Final answer:
When a 10.0µF capacitor’s plate radius is doubled and connected to a 12.0V battery, the charge on the plates remains 120µC, as capacitance and voltage are unchanged.
Step-by-step explanation:
The question asks how much charge will be on the plates of a 10.0µF parallel-plate capacitor connected to a 12.0V battery, after the radius of each plate is doubled. The capacitance of the capacitor is given as 10.0µF, and the voltage from the battery is 12.0V.
The charge on a capacitor is calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Since the radius of each plate is doubled and the separation between them is unchanged, the area of the plates increases by a factor of four (because area is proportional to the square of the radius).
However, in this case, since the capacitor is connected to a constant voltage source (the battery), the capacitance remains unchanged at 10.0µF. Thus, the charge stored on the capacitor is calculated as:
Q = CV
= 10.0µF × 12.0V
= 120µC
The amount of charge on the plates remains 120 microcoulombs (µC) because the capacitance and voltage are constant.