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7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?
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Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors