Answer :
Final answer:
When the area of the plates in a parallel-plate capacitor doubles, the capacitance also doubles. The charge on the capacitor can be calculated using the formula Q = C×V, where Q is the charge, C is the capacitance, and V is the voltage. The voltage across the capacitor remains the same when the area of the plates is doubled.
Explanation:
In this case, the initial capacitance of the capacitor is 6.0 µF and the initial voltage across the capacitor is 1.5 V. When the area of the plates doubles, the capacitance will also double. So, the new capacitance will be 12 µF (F for Farad).
The charge on the capacitor can be calculated using the formula Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage. Therefore, the charge on the capacitor will be 12 µF × 1.5 V = 18 µC (C for Coulomb). The voltage across the capacitor remains the same when the area of the plates is doubled. So, the voltage across the capacitor will still be 1.5 V.
The capacitance of the capacitor is given by the formula C = ε × ε0 × A / d, where ε is the permittivity of the material between the plates, ε0 is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. Since the area of the plates has doubled, the new area will be 2A and the new capacitance will be 2C. Therefore, the new capacitance will be 12 µF × 2 = 24 µF.
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