# Tricky Capacitors Question

• [sauce]
In summary, the initial charge on the 25 uF capacitor is 1250 uC and on the 40 uF capacitor is 2000 uC. When they are connected, there is a movement of charge until the net charge is 750 uC. The final charge on the 25 uF capacitor is 288 uC and on the 40 uF capacitor is 426 uC. The potential difference across the 40 uF capacitor is 11.6 V. The capacitors are connected in parallel, and the voltage on each capacitor is constant. The electrons move from the negative plate to the positive plate, neutralizing the charges and resulting in an equal voltage across both capacitors.

#### [sauce]

A 25 uF and a 40 uF capacitor are charged by being connected across separate 50 V batteries. The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge on each capacitor, and what is the final potential difference across the 40 uF capacitor?

I know capacitance is given by C = Q/V.

1) I find the charge initially on the 25 uF and 40 uF capacitors (call them C1 and C2, respectively) to be 1250 uC and 2000 uC.

2) The plates are now connected, so there must be some movement of charge.

3) I take the difference of these charges (for some reason I don't know) and find Qtotal = 750 uC. This is apparently the net charge. Why?

4) Since the voltage is constant on both (again, why?), I find
Q1/C1 = Q2/C2 => Q1 = (C1/C2)Q2
Qtotal = Q1 + Q2 = (C1/C2)Q2 + Q2 = Q2(C1/C2 + 1) => Q2 = Qtotal/(C1/C2 + 1) = 426 uC.

5) Therefore Q1 = Qtotal - Q2 = 288 uC.

6) The potential difference across the 40 uF capacitor (C2) is then Q2/C2 = 11.6 V

In particular, I don't understand how we found the net charge. I don't understand why we set the voltage of each capacitor equal to each other in step (4) but then find something different in step (6)

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Draw a picture of the connected capacitors. What is common for both of them?

ehild

Is it the voltage? I can't tell if this combination is in series or in parallel.

[sauce];3607468 said:
Is it the voltage? I can't tell if this combination is in series or in parallel.

It's pretty easy to determine what kind of combination it is if you just draw a diagram as ehild suggested.

I've discussed this problem with my teacher briefly, she says it's in parallel. How do you know?

Both plates are connected, positive to the negative so the charges can move and partly neutralize . How much net charge remains on the connected plates?
The voltage is constant on both. The capacitors together constitute an equivalent new one, with some charge on the plates. What is the charge of this resultant capacitor?

ehild

In case of series connection, one plate of each capacitor is connected, the other plates are connected to something else. The charge is the same on both capacitors.
When connected in parallel, both plates of each are connected. The voltage is the same across both capacitors.

ehild

[sauce];3607487 said:
I've discussed this problem with my teacher briefly, she says it's in parallel. How do you know?

Erm, it is NOT a parallel connection. See the attached file.

#### Attachments

• caps.jpg
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My teacher insisted it was in parallel. Can you explain to me the unclear parts of the solution?

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The negative charges are free electrons, the positive charges are ions with missing electrons. When the plates are connected, the electrons are attracted by the positive ions, go there, and neutralize them: they will be neutral atoms.

Draw a picture and then I explain the other things tomorrow if you still need it. Now I go to sleep.

ehild

cepheid said:
Erm, it is NOT a parallel connection. See the attached file.

It is no sense of series or parallel when only two capacitors are connected. Series and parallel is defined with respect to the other parts of a circuit.

ehild

Thank you for your input. I do need help today however. Can anyone else help explain to me the solution though?

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ehild said:
It is no sense of series or parallel when only two capacitors are connected. Series and parallel is defined with respect to the other parts of a circuit.

ehild

Not totally sure whether I agree or not, but fine. I'll rephrase what I said: the connections are not as shown on my left diagram. The connections are as shown on my right diagram.

EDIT: and even with ehild's statement above, what the teacher said is still incorrect, which is the point I was trying to get across.

Right. Can you at least explain then the solution I posted?

See my picture. At the beginning, the capacitors are charged, C1 with Q1 and C2 with Q2. The blue dots are excess electrons, the red circles are positive ions, missing one electron. Q1=8, Q2=5.

The plates connected, plus to minus. The electrons are attracted by the positive ions, and as they can move freely through the wires, they go to the other side and happily unite with the ions making neutral atoms, until the driving force vanishes. The process is illustrated in the figure on the right. The electrons stop moving when the driving force vanishes, that is the connected plates are at equal potential. At the end, the 5 electrons neutralized 5 ions and 3 positive ions remained on one side altogether and Q1-Q2=3 negative electron on the other side. These charges distribute between the connected plates to result in equal voltage across the capacitors: U=q1/C1=q2/C2 , where q1+q2=3 (the figure below the other two).

@cepheid: What you draw, that one plate is negative, the other positive is only at the instant when the plates are connected. In equilibrium, both connected plates are either positive or negative, according to your first drawing.

ehild

#### Attachments

ehild said:
@cepheid: What you draw, that one plate is negative, the other positive is only at the instant when the plates are connected. In equilibrium, both connected plates are either positive or negative, according to your first drawing.

ehild

Understood and accepted! I was actually planning on drawing out something similar to what you just did, but now I don't have to. If I had done so, I would have realized my mistake.

Drawing them side by side like you did also helps illustrate why the voltages on them have to equalize after they are connected. If they aren't, then the cap with the higher voltage will drive charges onto the other one until the two voltages are equal.

cepheid said:
Understood and accepted! I was actually planning on drawing out something similar to what you just did, but now I don't have to. If I had done so, I would have realized my mistake.

I am pleased cepheid said:
Drawing them side by side like you did also helps illustrate why the voltages on them have to equalize after they are connected. If they aren't, then the cap with the higher voltage will drive charges onto the other one until the two voltages are equal.

You see why a sketch is essential. It tells more than a hundred words...

Now think what happens if those charged capacitors are really connected in series to a battery, but having net charge at the common plates. Usually the series connected capacitors are neutral at the beginning, so the net charge on the common plate is zero. Connecting the voltage source, the positive and negative charges separate, but their sum stays zero. What are the voltages and charges of the capacitors if there is a net charge at the junction?

ehild