# Finding charge on a capacitor given potential difference across two points

• ChiralSuperfields

#### ChiralSuperfields

Homework Statement
Relevant Equations
For this part(b) of this problem, The solution is 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!

What do you label as Q1?

• ChiralSuperfields
What do you label as Q1?

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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• ChiralSuperfields
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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Thank you for your reply @nasu! I forgot capacitor in series have the same charge!