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Suppose I have a charged capacitor, meaning there's a voltage (electric potential difference), say, $2V$, between its two plates. We don't know the electirc potential of the individual plates, right? It might be $2V$ and $0V$ or $5V$ and $3V$.

If I connect one plate of a capacitor ($2V$) to an object with electic potential $0V$, the voltage accross the objects will be $2V$. Will there be any current flow? I know, the circuit isn't closed. But it doesn't mean anything - the voltage across objects will cuase a very short current, because the electric charge will move from one object to the other (touch two differently charged spheres, their charge will distribute evenly across both of them, if I'm not mistaken).

You will probably say the other plate is holding the charge the plate, so it won't go to the ground even though there's a potential difference across the plate and the ground. But that implies that voltage doesn't always mean movement of charges (if that's how it works, fine).

Let's see if I understand capacitors correctly - does connecting both plates of a charged capacitor always discharge the capacitor?

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To know the electric potential of the plates you first need to define where and if there is a point with zero potential. This point is arbitrary. Once you defined that, you can calculate the potential of the plates. For instance, if one of the plates of the capacitor is defined to be at zero V, then the other plate will be at 2V. If you define it that way and one plate is at 2V it will discharge a little as you connect it with an object at zero volts. How much, it would depend on the object details, but charge will flow until both potentials are the same (likely not longer 2V).

The other plate will remain charged in your setup, as it cannot get rid of the charge, but the potential will change, it will adjust to the new distribution of charges across space.

And finally yes, if you connect the two plates of a capacitor you are closing a circuit and the charges will flow until the potential in both plates is the same. Because of symmetry (assuming both plates are the same size and shape), the final configuration will be a zero charge on each plate.

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We don't know the electirc potential of the individual plates, right?

Yes, we do. Or, that depends on what we want to know. Electric potential is just the potential compared to some other point - some arbitrarily chosen reference point.

  • If for example, you chose one of the capacitor plates as the reference, then this plate has an electric potential of $0\;\mathrm{V}$ and the other plat $2\;\mathrm{V}$.
  • If you compare with ground as the reference, then one plate might have $3\;\mathrm{V}$ and the other $5\;\mathrm{V}$. This is not known, no, but it isn't important unless we need that connection.

If I connect one plate of a capacitor ($2\;\mathrm{V}$) to an object with electic potential $0\;\mathrm{V}$, the voltage accross the objects will be $2\;\mathrm{V}$.

Now, be accurate here. It depends on which plate, since they don't have the same potential.

If you pick the plate at a potential of $2\;\mathrm{V}$, then yes, the potential difference or voltage across this plate and ground at (if ground is $0\;\mathrm{V}$ compared to the same reference - in other words, if ground is the reference) will be $2\;\mathrm{V}-0\;\mathrm{V}=2\;\mathrm{V}$. This is the case no matter if they are connected or not.

Will there be any current flow? I know, the circuit isn't closed. But it doesn't mean anything - the voltage across objects will cuase a very short current, because the electric charge will move from one object to the other

Correct. Yes, there will be a current flowing if there is a potential difference across two connected points if it is is not restricted or resisted by anything along the way!. Charge will flow from the place of high to low potential until the potentials are equal (the higher one is lowered and the lower one is raised).

At steady conditions in a circuit there will always be equal potential along points on conductors, if there are no components in between.

If there is a component in between them, then the current might be resisted and less will flow. If - as the extreme case - the resistance is very large then it acts like the points are not connected at all and no current will flow, even though there is a potential difference across them.

You will probably say the other plate is holding the charge the plate, so it won't go to the ground even though there's a potential difference across the plate and the ground.

Not understood. Both plates hold charge (maybe), one just holds more than the other (and thus has higher potential).

The charge will never go to ground unless there is a connection to ground. Potential difference or not. What do you mean with this?

[...] voltage doesn't always mean movement of charges

True. The voltage across the two capacitor plates also doesn't mean any charge moving - they want to move (therefore the word "potential") and they will if allowed (if they are suddenly connected).

does connecting both plates of a charged capacitor always discharge the capacitor?

Yes. Because by "charging" the capacitor we mean that there is a voltage across it's plates. Because then, if the plates are connected through a circuit, current will flow.

Charge will always want to move towards lower electric potential. That is the whole point in electric circuits. So if the potential difference is destroyed, then no charge will want to flow anymore which means no current.

I hope this makes everything more clear.

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  • $\begingroup$ I was always taught that electic potential isn't relative, it's just a quantity. However the voltage is defined as difference in electic potential between two points. Ok. So in short - connecting one plate of a charged capacitor to an object such that there's a voltage across the object and plate will discharge the plate a little. Can it discharge the plate completely? $\endgroup$ – user4205580 Jul 2 '16 at 6:05
  • $\begingroup$ @user4205580 Electric potential is not an absolute value anymore than gravitational potential is. Gravitational potential energy is $U=mgh$, where the $h$ is the height from some reference level. Electric potential is $U=kq/r$, where $q$ is the charge of the reference point and $r$ the distance from it. There is no such thing as absolute values of potential. It is always compared to some reference. $\endgroup$ – Steeven Jul 2 '16 at 8:51
  • $\begingroup$ @user4205580 As explained, if you connect a point of lower (or higher) potential with one capacitor plate, charges will flow away from (or towards) the plate. This reduces (or raises) the potential on this plate. Charges will flow until the potentials of the plate and the connected object are equal. We call a capacitor charged when there is a potential difference across it's two plates. So if the capacitor is totally or only partly discharges, depends on how much the potential is reduced (or raised) on the plate, which depends on the potential of the connected object. $\endgroup$ – Steeven Jul 2 '16 at 9:00
  • $\begingroup$ Then what is the difference between electic potential and voltage? Voltage is the difference in electic potential, and the electic potential itself is relative as well? $\endgroup$ – user4205580 Jul 2 '16 at 9:36
  • $\begingroup$ @user4205580 Correct. The difference is that electric potentials are calculated compared to a common reference, while voltage is a difference between any points, not necessarily the same reference. If the electric potential is calculated in two points relative to the same reference, then the difference between them is a voltage. The idea is that voltage compares potentials, while potentials themselves just are values. $\endgroup$ – Steeven Jul 2 '16 at 9:40

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