Question

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?

Answer 1

42.9 μF

Step-by-step explanation:

V = 3.50 V, Q = 150 μC

C = Q/V = 150/3.50 μF = 42.9 μF