How Does Changing the Battery Voltage Affect the Charge on a Capacitor?

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SUMMARY

When a capacitor is connected to a 6.0 V battery and then to a 12.0 V battery, the charge on its plates increases. This is based on the formula Q=CV, where Q represents charge, C is capacitance, and V is voltage. The increase in voltage from 6.0 V to 12.0 V directly results in a proportional increase in charge, assuming the capacitance remains constant. The polarity of the voltage does not affect the amount of charge stored on the plates in this context.

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A capacitor is first connected to a 6.0 V battery and then disconnected and connected to a 12.0 V battery. How does the charge on one of its plates change: (a) It increases, (b) it decreases, or (c) it 1stays the same?








(c) (because the charge on the plate is
depending upon the area of the surface of the plate)

Right?? Not right?
 
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Q=CV
but one thing is not clear, because polarity matters
 
Now I think it's A) increases...Is this right
 
several assumptions would have to be made, but A) seems the most logical for this question... unless it is part of a bigger problem which tells you more about the setup.
 

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