- #1

member 731016

- Homework Statement
- Please see below

- Relevant Equations
- Please see below

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!

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!