Question

A 1.50-µF capacitor charged to 40.0 V and a 2.40-µF capacitor charged to 15.0 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 1.50-µF capacitor?

Answer 1

The final configuration is actually in parallel than in series. This is because the wire makes the attached components equipotential.



Looking at this diagram of how things are connected, the charges will distribute in such a way that potential are the same and not equally distribute themselves. However, since the positive is connected to the negative of the other plate, some charges will be neutralized. Due to the symmetry of the system, The potential at V = -U.

First I have to find the charge deposited on each plate. Then I will filled with the amount of net charge that is available on one side of the capacitor. Again, due to the symmetry the potential at V=-U. So it doesn't really matter if I consider side V or side U.

So for 2.4μF:
\[Q=2.4*10^{-6}*15V=3.6*10^{-5}C\]and for 1.5μF:\[Q=1.5*10^{-6}*40V=6*10^{-5C}\]

So the magnitude of charges left to distribute themselves on either side it:
\[(6-3.6)*10^{-5}C=2.4*10^{-5}C\]

Having equipotential also means that they are connected in parallel, and thus the formula is:
\[Q_{tot}=2.4*10^{-5}=C_{eq}*V\]
\[C_{eq}=C_{1}+C_{2}= (2.4+1.5)*10^-{6}F=3.9*10^{-6}F\]

So the potential is:
\[V=\frac{2.4*10^{-5}}{3.9*10^{-6}}=6.15V\]

And thus the charge on the 1.5μF capacitor is:
\[Q=1.5*10^{-6}C*6.15V=9.231*10^{-6}C\]