How does instant charging of one plate affect the potential of the other plate of a floating capacitor? If I have an uncharged floating capacitor and I instantaneously connect one plate to some potential, then that plate will acquire some charge. In practice, the other floating plate will instantaneously rise to the same potential as the first one. But wouldn't this require the second plate having the same charge as the first?
 A: Everything you've probably learned about capacitors, especially including the statement that opposite plates of the capacitor carry opposite charges, applies only to a capacitor in a circuit. If your capacitor is floating, so that the plates are not connected to anything, the charge on the plates is not going to change. If you hook up only one plate to a battery or something that pumps charge into it, the other plate is not magically going to acquire any charge. All that does happen is that the electric field from the first plate will push the electrons of the second plate toward one side or the other. You'll get a separation of the charge along the thickness of the second plate, but the total amount of charge on the plate doesn't change.
A: The charge will redistribute as follows:
$$\frac{+Q}{2}\begin{vmatrix}
 \frac{+Q}{2} & \frac{-Q}{2} \\
  \text{charged plate} &  \text{free plate}
\end{vmatrix}\frac{+Q}{2} $$
Note that the charges on the inner and outer sides of each plate are distinct. In a capacitor (even one not in a circuit), it is mandatory that the opposing faces have opposite charge, but the charge on the outer surfaces can have any value. This comes from the fact that field inside a metal must be zero (as it cannot have a current through it when considering electrostatic situations). The remaining charge distribution can be found through charge conservation.
Whenever you have a set of n parallel plates, and you give some charges to each of the plates, the easiest way to calculate the distribution is to take the net charge, divide by two, and assign this charge to both of the outer faces of the outermost plates individually. Now use charge conservation per plate and the "opposing faces opposite charge" rule to calculate charges on remaining plates. For example, if I have three plates, and I give a charge Q,2Q,3Q to the plates, the distribution is first written as ($\frac{Q+2Q+3Q}{2}=3Q$):
$$3Q | \qquad|\qquad | 3Q$$
Applying charge conservation on outermost plates (their net charge should be Q and 3Q respectively,
$$3Q |-2Q \qquad|\qquad 0| 3Q$$
Using the opposite charges rule:
$$3Q |-2Q \qquad 2Q|0\qquad 0| 3Q$$
Now, back to your original question. As you can see, the the second plate acquired a charge distribution but not a net charge. Now, the potential of the second plate will be $V_0-\frac{Qd}{2\epsilon_0 A}$ with the usual meanings for symbols ($V_0$ is potential of first plate). This is easily obtained from the fact that net field due to the first plate is$\frac{Q}{2\epsilon_0 A}$. Yes, the second plate has acquired a potential, but there's no harm in that. You do not need a wire to transmit potential; it is transmitted by the field (which in turn is transmitted by EM waves). For example, in an empty region of space, bring a charge q to a distance r from some point. That point just acquired a potential $\frac{q}{4\pi\epsilon_0 r}$, without any physical contact.
Summing up:


*

*The second plate does NOT acquire the same potential as the first one

*The second plate will NOT acquire a net charge, but there will be a redistribution of charges

*Potential needs no medium to be transmitted.

*The potential of a plate or a point can change without the charge of the plate/point changing.

A: You have three conducting systems, and therefore three relevant potential levels. (A) The potential of the disconnected capacitor plate. (B) The potential of the connected capacitor plate and its wire lead. (C) "Ground" ... When you say that you change the potential of (B), you must be talking about "relative to ground", let's say earth ground.
If you model this as an electronic circuit, you have a large capacitance between (A) and (B) and a negligible (stray) capacitance between (A) and ground. Therefore, it's a voltage divider, and the voltage of (A) is a weighted average between the voltages of (B) and ground. But actually ~100% of the weight is in (B), and ~0% in ground. So (A) will follow the voltage of (B) whatever it is. (A) will not particularly care about the voltage of ground. Why would it? Ground is nothing special, it's just the wiring in the walls.
To sum up, the capacitor is uncharged (no charge on either side of either plate), and (A) follows the voltage of (B). Or if you want to be pedantic, the capacitor is almost uncharged, and just a tiny bit of charge moves off the inside of the plate in (A) and into the (stray) capacitor connecting (A) and ground (i.e. the displaced charge would be coating the outside surface of the plate and all the surfaces of the disconnected lead wire, and meanwhile an equal and opposite charge would be coating pipes and other metal objects in the room).
One aspect of this answer is, changing the voltage of something does not have to be particularly difficult. Unless that something has a large capacitance to ground, it takes only the slightest "nudge" to make its voltage jump up or down by any amount.
A: Capacitor plates always have nearly same potential, because the plates are at nearly same position in space. Very simple :)
If capacitor plates have large potential difference, then there must be HUGE opposite charges on the plates. 
