Calculating Charge Flow in Capacitor: 1.02V to 6.78V Battery Swap

[BuBbLeS01]

Capacitors...Please HELP!

Homework Statement

An uncharged capacitor is connected to the terminals of a 1.02 V battery, and 1.69 μC flows to the positive plate. The 1.02 V battery is then disconnected and replaced with 6.78 V battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate?

Homework Equations

The Attempt at a Solution

I am not really sure how to start this problem without having the capacitance. Can someone help me get started..thanks!

 

[BuBbLeS01]

Can I set these up as ratios?

 

[Ronnin]

Capacitance is just charge/voltage. You could set up two ratios equal (capacitance isn't changing with additional voltage, but as voltage increases you must increase charge to hold the capacitance constant.

 

[BuBbLeS01]

So I could do...
1.69x10^-9 C/1.02 V = Q/6.78 V
Q = 1.12x10^-8 C
Then subtract them to get my answer being...
9.54x10^-9 C

 

[BuBbLeS01]

No that's not right...ahh what am doing wrong??