Question about the voltage between two charged capacitors

[Skaperen]

TL;DR Summary

voltage between two charged capacitors

suppose you have two capacitors with a 0.1 Farad value and 12 VDC rating. label these as A and B. label the terminals of each as 1 and 2. you also have a voltmeter with a 40 volt linear range for DC. you also have a 9 volt DC power supply fed by mains. you charge each capacitor to 9 volts with terminal 1 being - (negative) and terminal 2 being + (positive). you connect the voltmeter to terminal A2 and to terminal B1. does it read any voltage? can - of one capacitor discharge + of the other?

 

[.Scott]

Are you also connecting A1 and B2 together?
If it's not a full electrical circuit, the voltmeter will likely read zero.

Also, this should probably be put into "homework help" form.

 

[.Scott]

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.

 

[Skaperen]

there is no A1-B2. i am wondering if a charged capacitor can be treated as having an electrostattc charge. not the whole capacitor, just one terminal on each (A2 and B1).

 

[.Scott]

there is no A1-B2. i am wondering if a charged capacitor can be treated as having an electrostatic charge. not the whole capacitor, just one terminal on each (A2 and B1).

Any atomic material can carry an electrostatic charge. Each side of the capacitors you described will have charge - so there are four sides A1, A2, B1, and B2. Since these are rated at 12V, the differences between A1 and A2 will normally be no greater than 12V - and so with B1 and B2.

In my previous post, I described an "ideal" voltage meter. For the situation you have described, a meter could be made that is close enough to ideal to show the effects I will describe:
1) You charge both A and B to 9V. Since this is a 0.1F capacitors, these are actually very substantial charges. If you were to accidently drop a piece of aluminum foil across the terminals, the result would be explosive with the rapid evaporation of some of the aluminum.
2) You set both capacitors down onto a glass plate in a room with a relative humidity of 0% and measure the voltages A1-A2 and B1-B2. Both are 9V.
3) Before you make the next measurement, let's think about the charges for a moment. In the simplest case, each capacitor consists of two conducting elements. There could also be a conductive "can" holding them, and to keep things simple, we will presume that the "can" is connected to the terminal 1. Each of those two elements has some charge value which represents the number of excess or deficient electrons. As these two capacitors sit on the glass plate, there is a can-to-can (A1-B1) capacitance. But this is such a small capacitance (I said 1 fF earlier) that your typical meter would be unable to measure it.
4) So now you pull out your specially engineered voltmeter. This is a very high-impedance, low-capacitance device and is therefore very susceptible to ESD damage. You would probably need about 30 minutes of study to get good readings from it without destroying it. We will give it a voltage rating of 2KV. When you connect it from A1 to B2, it reads 9V. Between B1 and A2, it also reads 9V. Any normal meter would read 0V - but normal meters act as conductors across such trite charges.
5) You momentarily allow B1 to come into contact with a sweater which immediately draws some electrons away from it.
6) Your take new readings: A1 to A2: still 9V; B1 to B2: still 9V; A1 to B1: 1000V; A1 to B2: 1009V; A2 to B1: 991V.
7) Now you move one of the capacitors to a new location on the glass plate. You have not changed the charge, but the capacitance is down to 0.2fF.
8) You repeat the measurements and destroy your meter in the process. (ooops)