How much charge does the battery deliver?

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Homework Help Overview

The discussion revolves around calculating the charge delivered by a battery in a circuit involving capacitors in series. Participants are exploring the concept of equivalent capacitance and its application in determining charge using the formula Q=CV.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method for calculating equivalent capacitance for capacitors in series and question the correct application of the formula. There are attempts to clarify misunderstandings regarding the reciprocal relationship in series capacitance and the implications for charge calculations.

Discussion Status

There is ongoing clarification regarding the calculation of equivalent capacitance, with some participants providing corrections and guidance. Multiple interpretations of the problem are being explored, particularly around the physical understanding of voltage and charge in capacitors.

Contextual Notes

Participants note previous mistakes in similar problems, indicating a potential pattern in misunderstandings. There is a discussion about the terminology used in describing charge on capacitors, suggesting a need for clearer conceptual understanding.

Kelsi_Jade
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Homework Statement


upload_2017-3-28_1-26-56.png


Homework Equations

/ 3. The Attempt at a Solution [/B]
Here is what I attempted for the HW. I only got 2/10 for this here so I'm wanting to correct my mistakes.
upload_2017-3-28_1-25-22.png
 
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How do you get the equivalent capacitance of capacitors in series?
 
1/Ceq = 1/C1 + 1/C2 = 5/6mF
 
Kelsi_Jade said:
1/Ceq = 1/C1 + 1/C2 = 5/6mF
It's 6/5 mF (you forgot to take the reciprocal).
You have added them directly in your attempt.
 
So then the circuit becomes one with one capacitor Ceq= 5/6mF
And we know Q=CV
So, Q=(5/6mF)(10V) = 25/3 mC = 8.33x10^-6F?
 
cnh1995 said:
It's 6/5 mF (you forgot to take the reciprocal).
You have added them directly in your attempt.
Whoops! So I did.
So Ceq=6/5mF and then plug that into the Q=CV equation?
 
Ceq = 1/(1/C1 + 1/C2) = C1*C2/(C1+C2)

nevermind, you picked it up, and I can't delete this
 
Kelsi_Jade said:
Whoops! So I did.
So Ceq=6/5mF and then plug that into the Q=CV equation?
Yes.
 
I see that in at least two problems in a short interval you make a similar stumbles, though you finally get it right. But since you seem to think in formulae and you forget them, even if you get it right now in six or 12 months time when you need it you're likely have forgotten and could stumble again. Cure IMO is to think physically not formulaically.

Physically, in the nature of what voltage is, voltages across elements in series add up (be the elements resistors, capacitors or inductors). Whereas for elements in parallel the voltage across them is the same for each almost by definition. By the nature of charge electric charge, the charge on capacitors in parallel add up to total charge, while capacitors in series each have the same charge.

I blame a lot of stumbles of students on the misleading slangy phrase "charge on a capacitor". Creating a charge - well actually you can't create any - but separating electrical charges is energetically expensive. So a capacitor does not have a charge. What it has is a charge separation – equal and opposite positive and negative charges close to each other, typically only a micrometer distant from each other, so overall neutrality. Known, elementary and obvious, maybe but I think students need to actively remind themselves when doing problems in order not to get confused.
 
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Thank you guys! I really appreciate all of the help, I know there are a few of you that have commented a few times on my posts - It's been a big help!
 

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