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2 years ago

5

When the potential difference between the plates of a capacitor is increased by 3.50 V , the magnitude of the charge on each pla

te increases by 15.0 μC . What is the capacitance of this capacitor in μF?

1 answer:

hodyreva [135]2 years ago

5 0

Answer:

42.9 μF

Explanation:

V = 3.50 V, Q = 150 μC

C = Q/V = 150/3.50 μF = 42.9 μF

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