Circuits with capacitators connected parallel

In summary, when C1 (with a capacitance of 0.05 μF) is connected to a 0.4 V battery and fully charged, and then disconnected and connected to C2 (with a capacitance of 0.1 μF), the total charge on both capacitors will be distributed evenly due to the parallel connection. The final potentials of both capacitors will also be the same. Using the equation Q = CV, the charge on each capacitor can be calculated by multiplying its respective capacitance by 0.2 V. The sum of the charges on both capacitors will remain the same as the initial charge on C1. Therefore, the percentage of charge transferred from C1 to C2 is 50
  • #36
mbkau00 said:
Thanks for your help and persistence.

Unfortunately I don't get an answer near 70% which is the correct answer. I think i must admit defeat, considering time spent.
Thanks again :)

Well, you're very close to the final formulas now. What did you arrive at for the charge on C2?
 
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  • #37
1.5*10^-7Q1 = 1*10^-15 - 5*10^-8Q1
Q1 = 6.67*10^-9

2*10^-15 - 1*10^-7Q2 = 5*10^-8Q2
Q2 = 1.33*10^-8

I am honestly as lost as when I started as I don't have a definitive approach in why certain variables are, or aren't related.
 
  • #38
mbkau00 said:
1.5*10^-7Q1 = 1*10^-15 - 5*10^-8Q1
Q1 = 6.67*10^-9

2*10^-15 - 1*10^-7Q2 = 5*10^-8Q2
Q2 = 1.33*10^-8

I am honestly as lost as when I started as I don't have a definitive approach in why certain variables are, or aren't related.

Well, your values for Q1 and Q2 look fine!

Perhaps a crude analogy would help. You have two containers with different dimensions (capacities). One contains a certain volume of water (charge) while the other is empty. You want to know how much of the water needs to end up in each container so the water will reach to the same height (voltage) in each. Water volume is conserved -- it doesn't disappear when you move it around. Charge is likewise conserved.

For the capacitor the underlying relationship is Q = C*V. For the containers you'd have something like Volume = A*h, where A is the area of the base of the container and h the height of the fluid.

For the fluid case, if the heights of the fluid is the same then V1/A1 = V2/A2 and V1 + V2 = V, the total volume of fluid you're working with. The similar equations you arrived at for the capacitors are Q1/C1 = Q2/C2 and Q1 + Q2 = Q.

When solved the expressions for Q1 and Q2 have the simple form:

##Q1 = \frac{C1}{C1 + C2} Q ##

##Q2 = \frac{C2}{C1 + C2} Q ##

So each capacitor gets a fraction of Q that is equal to that capacitor's portion of the total capacitance.
 
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  • #39
Your analogy definitely adds a great perspective. Thanks so much.

I still can't see how the answer is 70% of charge went from C1 to C2.
 
  • #40
mbkau00 said:
Your analogy definitely adds a great perspective. Thanks so much.

I still can't see how the answer is 70% of charge went from C1 to C2.

Remember that C1 started out with charge Q (it ended up with charge Q1). Make sure that you're comparing Q2 with Q.
 
  • #41
Thank you.
 

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