# Capacitors - Finding the charges and potential at junction

1. Jan 26, 2014

### Pranav-Arora

1. The problem statement, all variables and given/known data
Three capacitors of 2μF, 3μF and 5μF are independently charged with batteries of emf's 5V, 20V and 10V respectively. After disconnecting from voltage sources, these capacitors are connected as shown in figure with their positive polarity connected to A and negative polarity earthed. Now a battery of 20V and an uncharged capacitor of 4μF capacitance are connected to junction A as shown with a switch S. When switch is closed, find:

a)the potential of the junction A.

b)final charges on all four capacitors.

(Attachment 1)

2. Relevant equations

3. The attempt at a solution
Honestly, I have no clue what happens to the initial charges of capacitors when the switch is closed. I tried to solve the problem as if there are no charges on capacitors initially. Attachment 2 shows the simplified circuit I made. I found the potential at A and got 40/7 V which is incorrect, so I think there is some role of the initial charges which I am unable to figure out. :(

Any help is appreciated. Thanks!

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• ###### simplified circuit.png
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2. Jan 26, 2014

### SammyS

Staff Emeritus
Let's display those circuits as images.

How much charge is on each capacitor immediately after charging?

Once the three capacitors are connected, they are in parallel with with their positive plates connected by conductors. What is the total charge on their positive plates?

3. Jan 26, 2014

### Pranav-Arora

Hi SammyS! Nice to see you. :)

The charges on 2μF, 3μF and 5μF are 10μC, 60μC and 50μC respectively.
The total charge is 120μC but I don't see how calculating the charges helps.

4. Jan 26, 2014

### ehild

You can determine the potential before closing the switch from that 120 μC.

What happens when the switch is closed? What happens in a very short time after closing the switch? Does the charge stay on the parallel connected capacitors?

ehild

Last edited: Jan 26, 2014
5. Jan 26, 2014

### Pranav-Arora

Hi ehild! :)

How?

I have no clue about this but I guess the charges won't stay on the plates. They should change due to battery.

6. Jan 26, 2014

### ehild

The three capacitors are connected in parallel, and you know the charge of the resultant capacitor, so you can determine the voltage. But the problem does not ask that.

Yes, the charge on the parallel connected capacitors won't stay there. The circuit is grounded, and there is the battery with fixed emf. The charge can flow through the battery and into or out of the earth .

ehild

7. Jan 26, 2014

### Pranav-Arora

Does that mean that there is no role of the initial charges and I can simply go about solving the circuit as if there were no initial charges?

I am honestly lost. :(

8. Jan 26, 2014

### ehild

I am not sure, either, and without the grounding, the situation would be different.
What would be the potential of A and the charges on the capacitors if it was an ordinary circuit without initially charged capacitors?

What other possibility can you think of?

ehild

9. Jan 26, 2014

### Pranav-Arora

40/7 V would be the potential of A. The charge on 4μF would be 400/7 μC.

I can have one of them uncharged initially and others charged, do you ask me this kind of possibility or did I misinterpret your question?

10. Jan 26, 2014

### ehild

What would be the situation if the initial charge stayed on the three capacitors?

ehild

11. Jan 26, 2014

### Pranav-Arora

With switch S open, the charges on three capacitors would rearrange because the three capacitors are at a different potential from each other.

12. Jan 26, 2014

### ehild

Of course, the charges of the three connected capacitors rearrange, so the potential difference is equal across all of them. You can treat them as a single capacitor, with charge Q=120μC. Now you connect a battery and a 4 μF capacitor across. What will be the potential of A?

You should note that charge can flow through the battery, but can not accumulate in it. So the sum of charges on the left plates of the capacitors is equal to Q=120 μC, x on the lower capacitor and Q-x on the upper one. The right plates will be negatively charged, as the ground provides charge.
Between the ground and point A, the potential changes by equal amount both across the upper capacitor and along the lower branch of the circuit.

ehild

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• ###### charcap.JPG
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13. Jan 26, 2014

### Pranav-Arora

The potential change in the lower branch is: $20-x/4$.
The potential change in the upper branch is: $(Q-x)/10$.

Equating them doesn't give me the correct answer. :(

If I write the potential change in the lower branch as $20+x/4$ and equate, I do get the right answer.

Is this because we assume x to be positive (though it turns out to be negative)?

