The net charge of any of those internally connected pairs of plates is always zero. That is, when you charge the capacitors, charge doesn't leave the wire between C and D, it only moves along it, and is held in place by the electric field of the adjacent plates. If a circuit is completed that allows charge to flow from D's negative plate to A's positive plate, the charges will move back to the right place, but the net charge of the 4 capacitors will always be the same.
Connecting the positive terminal of A will not allow charge to flow back from D, so nothing will happen. Similarly, connecting the wire between C and D won't make charge flow in or out of it, at least not in any way significant to the circuit. It only changes the reference for where we make our measurements from.
Suppose we take some object, for example a conducting sphere, and start with it at the same electrical potential as its surroundings. Now we add one electron the sphere, and because the sphere now has a higher negative charge that its surroundings the potential of the sphere will be slightly lower (i.e. more negative) than its surroundings. Adding the electron has created a potential difference. Now add a second electron and the potential difference gets bigger. Add more and more electrons and the potential difference keeps getting bigger.
A side note: this is basically how a Van de Graaff generator works. Charge is added to the sphere at the top by mechnical means, and this can create a big enough potential difference to generate some impressive sparks.
Anyhow, if we transfer a charge $Q$ to our metal sphere the resulting voltage change is given by:
$$ V = \frac{Q}{C} \tag{1} $$
where the constant $C$ is called the capacitance of the sphere (strictly speaking it's the self-capacitance). For spheres we can work equation for the out the capacitance fairly easily, and in fact it's given by:
$$ C = 4\pi\varepsilon_0 r $$
where $r$ is the radius of the sphere. For objects with different shapes the equation for the capacitance will be different, but the key point is that any object of any size and shape has some capacitance and we can use equation (1) to work how what voltage difference is created when we add charge to it.
The point of all this is that the ends of your battery also have a capacitance. Suppose the chemical reaction in the battery transfers a charge $Q$:
The if the capacitance of the anode is $C_-$ and the cathode $C_+$, the voltage change due to transfer the charge will be give by equation (1) as shown on the diagram. The total voltage will be:
$$ V = V_+ - V_- = \frac{2Q}{C_b} $$
where I've assumed the capacitances of both ends are the same and I've used $C_b$ for them both. What actually happens in a battery is that the reaction runs and transfers charge until the voltage $V$ builds up to the battery voltage. At that point no more charge can be transferred and the reaction stops.
This probably all seems a bit long winded, but having gone gone through all this we can answer your questions really easily:
- when you attach a long wire to the ends of the battery you will increase the capacitance. The battery voltage is constant, and we can rearrange equation (1) to give:
$$ Q_T = \tfrac{1}{2} C_T V \tag{2} $$
where $C_T$ is the new bigger total capacitance with the wire attached. So when you increase the capacitance, $C_T$, you increase the charge $Q_T$. That means the reaction in the battery will restart and transfer more electrons until the total charge rises to $C_T$.
- the electrons spread out across the ends of the battery and the wire. If we call the capacitance of just the wire $C_w$ and the capacitance of the end of the battery $C_b$, then the total capacitance is $C_T = C_w + C_b $. If we put this into equation (2) we get:
$$\begin{align}
Q_T &= \tfrac{1}{2} (C_w + C_b) V \\
&= \tfrac{1}{2} C_wV + \tfrac{1}{2} C_bV \\
&= Q_w + Q_b \tag{3} \end{align}$$
where $Q_w = \tfrac{1}{2} C_wV$ is the charge on the wire and $Q = \tfrac{1}{2} C_bV$ is the charge on the battery. So if you disconnect the wire it keeps a charge $Q_w$. The size of the charge is given by $Q_w = \tfrac{1}{2} C_wV$.
- The potential difference between the wires is just the battery voltage $V$. If you connected them together with some suitable voltmeter, that meter would show a voltage of $V$.
In practice the capacitance of a piece of wire is going to be very small, and the charge you'd build up on the wires will be tiny. Nevertheless, the capacitance will be greater than zero so there will be some charge.
Best Answer
Here is the thing about potential/voltage. Potential/voltage is a measure of difference in the, say, potential energy between two points in space. So the correct term is actually: "potential/voltage difference" when we talk about that stuff. So when you talk about batteries, the higher potential side of the battery is higher, with respect to, the lower side. It has nothing to do with anything else, you, me, your computer, the Sun or Earth. Technically, the higher potential side of the battery is not higher/lower than earth.
Here is what happens when you touch the "+" side of the battery to ground:
Soil resistance might be very low depending on the type of soil. However, air resistance is much higher which will prevent battery discharge.
By the way, both soil and air can conduct current with free ions. (Higher voltages cause big ionization i.e. lightnings) (Pouring salt water in soil also helps lower soil resistance)
Now, if you throw your battery in soil and pour some saltwater in there, you'll definitely be seeing a discharge!