Why does the potential difference between two capacitors seek equilibrium?

[x86]

Suppose we have a circuit with two capacitors connected to each other, C1 and C2.

C1 is charged and has a potential difference of 5 V (V1=5V), but C2 is not charged (V2=0)

Apparently it is known that the potential difference of these two capacitors will want to reach equilibrium. So V1=V2.

Why is this? I don't understand why this happens.

 

[Simon Bridge]

Its a bit like having two water tanks connected by a pipe at the bottom.
If they start out at different levels, they are going to want to have the same level.
Why is that?

In electronics, a conductor is just a container for charge, which allows the charge to move freely.

 

[x86]

Its a bit like having two water tanks connected by a pipe at the bottom.
If they start out at different levels, they are going to want to have the same level.
Why is that?

In electronics, a conductor is just a container for charge, which allows the charge to move freely.

In the water tank example, I assume it would be because both bodies of water exert a pressure on the water tank pipe. If one tank has more volume than the other, then it exerts more pressure on that side of the pipe (and the pressure from the other tank isn't enough to counteract this force). Then I assume both sides of the pipe's pressure would want to reach equilibrium, so that there will be no more net movement of water.

 

[Simon Bridge]

Well done... in an uneven level, there would be a higher pressure at one end of the pipe than at the other, so water will flow along it.

Similarly, if the voltage on one end of a wire is higher that at the other end, then what does the charge do?

 

[Simon Bridge]

Well done... in an uneven level, there would be a higher pressure at one end of the pipe than at the other, so water will flow along it.

Similarly, if the voltage on one end of a wire is higher that at the other end, then what does the charge do?

 

[x86]

Well done... in an uneven level, there would be a higher pressure at one end of the pipe than at the other, so water will flow along it.

Similarly, if the voltage on one end of a wire is higher that at the other end, then what does the charge do?

The electrons would all go to the higher potential? And this would also probably change the potential, because of all the electrons going there. Then the potential would get lower and disperse along the wire evenly?

 

[LvW]

Suppose we have a circuit with two capacitors connected to each other, C1 and C2.
C1 is charged and has a potential difference of 5 V (V1=5V), but C2 is not charged (V2=0)
Apparently it is known that the potential difference of these two capacitors will want to reach equilibrium. So V1=V2.
Why is this? I don't understand why this happens.

...connected to each other ?
I suppose you mean "in parallel" - right?
My answer:
Assuming that C1 is charged with Q1 (equivalent to V1=5V) and is connected at t=0 to another capacitor C2 in parallel, the total charge remains constant (Q1).
As a result we have a new charge distribution between both capacitors - proportional to their capacity C1 resp C2.
Both have, of course, the same voltage (because they share the same nodes) which is V=Q1/C with C=C1+C2.

 

[Baluncore]

Or to put it another way;
Before connection, C1 has a voltage of 5V, while C2 has a voltage of zero.
Capacitance is defined as the ratio of charge to voltage, C = Q / V.
The charge on C1 is Q = C1 * 5V. That is also the total charge available for redistribution.

At the instant of parallel connection, there will be a voltage gradient along the wires used to make the connection. That will allow charge to flow until the voltages reach the same value when the voltage gradient is no longer available to encourage the charge to move.

Two capacitors C1 and C2 when connected in parallel will have a total capacitance of Ct = C1 + C2.
Once connected in parallel the voltage on both capacitors will settle to V = Q / Ct, which is; V = 5V * C1 / Ct.