How Is Capacitance Calculated in a Charged Parallel-Plate Capacitor?

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Homework Statement

The two plates of a parallel-plate capacitor initially carry equal amount of positive charge. If some charges are transferred from one plate to another, the charges on the plates are respectively +900uC and +100uC. The potential difference across the plates becomes 4V. What is the capacitance of the capacitor?

Homework Equations

C=Q/V

The Attempt at a Solution

I just pick randomly pick +900uC instead of +100uC:
C=900/4=225uC

4. Then I check the answer:

Totally 400uC moves away from one to the metal plate and move to another plate, so:
C=400/4=100uC

According to what I have learn, Q is the charge of one of the metal plate. In here, why should I subsitute 400uC, the amount of charge that is transferred?

 

[Hootenanny]

Consider the relative charge...

Have you met Gauss' law?

 

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I haven't heard about it before. I will try to learn it by myself. Thank you