# [Physics] Using a charged capacitor to charge two others

capacitanceelectrostaticshomework-and-exercises

Here is the homework question in question:
The figure (below) displays a 13.1 V battery and three uncharged capacitors of capacitances C1 = 4.08 μF,C2 = 6.19 μF and C3 = 3.30 μF. The switch is thrown to the left side until capacitor 1 is fully charged. Then the switch is thrown to the right. What is the final charge (in C) on capacitor 3?

Here is what I did: I first found the final charge on C1 before flipping the switch right, by doing Q=CV and I got 53.488 microcoulombs. Easy

Then I assumed that after the switch is flipped, 2q coulombs of charge have flowed out of the top plate of C1, and that you end up with a charge q+a on C2 and q-a on C3, so that total charge is conserved where q and a are unknown charges. Since at the end the potential of C2 and C3 must be equal, I wrote

$\frac{q+a}{6.19}=\frac{q-a}{3.3}=V=Q/C$

I solved for a and found that a=.3045q, so that C2 will attain a charge of 1.3045q and C3 will attain a charge of .6955q.

To find q, I did the same thing but with C1 and C2:
$\frac{1.3045q}{6.19}=\frac{53.488-2q}{4.08}$ and found that q=18.7 microcoulombs, and then some arithmetic gave me the answer: C1 has a charge of 16.08 microcoulombs, C2 24.39 microcoulombs, and C3 13 microcoulombs. However, my Wiley says its wrong. Confirm that the voltage difference across all of the capacitors is 3.94 V.

You're answer cannot be correct since it does not satisfy KVL.

For your final charge solution, the final voltages across the capacitors are given by

$$V_{C1} = \frac{16.08}{4.08}\mathrm V = 3.94\mathrm V$$

$$V_{C2} = \frac{24.39}{6.19}\mathrm V = 3.94\mathrm V$$

$$V_{C3} = \frac{13}{3.3}\mathrm V = 3.94\mathrm V$$

But, by KVL, the voltages must satisfy

$$V_{C1} = V_{C2} + V_{C3}$$

so, back to the drawing board!