Capacitance question A capacitors capacitance, C is equal to Q/V right?
If Q was somehow cut in half, would the potential difference also be cut in half, since V is directly proportional to Q?
And if so, Its C would be the same?
But the energy it can hold won't be, as it is =1/2 Q2/C or 1/2 QV, it would be 1/4 of the original energy.
Now lets pretend I have a capacitor which has a charge Qtot and voltage Vtot
If I then hooked this up to another capacitor that is initially uncharged with no voltage across it via conducting wires, I think each capacitor would then have a charge Qtot/2 and voltage Vtot/2?
So the ratio of final energy to initial energy would be 1/2 I believe.
Is there some flaw in my logic here? I tried to talk about this to my professor and he said it was impossible to come to the conclusion I have above.
Also, where did the rest of the energy go?
 A: Everything you say is correct in the steady state.
The problem you run into is that when you remove charge from a charged capacitor to an uncharged capacitor, there is a potential difference. And somehow, you have to remove the energy from the electron that moves from one to the other. It turns out, as you calculated, that you in fact remove half of the total energy.
I'm not sure how advanced your knowledge is, so I will first show this in a semi-quantitative way - not quite accurate, but enough to get the idea. Imagine I have two capacitors, and I move little bits of charge across from one to the other - each bit of charge is sufficient to change the voltage on the capacitor by one volt:
Starting with one capacitor at 10 V, and the other at 0 V, the first bit of charge moves across a potential difference of 10 V. Now the capacitors are at 9 and 1 V - so the next bit of charge moves through 8 V. 8 and 2 - move 6. 7 and 3 - move 4; 6 and 4 - move 2. And now the voltages on the two capacitors are equal.
The total amount of work extracted from the charge was 10+8+6+4+2 = 30 'units'.
If we wanted to charge one capacitor to 10 V, we would have to charge first to 1 V, then to 2 V, etc. Work done would be 1+2+3+...10 = 55 'units'. Now you see that 30 is almost half of 55. The reason it's not exact is because I was using finite steps. To get this right we would use integration.
Then in the first case, 
$$V_1 = \frac{Q_0-Q}{C}\\
V_2 = \frac{Q}{C}\\
dW = dQ\cdot (V_1-V_2) = dQ\cdot\frac {Q_0 - 2Q}{C}$$
When $Q = \frac12 Q_0$, the voltage on the two is equal and no further work is done. The total work is
$$\int dW = \int_0^{Q_0/2} dQ\cdot\frac {Q_0 - 2Q}{C}$ = -\frac{Q_0\cdot Q_0/2 - (Q_0/2)^2}{C}=\frac{Q_0^2}{4C}$$
which is half the stored energy, exactly as expected.
Now if we have a resistor between the two capacitors, it is easy to see where the energy went. It's not so easy when there is just a wire.
In fact every wire has some inductance - and when you try to send a lot of current through an inductor, it will resist that change in current as it builds up a magnetic field. In fact, you would end up with an oscillator: at first the current builds up, until it reaches a maximum when the two capacitors are equally charged. At that point, the excess energy is stored in the magnetic field around the wire. But wait- there's more! The inductor won't let the current stop flowing when the voltages are the same and so it will "push" more charge into the second capacitor, until it is fully charged and the first is fully discharged. At which point the process starts again, in reverse.
In practice there will be some resistivity in the system, and this oscillating current will slowly die down. But just in case you thought I was making this up - this is in fact similar to the mechanism behind the original spark transmitter of Marconi, see for example http://www.hammondmuseumofradio.org/spark.html (note they show just one capacitor - but in a circuit as we described, you can consider the two capacitors as being in series, and therefore "equivalent" to a single capacitor).
