If the capacitor is connected across a battery,the voltage will stay constant at the battery voltage.
Capacitance C=Q/V1−V2 (V1−V2 is the potential difference between the plates of the plates of the capacitor)
when plates are connected by a wire than V1=V2 so V1−V2=0
C=Q/V1-V2 =Q/0 = Infinite Hence, C=Infinite
If the capacitor is charged and disconnected,the voltage will rise.Assume the capacitor holds charge Q then Q= CV or V=Q/C
separating the plates reduces C,so if Q remains constant V must be increase proportionally.
The capacitance of a parallel plate capacitor is proportional to the plate area and inversely proportional to plate-plate spacing. So, doubling the area doubles the capacitance. And, halving the distance doubles it again. So the capacitance of the new capacitor will be 4-times larger then the original.
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