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

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An uncharged capacitor initially connected to a 1.02 V battery accumulates 1.69 μC of charge on its positive plate. When the battery is replaced with a 6.78 V battery, the charge on the capacitor must increase to maintain its capacitance. Using the relationship between charge and voltage, the additional charge can be calculated by setting up a ratio based on the initial charge and voltage. The calculation shows that the total charge at 6.78 V is 1.12x10^-8 C, leading to an additional charge of 9.54x10^-9 C. The discussion emphasizes understanding the relationship between charge, voltage, and capacitance in capacitors.
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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!
 
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Can I set these up as ratios?
 
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.
 
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
 
No that's not right...ahh what am doing wrong??
 

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