Two charged capacitors cross connected to each other

[PhysicsUnderg]

Homework Statement

Capacitors C1 = 6.99 mF and C2 = 1.72 mF are charged as a parallel combination across a 219 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on capacitor C1

Homework Equations

V=Q/C, Q=CV

The Attempt at a Solution

I really do not know how to start this problem. I guess I don't know what is meant by the connection of positive plate to negative plate, visa versa... how does this effect the voltage? How do I reason out this problem?

 

[rl.bhat]

When thew capacitors are connected in parallel, voltage across each capacitor is the same and tha charge on each plate is
Q1 = C1V and Q2 = C2V.
When you connect positive plate to negative plate, charges will flow from one plate to the other until both the plates acquire the equal charges. And the charge on each plate will be (Q1 - Q2)/2.

 

[collinsmark]

I really do not know how to start this problem.

Finding charge on each capacitor, when charged by the 219 V battery, might be a good start.

I guess I don't know what is meant by the connection of positive plate to negative plate, visa versa...

Perhaps I can try to explain. Imagine for a moment that everything is color coded. Each capacitor has a red terminal and a blue terminal. The battery also has a red terminal (say, positive), and a blue terminal (negative).

When the are capacitors are charged by the 219 V battery, all the colors match

red..red..red +(positive)
.|...|...|
___...___...|
___...___...Batt
.|...|...|
blue blue blue -(negative)

Once charged, the battery is removed.

red..red
.|...|
___...___
___...___
.|...|
blue blue

Then, the capacitors are unconnected from each other. One is flipped upside down, and connected back to the other capacitor.


red..blue
.|...|
___...___
___...___
.|...|
blue red