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Alja [10]
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?
Physics
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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