# 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?

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Your presumption is wrong. The other plate will in fact retain its potential and charge if disconnected. –  C.R. Feb 5 '12 at 0:30

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.

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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.

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This answer is simply wrong. Read Manishearth's. –  C.R. Feb 5 '12 at 12:37

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:

1. The second plate does NOT acquire the same potential as the first one
2. The second plate will NOT acquire a net charge, but there will be a redistribution of charges
3. Potential needs no medium to be transmitted.
4. The potential of a plate or a point can change without the charge of the plate/point changing.
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"This comes from the fact that field inside a metal must be zero." You may want to clarify that. I personally come to the conclusion with Gauss' law, but maybe there are other ways to look at it. –  C.R. Feb 5 '12 at 14:48
Well, if it wasn't zero, then we would have a current. Field at any point inside a conducting material is zero (but not in cavities in the metal). This is pretty much assumed while dealing with electrostatics as we only deal with currentless situations. Even if we considered current, the conductor would have to polarize and stop the current at some point or the other. –  Manishearth Feb 5 '12 at 14:57
@KarsusRen How did you come to the conclusion via Gauss' law? If you used any formula for field and worked backwards, that's most probably circular reasoning, as most of the field formulae (as well as the fact that charges distribute themselves on the surface of conductors) come from the conjunction of this fact and Gauss' law. –  Manishearth Feb 5 '12 at 14:58
Added clarification.. –  Manishearth Feb 5 '12 at 15:02
Potential is just energy. The field of the first plate will polarize the second and therefore raise its potential. –  John McVirgo Feb 6 '12 at 6:23