Charge and stored energy for capacitors

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Homework Statement


Two capacitors (C1 = 3.4 uF, C2 = 17.5 uF) are charged individually to (V1 = 19.2 V, V2 = 6.6 V). The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.

A. Calculate the final potential difference across the plates of the capacitors once they are connected.

B. Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together.

Hint: Pick one capacitor, you know the charge on this capacitor before they were connected. Now that you know the potential difference after they are connected - remember the potential drop is the same for both of them - you can calculate the charge: Q=CV. You have to calculate the difference of Q before and after.

C. By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

Homework Equations


C(total)=C1 + C2
C=Q/v


The Attempt at a Solution


I found part A by adding the capacitances and the charges together, and got 8.65 V. I've tried several different things for part B, but I just am unsure exactly what the hint is saying. I subtracted one charge from the total charge, but that seemed (and was) wrong.

Any guiding lights? Much appreciated, guys.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
collinsmark
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I found part A by adding the capacitances and the charges together, and got 8.65 V.
'Looks like the right answer to me. :approve:
I've tried several different things for part B, but I just am unsure exactly what the hint is saying. I subtracted one charge from the total charge, but that seemed (and was) wrong.
Pick one of the two capacitors (either one -- it doesn't really matter which). You've already calculated the amount of charge on that capacitor before they were connected together.

Now consider that instead of connecting them together, calculate how much charge would be on that capacitor if you merely change the potential to 8.65 V. What's the difference in the amount of charge?

To double check your work, do the same thing with the other capacitor.
 

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