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Answer :
Final answer:
The charge on the 6.0 μF parallel-plate capacitor after the battery is disconnected and the plate area is doubled remains at 9 μC, which is option I.
Explanation:
When a 6.0 μF parallel-plate capacitor is fully charged by a 1.5-V battery and then disconnected from it, the charge on the capacitor remains constant, because there is no path for the charge to leave the plates. The charge (Q) on a capacitor is given by Q = C×V, where C is capacitance and V is voltage. For the given capacitor, the charge Q is 6.0 μF × 1.5 V = 9 μC. Once the battery is disconnected, and the area of the plates is doubled, the capacitance changes but the charge remains the same, as there is no path for the charge to go. Thus, the charge on the capacitor after the area is doubled is still 9 μC, which corresponds to option I in the list provided.
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