How Does Flipping a Capacitor Affect Charge and Capacitance in a Circuit?

[DrIxn]

Homework Statement

So 4 capacitors are hooked up as shown in the picture on the left
http://i.imgur.com/RiNcYz4.png
each with C=0.8 F and the voltage from the battery V=16.6 V

Then they are disconnected and hooked up as shown on the left with one capacitor flipped. What is the effective capacitance on the configuration shown left, and the charge on the final system shown right?

Homework Equations

CV=Q

The Attempt at a Solution

I got the effective capacitance worked out to 0.2 F. I also know that the charge on each of the capacitors initally will be equal to the charge on the effective capacitor, which using CV=Q I got 3.32 C on each.

What's throwing me off is the flipped capacitor in the final configuration. It looks like they will each have the same voltage because in any given loop you cross two capacitors. And if they all have the same capacitance then shouldn't they all have the same charge? What exactly changes?

 

[gneill]

I got the effective capacitance worked out to 0.2 F. I also know that the charge on each of the capacitors initally will be equal to the charge on the effective capacitor, which using CV=Q I got 3.32 C on each.

What's throwing me off is the flipped capacitor in the final configuration. It looks like they will each have the same voltage because in any given loop you cross two capacitors. And if they all have the same capacitance then shouldn't they all have the same charge? What exactly changes?

Think of it as each capacitor carrying the same charge and being added in one at a time. Three of them are connected with the same polarity, so the total charge adds up. The last one has the opposite polarity. What happens when its charges meet the existing charges?

 

[DrIxn]

Ah, thank you again :) I thought I had tried that answer already