Find the unknown capacity of two capacitors

In summary: Actually I felt a bit annoyed at the question-setter for that, but I suppose we should be used to that sort of thing.In summary, when considering two capacitors with unknown capacitances C1 and C2, connecting them in series and applying a voltage of 100 V results in a total charge of 1.5 · 10-4 C on the two positively charged plates. When the capacitors are switched to parallel and the same voltage is applied, the total charge on the plates is 4 · 10-4 C. By setting up and solving equations using the equation Q=C*V and considering the ratio of capacitances, the values of C1 and C2 can be determined. However, it is not
  • #1
Kirill Vlaksov
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Homework Statement



We consider two capacitors with the unknown capacitances C1 and C2. If
one connects the capacitors in series and applies a voltage of 100 V, finds
that two positively charged plates in total have the charge 1.5 · 10-4 C. If
one switches the capacitors in parallel and again applies a voltage of 100 V, finds
that two positively charged plates, in total have the charge charge 4 · 10-4 C. Which
capacities have the two capacitors?

Homework Equations


Q=C*V

Q=Charge
C=Capacitance
V=Voltage

The Attempt at a Solution


[/B] My attepmpt to solve this is attached in the photos bellow. In general , I was trzing to apply Q=C*V to the first and the second case and then tried to combine my answers
 

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  • #2
Hello.

I don’t think you are using the correct charge for each capacitor when they are in series. Read the second sentence of the problem statement carefully. Otherwise, your work looks good so far.
 
  • #3
It seems to me OK. You realize that in series the charge is the same on both capacitors; and that the potentials add up to the known totaL potential. The rest is an algebraic excercise. I would now express in terms of C1 and the ratio r of capacitances. You should be able to eliminate C1 and find two capacitances, but from these measurements you will not know capacitor has which capacitance.
 
  • #4
epenguin said:
It seems to me OK.

Hi, epenguin.

The problem states that when the two capacitors are in series, the "two positively charged plates in total have the charge 1.5 · 10-4 C".
To me, this is different than saying that "each positively charge plate has a charge of 1.5 · 10-4 C".

[If each positively charged plate has 1.5 · 10-4 C, then I get that there is no real solution to the set of equations for C1 and C2.]
 
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  • #5
TSny said:
Hello.

I don’t think you are using the correct charge for each capacitor when they are in series. Read the second sentence of the problem statement carefully. Otherwise, your work looks good so far.

epenguin said:
It seems to me OK. You realize that in series the charge is the same on both capacitors; and that the potentials add up to the known totaL potential. The rest is an algebraic excercise. I would now express in terms of C1 and the ratio r of capacitances. You should be able to eliminate C1 and find two capacitances, but from these measurements you will not know capacitor has which capacitance.

Thank you for your replies. I indeed missed a part concerning a total charge. Attached is the new calculations that I did. So in my calculations instead of 1.5 · 10-4 C i used 7.5 · 10-5 C ( which I got by deviding a total charge when capacitors were connected in series by 2 ). Then all calculations work out just fine. Thank you again for your help.
 

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  • #6
TSny said:
Hi, epenguin.

The problem states that when the two capacitors are in series, the "two positively charged plates in total have the charge 1.5 · 10-4 C".
To me, this is different than saying that "each positively charge plate has a charge of 1.5 · 10-4 C".

[If each positively charged plate has 1.5 · 10-4 C, then I get that there is no real solution to the set of equations for C1 and C2.]

You have to be right about that what the question means. I guess I had resistance in accepting that because in that case that part of the question seemed so artificially contrived - testing examsmanship :devil: rather than physics understanding. Why would anyone add those charges, to what physics, or what straightforward measurement would that correspond? The pair combines as one equivalent capacitance of 0.75 μF.

More constructively at least this made me realize something I had never heard of or thought, nor had occasion to: that two capacitors In series cannot have more than a quarter of the capacitance as the same two in parallel (the ≤ becoming = when the two capacitances are equali. Likewise two resistors in parallel cannot have more than a quarter the resistance of the two in series, which I don't remember having particularly heard of. )
1.5 is more than quarter of 4 so without needing to solve equations we should have been able to see immediately that the physically meaningful charge is not meant to be 1.5.
 
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1. What is the formula for finding the unknown capacity of two capacitors?

The formula for finding the unknown capacity of two capacitors is C = C1 + C2, where C represents the total capacitance and C1 and C2 represent the capacitance of the two individual capacitors.

2. How do I determine the capacitance of each individual capacitor?

The capacitance of each individual capacitor can be determined by using the formula C = Q/V, where Q is the charge stored on the capacitor and V is the voltage across the capacitor. This information can usually be found on the capacitor itself or in the manufacturer's specifications.

3. Can I use this formula for capacitors in series or parallel?

No, this formula only applies to capacitors in parallel. For capacitors in series, the formula is C = 1/(1/C1 + 1/C2).

4. What units are used for capacitance?

Capacitance is typically measured in Farads (F), but it can also be expressed in microfarads (μF) or picofarads (pF).

5. Are there any other factors that may affect the capacitance of a circuit?

Yes, the material and distance between the conductive plates of the capacitor can also affect the capacitance of a circuit. Additionally, the temperature and frequency of the current passing through the circuit can also impact the capacitance.

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