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If a fully charged parallel plate capacitor is disconnected from a battery and its plates are pulled apart to increase the distance between the plates, what is the value of charge on the capacitor?
I know the charge and the electric field will both remain the same, but I don't know why or how?
Since d increases, the capacitance C will decrease. Also since positive work is done on the system to seperate the attracted plates, the potential difference also increases.
I know the charge and the electric field will both remain the same, but I don't know why or how?
Since d increases, the capacitance C will decrease. Also since positive work is done on the system to seperate the attracted plates, the potential difference also increases.