A capacitor stores charge Q = 3 μC at V = 6 V. What is its capacitance?

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Multiple Choice

A capacitor stores charge Q = 3 μC at V = 6 V. What is its capacitance?

Explanation:
Capacitance is defined by Q = C V, which means the charge stored is proportional to the voltage with the proportionality constant C. To find C, divide the charge by the voltage: C = Q / V. Here, Q = 3 μC = 3 × 10^-6 C and V = 6 V. So C = (3 × 10^-6 C) / (6 V) = 0.5 × 10^-6 F = 0.5 μF. So the capacitor’s capacitance is 0.5 μF. If it were 1 μF, it would store 6 μC at 6 V; if it were 0.3 μF, it would store 1.8 μC; if it were 2 μF, it would store 12 μC.

Capacitance is defined by Q = C V, which means the charge stored is proportional to the voltage with the proportionality constant C. To find C, divide the charge by the voltage: C = Q / V.

Here, Q = 3 μC = 3 × 10^-6 C and V = 6 V. So C = (3 × 10^-6 C) / (6 V) = 0.5 × 10^-6 F = 0.5 μF.

So the capacitor’s capacitance is 0.5 μF. If it were 1 μF, it would store 6 μC at 6 V; if it were 0.3 μF, it would store 1.8 μC; if it were 2 μF, it would store 12 μC.

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