Where are exactly the charges of charged capacitor?

If a capacitor is connected to a battery and is charged, are there charges inside the wires or do they just accumulate on the surfaces of capacitor?

• A charged capacitor is not electrically charged but is electrically neutral. To charge a capacitor, free electrons are removed from one plate while being added to the other. electronics.stackexchange.com/q/35556 Apr 12 '13 at 17:48
• still you can ask how are they distributed Apr 12 '13 at 17:50
• @AlfredCentauri: We can always charge one plate of capacitor and leave the other neutral or charge it differently ;-) (not with a battery, of course). Apr 12 '13 at 21:19
• @VladimirKalitvianski, you are of course correct. And, if the context were not that of electric circuits but, say, the "physicist's capacitor", then your observation would even be relevant ;-) Apr 12 '13 at 22:13
• @VladimirKalitvianski, seriously though, the point of my original comment was to emphasize that, when we say a capacitor is "charged", we don't mean electrically charged, but energy "charged", i.e., the capacitor can supply energy to a connected circuit. Now, let's say that we did as you suggest and placed charge on one plate of a capacitor. Then, connect a wire between the plates. It would then be the case that the capacitor is no longer "charged" even though it would be electrically charged, i.e., non-neutral. Apr 12 '13 at 22:50

Yes, the surfaces of the capacitor have a net negative or positive charge. The wires are also slightly charged because they also have a very small capacitance, but this is usually neglected.

The ideal capacitor contains charge only in the plates and the charge distribution is uniform. In reality though :

1.The charge distribution is not uniform, the edges having more charge density.

2.The induced field is not perfectly uniform

3.There is always charges on the wires (they carry current) but that does not affect the net capacitence

It helps to think about Maxwell's equations here. The first one of which is

$$\nabla \cdot \mathbf{E} = \rho / \epsilon_{0}$$

where $\rho$ is the charge density.

Electrostatics is always like a chicken and egg problem: charges make fields, but fields move charges around, so it can lead to some confusion. It turns out that in the steady state we do not care about which came first. We just know for a fact that Maxwell's first equation will be true at every point in space.

Let's start with a point inside a conductive part, it can be the connecting wire, or the plate of the capacitor itself. The definition of conductive is that no potential difference can exist across it. No potential difference means no field (at each point inside a conductor). So the left hand side of the equation above is zero, which implies that the charge density $\rho$ is zero everywhere inside any conducting part. A corollary of this deduction is that if there is any charge present at all it will be located only on the surfaces of the wire or the capacitor plate.

The other piece of information that is generally key in the capacitor situation is that:

There is a constant voltage difference $V$ across the two plates, usually determined by a battery.

I won't prove it here for lack of space, but that, in fact, defines the situation completely. Given the geometry of the conductors (that includes the plates of the capacitor and the wires connected to them) there is a unique arrangement of charges on their surfaces that will guarantee that this voltage difference difference $V$ is consistent with the electric fields generated by the charge arrangement.

Those fields are such that if we pick any possible path $\Gamma$ from one conductor to the other, then we will have

$$-\int_{\Gamma} \mathbf{E} \cdot \mathrm{d} \mathbf{l} = V$$

It is really quite magical that the charges themselves arrange themselves in such a way, but that is what Nature does. If you bend the connecting wire a little bit you will change the geometry of the conductors and the charges will just find a new arrangement to make Maxwell's equations work out.

With all that said, due to the fact that most of the surface of the conducting parts is on the plates of the capacitor, that is where you will find the majority of the charge.