Archived Capacitor Charge Redistribution in Series and Parallel Circuits

[snoweangel27]

Homework Statement

I am trying to study for an upcoming exam, and came across a problem that I can't seem to solve.

I am given three capacitors, where C(2) and C(3) are in series, and C(1) is parallel to C(2,3).

C(1) is connected alone to a battery and charges until the magnitude of the charge is 4.0x10^-8 C. Then it is removed from the battery and connected to two other capacitors, C(2) and C(3). The charge on the positive plate of C(1) after is is attached is reduced to 1.0 x 10^-8 C. What are the charges on each plate of C(2) and C(3)?

Homework Equations

capacitor is series: V(1) = Q/C(i)
capacitor is parallel Q(i) = C(i)V

The Attempt at a Solution

I know that the charges of C(2) and C(3) are going to be equal, since they are in series, and that the Potential difference of C(1) is going to equal the potential difference of C(2,3), so I tries to solve for V in the second equation then set the two equations equal, but I get confused trying to figure out what is the Capacitance of each of the Capacitors.

 

[gneill]

A complete solution is offered.

The situation can be depicted as follows:

Capacitor C1 begins with 4.0 x 10^-8 C or 40 nC of charge supplied by a battery (not shown). We assume that C1's positively charged plate is on top in the figure. That would mean +40 nC on its upper plate and -40 nC on its lower plate (capacitors maintain equal and opposite charges on their plates).

We are told that after being connected to the other capacitors, C1's charge is reduced to 10 nC. That represents decrease of 30 nC, which must have moved to the other capacitors. Since C2 and C3 are in series they will both receive the same 30 nC of charge (Conservation of charge). So as depicted, there will be +30 nC on their upper plates and -30 nC on their lower plates.

Note that there is no need to determine capacitance values or voltage values. Conservation of charge suffices to answer the problem as posed.