Final answer:
When a parallel plate capacitor's plates are pulled apart to twice their original distance while connected to a constant voltage source, the capacitance is halved, but the charge remains unchanged.
Step-by-step explanation:
The question is about a parallel plate capacitor whose capacitance (C0) and magnitude of charge (Q0) are initially determined when connected to a battery with voltage (V0). When the separation between the plates is increased to 2d while the capacitor is connected to the battery, the new capacitance and magnitude of the charge on the plates need to be found.
The capacitance of a parallel plate capacitor is given by C = ε0A/d, where ε0 is the permittivity of free space, A is the plate area, and d is the separation between the plates. When the plates are pulled apart to a new distance of 2d, the new capacitance (Cnew) becomes half the original capacitance (C0/2). However, since the capacitor is still connected to the battery, the voltage across it remains constant. The charge on a capacitor is given by Q = CV, so if the capacitance is halved but the voltage remains the same, the charge (Qnew) remains unchanged (Q0).
Therefore, after the plates are 2d apart, the new capacitance of the capacitor and the magnitude of the charge on the plates are: b. Q0&0.5C0.