A capacitor stores 5.3 x 10^-5 C of charge when connected to a 6.0-V battery. How much charge does the capacitor store when connected to a 9.0-V battery?
Correct answer: A. 79.5 μC
- A. 79.5 μC
- B. 35.3 μC
- C. 79.5 pC
- D. 35.3 pC
Explanation
Q = 5.3 x 10^-5 CV = 6.0 VUsing the formula Q = CV, we get:C = Q/V = (5.3 x 10^-5 C)/(6.0 V) = 8.83 x 10^-6FNow, let's find the charge stored in the capacitor when it is connected to a 9.0 V battery. We have:V = 9.0 VC = 8.83 x 10^-6 FUsing the formula Q = CV, we get:Q = CV = (8.83 x 10^-6 F) x (9.0 V) = 7.95 x 10^-5 C
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About Capacitors
Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
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