The 800 pF capacitor was charged to a voltage of 36 V. a) What charge appeared on each of the capacitor plates?

The 800 pF capacitor was charged to a voltage of 36 V. a) What charge appeared on each of the capacitor plates? b) What is the energy of a charged capacitor? c) How will the energy of a capacitor change if the voltage between its plates is increased by 3 times?

Solution:
C = 0.0000000008 F
q = CU = 36C = 0.0000000288 C = 28.8 nC
W = qU / 2 = 518.4 * 10 (-9 degree) J
q2 = 3 * 28.8 = 86.4 nC
U2 = 3 * 36 = 108V
W2 = q2U2 / 2 = 4665.6 * 10 (-9 power) J
Answer: ± 28.8 nC; 518.4 * 10 (-9 degrees) J; 4665.6 * 10 (-9 degree) J