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When the potential difference between the plates of a capacitor is increased by 3.50 V , the magnitude of the charge on each plate increases by 15.0 μC . What is the capacitance of this capacitor in μF?
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33 votes
33 votes
When the potential difference between the plates of a capacitor is increased by 3.50 V , the magnitude of the charge on each plate increases by 15.0 μC . What is the capacitance of this capacitor in μF?

User Buddha
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1 Answer

13 votes
13 votes

Answer:

42.9 μF

Step-by-step explanation:

V = 3.50 V, Q = 150 μC

C = Q/V = 150/3.50 μF = 42.9 μF

User AdmiralWen
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3.6k points
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