Steven W. answered • 03/17/17
Tutor
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(4,304)
Physics Ph.D., college instructor (calc- and algebra-based)
Hi Megan!
I think you are on exactly the right track (as you may already know, if you have been able to check the answer for Q3). What you can do now is determine the potential difference across C3, using V = Q/C. Then, the difference between that voltage and the battery voltage must be the voltage dropped across either of the capacitors in parallel. Knowing the voltage across those capacitors and their capacitance, you can calculate the charge on each using Q = CV, as you mentioned. I did not go into more detail because it sounds like you may just need a little push in the right direction to keep going, but, if you would like to talk about it in more detail, just let me know. Hope that helps out!
Steven W.
tutor
Once you know the charge on C3, you can determine V3, the potential across Capacitor 3, using V3 = Q3/C3.
Then, you can reason that whatever of the battery voltage is not dropped across Capacitor 3 must be dropped across the parallel combination of Capacitors 1 and 2. We can call that V12, and it is calculated by:
V12 = 1.52 V - V3
Since Capacitors 1 and 2 are in parallel, this same voltage mist be dropped across each one. Therefore, you now know the potential difference across Capacitor 1 and across Capacitor 2.
Knowing the potential difference across each of those capacitors, and knowing their capscitances, you can then calculate the charge on each one, using:
Q2 = C2*V12
Q1 = C1*V12
Just let me know where the challenges are for you in this description, and we can talk about it more.
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03/17/17
Megan L.
I get V3 to be the same as my battery voltage, 1.52 V. That would mean V12 is 0 which would make both Q1 and Q2 0 as well. Am I doing something wrong?
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03/17/17
Steven W.
tutor
I think what you have to watch out for there is that the equivalent capacitance of all three capacitors is very close to the capacitance of just Capacitor 3 (because we are adding inversely and Capacitor 3 has such a smaller capacitance than the combination of 1 and 2.
I calculated the equivalent capacitance of all three to be 0.296 μF, which is almost equal to the 0.3 μF of Capacitor 3 alone. From that, I got a charge for Capacitor 3 of 4.51 x 10-7 C, and if I compute the potential difference across Capacitor 3 from that, I get 1.50 V, leaving a small potential difference to be dropped across the other two.
So I think it is just a matter of keeping a few more significant figures, so you can distinguish those small differences. Let me know if that helps.
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03/17/17
Megan L.
Okay, thank you so much. In the end, I just had to use more significant figures.
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03/17/17
Steven W.
tutor
My pleasure! I'm glad that helped some.
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03/17/17
Megan L.
03/17/17