Let me make sure you have some of the basics.
Let's say that instead of two capacitors, you have two metal bars A and B, each with terminals labelled 1 and 2.
And you have a voltmeter, with terminals labelled M1 and M2.
And you set them up this way: A2-M1, B1-M2 - and you leave A1 and B2 unconnected.
At this point we need to talk about this meter. In the ideal case, the meter will not pass any current. It will detect the difference in the electrical potential between its two terminals, report it, and be otherwise inert.
But that is not how real meters work. Real meters have a slight internal resistance of millions or billions of ohms. So, when nothing is connected, they read zero - because that slight internal resistance has discharged any difference in the electrical potential of their terminals.
Now lets say that in the process of picking up A to connect to your meter, being negligent of proper ESD protocols, you leave a slight charge on A. Let's say it's a negative charge, so now A has gazillions of extra electrons. When A connects to an ideal M, M will suddenly register a huge voltage change - perhaps 10Kv or more. If M is ideal, that reading will stay. But in a real meter, you may never see that 10Kv reading because the internal resistance will quickly balance the charge between M1 and M2 by allowing electrons to pass from M1 to M2. So, within a second (perhaps within millisecond), the meter will read zero.
And the same the same things will happen when B is connected.
First say that you connect A1 to B2 (while A2 and B1 are being metered). In a real meter, nothing much happens - you might get a quick little voltage reading because you accidently deposited an ESD charge on one or both of A and B - but the end result is that the meter starts out at zero (before the A1-B2 connect) and ends up at zero.
Now let's say that you are using an ideal meter when you connect A1-B2. Before the connection, the meter may be reading thousands of volts. Whatever the voltage, it is an electrostatic charge. Basically, A and B are jointly acting as a partially charged 1 femtoF, 40Kv capacitor. And when you connect A1-B2, you immediately discharge it - so, at that point, the meter goes to zero.
This is why you really need to connect A1-B2 in you experiment. Otherwise, you need to deal with how "ideal" your meter is and how careful you are about depositing an overall charge (exclusive of the internal 9v charge) on your capacitors.
So, try to work through the answer and if you need additional assistance, we'll keep an eye out.
