To operate a given flash lamp requires a charge of 305 mu C

To operate a given flash lamp requires a charge of 30.5 mu C. What capacitance is needed to store this much charge in a capacitor with a potential difference between its plates of 3 V?

Solution

Q = 30.5 muC = 30.5 * 10^-6 C

V = 3 V

C = Q / V

C = (30.5 * 10^-6) / 3

= 0.00001017 F = 10.17 muF