# Capacitor charged then disconnected

1. Mar 11, 2013

### Woopydalan

Hello,

I am confused as to what happens when a capacitor is charged by a battery, disconnected, and then connected to a second uncharged capacitor.

Is the situation different for whether the two capacitors are connected to each other in series or parallel?

If in series, shouldn't the charges be equal on both capacitors, and if connected in parallel, they are both the same voltage.

Of course, I was thinking is there such a thing as connecting the capacitors in series or parallel if they are the only part of the circuit, meaning that connecting something in series or parallel is exactly the same thing if there is no battery part of the circuit.

2. Mar 11, 2013

### Staff: Mentor

Indeed, I can't imagine more than one way of connecting the capacitors into a single circuit.

Well, you could take both leads from one capacitor and connect them to a single lead of the other capacitor, but that (a) doesn't make a complete circuit that contains both capacitors, and (b) instantly "shorts out" the first capacitor.

3. Mar 11, 2013

### DrZoidberg

As you already noticed, without the battery the two capacitors are in parallel and also in series at the same time. And since they are in parallel their voltage will be the same. The charge however isn't necessarily equal. The rule that capacitors in series have the same charge only applies if both capacitors were originally uncharged.

4. Mar 11, 2013

### rcgldr

Assuming no leakage or losses, I'm wondering if potential (voltage) should be conserved for both cases, when one plate from each capacitor is connected (series) or when both plates of each capacitor are connected (parallel).

5. Mar 11, 2013

### TurtleMeister

Assuming the capacitors are not polorized there should only be one way to connect them together. Your question about series or parallel makes no sense. Or maybe I'm misinterpreting your question. Whatever the problem you're having, this webpage may help clear some things up for you. It explains how energy is lost in the process of charging or discharging a capacitor.

http://www.smpstech.com/charge.htm

6. Mar 11, 2013

### rcgldr

In the diagram from the web page you linked to:

http://www.smpstech.com/charge.htm

Diagram (A) show the capacitors already connected in "series". If the switch in (A) included a circuit, then connecting up that circuit as shown in (A) would be using the capacitors in series. Getting back to what I think is the original question, using diagram (A), imagine that intially there is no connection between the capacitors, such as an open switch at the bottom of the diagram, the capacitor on the left is charged, and the capacitor on the right is uncharged. When the bottom connection is made, the bottom plates of both capacitors are connected, and depending on the initial states (charge bias) of the two capacitors, charge could be transferred. For example, the uncharged capacitor could have have any charge on both plates, as long as the charges were equal. For the charged capacitor, the only known is that the difference in charge x distance corresponds to it's voltage, but the actual amount of charge on each plate is unknown. In this circumstance, regardless of the initial charged state, I suspect that the total voltage after connecting the bottom two plates of the capacitors will remain constant.

Diagram "B" shows the parallel case, where voltage is conserved, which what I thought should happen. This can get complicated if the capacitors are unlike, such as a different distance between plates. To use the capacitors as a "parallel" source, you'd need to connect a circuit in parallel to the top and bottom halfs of the loop shown in diagram (B).

7. Mar 11, 2013

### Woopydalan

Ok here is why I am finding myself confused.

I had a homework problem stating

A 10.0 uF capacitor is charged to 15.0 V. It is next
connected in series with an uncharged 5.00 uF capacitor.
The series combination is finally connected across a 50.0-V
battery, as diagrammed in Figure P26.62. Find the new
potential differences across the 5 uF and 10 uF capacitors.

Well, the dang problem said that they are connected in series. This would imply that there is such a thing as connecting them in parallel if they would differentiate, so I figured that there were two different ways.

By the way I'm not asking this as a homework question, I was just pointing out that the problem stated that it is then connected in series, with the implications previously stated.

Also, I thought that being in series meant they had the same charge, but apparently that rule only applies if they were both initially uncharged (not mentioned in the textbook).

8. Mar 11, 2013

### the_emi_guy

I think what they mean here is that the capacitors first get connected with a single connection, i.e. no "circuit" yet. Then, second, the 50V battery connects the open ends of the caps to complete the new circuit.

9. Mar 11, 2013

### Woopydalan

That is correct, emi guy. How did you abstract that information from the way the question was worded? It sounds to me like you make a circuit with the two capacitors, let charges move around, then plug it to the 50 V battery.

10. Mar 12, 2013

### smatik

I think you're saying that we first charge a capacitor ' A' with battery 'B'. then we wait 5 secs for the capacitor to get charged and then we throw the battery 'B' and place another capacitor 'C' in its(B's) place,with our insulating gloves.
I think the charge would be conserved and potential difference across them should be same. By this we get two equations and can find charge stored on both the capacitor(in final situation) in terms of their capacitance and potential of battery

11. Mar 12, 2013

### Woopydalan

Sorry for a foolish question.

But I'm looking at 3 capacitors in series that have a charge of 150 uC each. When you make an equivalent capacitor, the charge on the equivalent capacitor is 150 uC. Why isn't it 450 uC? Where did those other 300 uC of charge go?

12. Mar 12, 2013

### rcgldr

At the instant the capacitors are connected with a single connection, there could be a brief moment of current depending on the charge bias of the capacitors. If the "uncharged" capacitor had a postive charge bias (postive and equal charge on both plates), and a single connection was made to the other capacitor to a plate with a negative charge, then there would be a brief moment of current between those plates. Depending on the charge of the other pair of plates, the equilbrium state of the two connected plates may be one where the charge on both plates is not the same, due to attraction or repulsion of charge due to the other pair of plates.

The issue is that after the single connection, connecting the pair of capacitors to a circuit could end up using the capacitors in series or in parallel.

Assuming the 3 capacitors are identical, the equivalent capacitor should be 50uC, with triple the voltage.

Wiki article:

http://en.wikipedia.org/wiki/Capacitor#Networks

Last edited: Mar 12, 2013
13. Mar 12, 2013

### TurtleMeister

Wow, I do not know how to solve that. If I have more time later I may try to figure it out. What book is it in?
The charge in a capacitor is between the two plates. Wouldn't you have to have a circuit between those two plates to change it? I guess there could be a very very brief current due to capacitance around the outside of the capacitor, but it would normally be so small as to be undetectable.

14. Mar 12, 2013

### rcgldr

I meant the actual charge on the plates. Imagine the plates are large enough and can hold enough charge that you can create a static spark just by bringing two plates with different amounts of charge close to each other. A sudden connection between plates of differing charges could involve some brief moment of current flow, depending on the charges on the other pair of non-connected plates.

Last edited: Mar 13, 2013
15. Mar 13, 2013

### DrZoidberg

Initial situation:
C1 = 10µF
V1 = 15V
Q1 = 10µF * 15V = 150µC
C2 = 5µF
Q2 = 0µC

Final situation:
V = 50V
V1 + V2 = V
Q1 = Q2 + 150µC
V1 = 15V + Q2/10µF
V2 = Q2/5µF

=> 50V = 15V + Q2/10µF + Q2/5µF
50V - 15V = 3Q2/10µF
350µC = 3Q2
Q2 = 116.67µC
Q1 = 266.67µC
V2 = 116.67µC/5µF = 23.33V
V1 = 266.67µC/10µF = 26.67V