Finding charge on a capacitor given potential difference across two points

AI Thread Summary
The discussion focuses on calculating the charge on capacitors in a series configuration. The total charge is determined to be 363 µC, with half of that charge (181.5 µC) assigned to the upper capacitor. The voltage across the upper capacitor is calculated to be 21V, leading to a stored charge of 42 µC for another capacitor in the circuit. Clarifications are made regarding the notation for charge labels, emphasizing that capacitors in series share the same charge. The importance of recognizing this principle in series circuits is highlighted.
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Homework Statement
Please see below
Relevant Equations
Please see below
For this part(b) of this problem,
1675993341801.png


The solution is
1675993422576.png

However, I tried solving (b) like this:

Since ##Q_{total} = 363 \times 10^{-6} C## then ##Q_1 = 181.5 \times 10^{-6} C ## since the equivalent upper capacitor is in series with the equivalent bottom capacitor so should store the same amount of charge.

Since ##C_{upper} = 8.67 \times 10^{-6} C## then voltage across upper equivalent capacitor is ##\frac {181.5}{8.67} = 21V ## then charged stored by ##C_3## is ##Q_3 = 2 \times 10^{6} \times 21 = 4.2 \times 10^{-5} ##

I don't understand why they use the total charge for the upper capacitors when they only store half the charge.
Many thanks!
 
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What do you label as Q1?
 
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nasu said:
What do you label as Q1?
Thank you for your reply @nasu!

I'm not sure, just some notation for the upper equivalent capacitors charge.
 
Then this is already labeled as Q total. This is the charge on the upper group of capacitors. And the same charge is on the lower group, which is in series with the upper group.
.
 
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nasu said:
Then this is already labeled as Q total. This is the charge on the upper group of capacitors. And the same charge is on the lower group, which is in series with the upper group.
.
Thank you for your reply @nasu! I forgot capacitor in series have the same charge!
 
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