Also, I don't get why did we solve it this way. Why is it wrong to simply solve it as if there were no charges initially? What would I do if the circuit was not grounded?

14. Jan 26, 2014

### ehild

Why do you write 20-x/4? Supposing the left plates of capacitors have the charge and the right plates have the opposite charge, UA=x/4+20, UA=(Q-x)/10.
x happens to be negative - it does not change anything.

The "usual" solution assumes series connection of the capacitors, with equal charges on them which is not true in this case. The initially given charge can not disappear, it only can be distributed between the connected plates.

ehild

15. Jan 26, 2014

### Pranav-Arora

Sorry for asking foolish questions but the circuit is earthed, the charge can flow into the ground, right?

Please look at the attached problem. This is a problem from the exam I gave last year.

When switch $S_1$ is pressed and released, the charge on the upper plate of $C_1$ is $2CV_0$. When switch $S_2$ is pressed and released, the charge redistributes between $C_1$ and $C_2$ leaving each of the capacitor's upper plates with charge $CV_0$.

I remember that during the exam, I selected D as one of the correct answers. And D (with B) is the correct answer but where does the charge go? I mean, on the upper plate of $C_2$, the charge was $CV_0$ before $S_3$ was pressed. When $S_3$ is pressed for long time, then according to the given answers, the battery puts new charges on the plates. What happened to the charges that were present before $S_3$ was pressed?

#### Attached Files:

• ###### JEE problem, paper 1, 15.JPG
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16. Jan 26, 2014

### SammyS

Staff Emeritus
from Post #7:
I'm going back to your question in post #7. Things get muddled after that.

When the switch is open, the battery has no effect on the charges on or voltage across the three initial capacitors.

What is the effective Capacitance of the three capacitors in parallel ?

What is the voltage that an equivalent capacitor with a charge of 120 μC would have across it?

17. Jan 26, 2014

### ehild

The "charge" is not particle, it is an attribute of a piece of material. Usually, materials are electrically neutral, and become charged if they get excess charge - electrons removed or added. A metal piece consists of positive ions and free electrons. The capacitor plate is charged to some positive q - that means it misses some electrons. If you add -q charge (electrons), the plate becomes neutral, all the ions have enough electrons around them. If it is a semiconductor, we say that the added electrons recombine with the holes.
If you add more electrons, the plate becomes negatively charged.
C2 was connected to a battery of opposite polarity. The positive plate got negative charge, the negative plate got positive charge. The added charge neutralized the previous one, and provided excess charge of opposite polarity.

What happens on one plate of the capacitor, the opposite happens on the other plate.

ehild

Last edited: Jan 26, 2014
18. Jan 27, 2014

### Pranav-Arora

Thanks ehild for the explanation! :)

So the charge stored or present in the battery is responsible for the new charges on plates, am I right in saying this?

I am still not sure about the no earthing case. Can I solve it by assuming as if there were no intial charges?

Hi SammyS!

The effective capacitance is 10μF and voltage is 12V.

19. Jan 27, 2014

### Staff: Mentor

I don't know if it will help you at this stage of the game, but note that for analysis purposes an initially charged capacitor can be represented by a model consisting of an uncharged capacitor of the same size in series with a voltage supply with a potential difference equal in value to the initial potential across the capacitor.

You can then use all the usual circuit rules to combine sources, etc., to simplify the circuit and determine any change in charge that will occur on the capacitor.

#### Attached Files:

• ###### Fig1.gif
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20. Jan 27, 2014

### ehild

The battery separates the opposite charges, otherwise the whole battery is neutral. The electromotive force of the battery drives positive charges to the three connected capacitors and the same amount of negative charge to the 4μF one.

Sorry, I misunderstood something. I did not read the original problem carefully. You get the same result either with earthing or without it. For the non-earthing case you can also apply Gneill's method to find the final charge of the 4 uF capacitor, assuming zero initial charges, but an extra battery.

What I thought was that charge was given only to one plate of all capacitors at the beginning (I saw such problem just about the same time https://www.physicsforums.com/showthread.php?t=734124). Here is a question for you to think about: What is the voltage you measure across the capacitor plates if you give Q charge to one plate, without earthing the other plate? And what is the voltage if the other plate is earthed?

Reading the original problem, the capacitors were charged by connecting them to voltage sources, so one plate got some Q charge, the other plate got -Q charge. The same as if the other plate would have been connected to earth.

ehild

Last edited: Jan 27, 2014