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A capacitor with capacitance 8.9 x 10^-9 F has a voltage of 3.00 x 10^3 V applied across it. Using Q = CV, how much charge is stored on the capacitor?

A3.7 x 10^11 C
B2.7 x 10^-5 C
C2.7 x 10^-8 C
D2.7 x 10^-2 C
Answer & Solution
Correct answer: B. 2.7 x 10^-5 C
1. The capacitance relation C = Q/V rearranges to Q = CV. 2. Q = (8.9 x 10^-9 F)(3.00 x 10^3 V) = 2.7 x 10^-5 C. 3. Dividing V by C instead of multiplying, as in option A, produces a value with the wrong units entirely and a nonsensical size. _Source: OpenStax Physics (CC BY 4.0), Ch 18 "Static Electricity", section 18.5 Capacitors and Dielectrics_
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