a problem about physics, the question of current vs. capacitance, and does

It's not that there is no electric current, but there is an electric current, which is the directional movement of free electrons in the conductor, and the positive charge in the atom is the nucleus, and the nucleus does not move, but the free electrons that can move.

Due to historical reasons, human beings did not have the concept of atomic structure when studying electricity, so the concept of defining the direction of positive charge movement as the direction of electric current was formed. But the physical essence is that the directional movement of free electrons forms an electric current, and the direction of the current is the opposite direction of the directional movement of free electrons.

The team will answer for you.

When the capacitor is charged, there is a current flow, even if the capacitor is equivalent to an open circuit, because the charge on the two plates of the capacitor moves back under the action of voltage, thus forming a current, but once the capacitor is fully charged, the charge on the plate will no longer move, and there will be no current.

It has nothing to do with whether the current is open or not, the current is formed by the directional movement of the charge, and there is an electric current as long as there is a directional movement of the charge.

So why is it said that there is no current in the open circuit? Because the charge can only move to both ends of the circuit after the circuit is opened, and cannot continue to move directionally, it is said that there is no current.

The teacher will say that the current can't flow" is also not rigorous, and "the charge can't pass" is more appropriate.

I guess it's the teacher's image statement, or maybe I didn't understand it, but I agree with you.

Hello classmates, I think you are confusing the transient process with the steady state process, when the capacitor is fully charged, the potential of version A&D is equal (this right can also directly say that the voltage is the same), but the capacitance is not a sudden quantity, with a buffering effect (that is, the above electricity is not filled all at once, nor is it released all at once, this process is the transient process). Then the distance between the two panels becomes larger, C becomes smaller, Q does not change immediately, according to Q=Cu, Q does not change, C becomes smaller, then U increases, UA is greater than UD, and the charge flows from B to C. After a certain amount of time (determined by the magnitude of RC), the capacitance reaches equilibrium again, and the potentials of A and D begin to equalize again.

So, don't mix the before and after states. I don't know if you're clear? It is not clear ...... can continue to communicate

Obviously, the first idea is correct, and this is the active case in the capacitive circuit.

There is a power supply that does not change so.

First of all, the equipotential of A and D requires equilibrium, and the current is unbalanced.

It can be assumed that after moving the plates, the charges do not move directionally for the time being, and according to the definition, the potential difference between the two plates increases. If there is a certain equipotential between E and F, and the potential is set to zero, then the potential at point A will be higher than D, and the current will be from B to C

After the capacitor is charged, the potential at both ends is equal to that at the two ends of the power supply, and the distance between A and E is extended, and the potential of A and E will not be changed, and no current will pass through without potential change.

Summary. When the capacitance of a capacitor decreases, the charge is lost from the capacitor and becomes heat. This is because when the capacitance of a capacitor decreases, the strength of the electric field inside it will also weaken, so that the charge originally stored in the capacitor will continue to lose, and with the loss of charge, the energy in it will also become heat, so that the charge in the capacitor will gradually decrease.

The circuit is connected to only one capacitor, so when the capacitance decreases and the voltage does not change, where does the charge go?

When the capacitance of a capacitor decreases, the charge is lost from the capacitor and becomes heat. This is because when the capacitance of a capacitor decreases, the strength of the electric field inside it will also weaken, so that the charge originally stored in the capacitor will continue to lose, and with the loss of charge, the energy in it will also become heat, so that the charge in the capacitor will gradually decrease.

I have a question here: a circuit is connected to only two electrode plates as capacitors, when the area of the two plates is reduced (that is, the capacitance is reduced), and the v is unchanged at this time, that is, the charge is reduced, then these reduced charges flow directly into the air or into the circuit, if to the circuit then how to flow.

When the area facing each other decreases (i.e., the capacitance decreases), and V does not change, that is, the charge decreases, and these reduced charges flow to other circuits. For example, when there are multiple capacitors in a circuit, the reduced charge will flow to the other capacitors, causing the capacitance values of the other capacitors to increase. In addition, when there are multiple transistors or components in a circuit, the reduced charge also flows to those transistors or components, causing the voltage value of those transistors or components to change.

Due to the nature and structure of the capacitor itself, in the circuit connecting the capacitor, the branch connected to the capacitor in series is an open circuit, that is, it can be understood that it is equivalent to being disconnected, and the current cannot pass through.

So 1When the amount of charge in the capacitor is constant, the line is in a steady state of open circuit and cannot generate current.

2.When the amount of charge in the capacitor changes, although the line is still in an open circuit state at this time, the charge and discharge (i.e., change) at the two poles of the capacitor, (according to the conservation of charge, the charge will not be generated and disappear out of thin air) will inevitably cause the temporary movement of positive and negative charges in the wire.

The temporary movement of the charge, which creates an instantaneous current.

So when the amount of charge in the capacitor is unchanged, it means that the electrons in the capacitor have not run out or into the capacitor: when the amount of charge in the capacitor is unchanged, there is no current in the circuit because the electrons stored in the capacitor do not move directionally, and the directional movement of electrons can form a current, simply put.

By Q=C*U, Q does not change, and the capacitance C does not change, so U does not change, that is, the potential difference between the two versions is a fixed value, so there is no current.

It's like a battery without wires, there is a potential difference between the positive and negative electrodes, but there is absolutely no current.

Because there is no potential difference between the two plates.

So there is no electron made.

Orienteering.

Motivation so. There will be no current.

That's right. Capacitance c = s 4 kd, i.e., the capacitance is proportional to the opposite area.

It is inversely proportional to the distance between the two plates, if there is a plate in which the sound wave vibrates.

The distance between the two plates changes.

Causes a change in capacitance.

Since the voltage does not change, the change in charge is obtained according to Q=UC.

An electric current is formed.

This is converted into an electrical signal.

c=εs/4πkd

where is a constant, s is the area opposite the capacitive plate, d is the distance of the capacitor plate, and k is the constant of the electrostatic force wheel.

An isolated conductor forms a capacitance with infinity, and the conductor is grounded to the infinity and connected to the earth as a whole.

What you are talking about is the capacitance problem of an isolated conductor, which is theoretically charged, infinite, so the capacitance is close to 0, so the charge of the plate q=uc is also close to 0

When S is disconnected, the current is 0 after stabilization

All resistors are then equivalent to wires. So the capacitor is connected directly to the power supply. The voltage between the two boards of the capacitor is equal to e

When S is closed, the current on R1 is 0 after stabilization

So R1 is equivalent to a wire. So the capacitor is connected in parallel with R2, where U2+U3=E, U2 E

I don't know how to continue asking.