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Homework Statement
The capacitors in the ﬁgure are initially uncharged and are connected as in the diagram (attached). Then switch S1 closed and switch S2 is left open.
After a long wait, what is the magnitude of the potential diﬀerence [tex]V_{cd}[/tex]? Answer in units of V.
Homework Equations
[tex]C = \frac{Q}{\Delta V}[/tex]
Effective Capacitance of Capactors in Series: [tex]\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}[/tex]
Effective Capacitance of Capacitors in : [tex]C_{eq} = C_{1} + C_{2}[/tex]
The Attempt at a Solution
From the diagram, we can see that the two capacitors on top and the two on both are each in series. Therefore, using the formula for capacitors in series, the effective capacitance of the two top is 1.84e5, and the same for the two on the bottom.
The top and bottom capacitors then form a  series, so we can add their capacitances to determine the total effective capacitance of the circuit = 3.68e5.
We know that the voltage of the battery is 137V and we know the effective capacitance of the circuit, therefore we can calculate that Q = C/\V = 0.0050416 C.
We can then use the formula [tex]C = \frac{Q}{\Delta V}[/tex] to calculate the potential at c and d. C is at the end of a 92 microF capacitor, so using the formula, /\V = 54.8V. D is at the end of a 23 microF cap., so /\V = 219.2V. Therefore, [tex]V_{cd} = V_{d}  V_{c} = 219.2  54.8 = 164.4V[/tex]. However, this answer is wrong.
I'm not sure where I went wrong in this problem. I believe I solved the effective cap. of the circuit correctly, but maybe my calculations for the potential at c and d were erroneous. Any direction in solving this problem would be really appreciated. Thanks!
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