A certain capacitor stores 21 J of energy when it holds 1592

A certain capacitor stores 21 J of energy when it holds 1,592 uC of charge. What is the capacitance in nF? Answer:

Solution

E = V* Q / 2

21 J = V * (1592 * 10^-6 C) / 2

V = 42 / 1592 * 10^-6= 26382 V = 26.38 KV

C = Q / V = Q / (2 E / Q)

= 0.5 * Q² / E = 0.5 * (1592 * 10^-6)² / 21

= 6.03 * 10^-8 F

= 0.0603 muF