Consider two identical metal plates of area A that are separated by a distance d
ID: 1871880 • Letter: C
Question
Consider two identical metal plates of area A that are separated by a distance d. The space
between the plates is filled with a non-conducting material (air, for instance). Suppose each
plate is connected to one of the terminals of a battery.
1.) Suppose you now double the area of each plate. Does the voltage between the plates change (recall that the plates are still connected to the battery)? Does
the amount of charge on each plate change? Since C = q=V, how must the capacitance change? Do this in prelab before coming to lab.
-The capacitance will decrease by x2 because to maintain the fixed potential difference the charge density should decrease, thus total charge will decrease
-The capacitance will increase by x2 because to maintain the fixed charged density the potential difference will have to increase
-The capacitance will increase by x2 because to maintain the fixed potential difference the charge density should remain the same, thus total charge will double
-The capacitance will decrease by x2 because to maintain the fixed charged density the potential difference will have to decrease
-The capacitance will decrease by x2 because to maintain the fixed potential difference the charge density should remain the same, thus total charge will decrease
-The capacitance will not change because both total charge and potential difference will remain the same
-The capacitance will increase by x2 because to maintain the fixed potential difference the charge density should double, thus total charge will double
Explanation / Answer
The voltage between the plates does not change as it is still connected to the same battery.
The amount of charge on each plate will change as the capacitance will change. And we know Q=CV
Capacitance being inversely proportional to the distance between the plates, decreases. It becomes half of the previous value.
First statement is true.