Answer:
0.636 kJ
Explanation:
The charge on any capacity, q = CV, thus,
The initial charge on the 70 pF capacitor is
q = Cv
q = 70*10^-12 * 3.6*10^3
q = 2.52*10^-7 C
The charge on the 280 pF capacitor is q = C*v
q = 280*10^-12 * 3.6*10^3
q = 1.008x10^-6 C
When they are connected as stated, the net total charge remaining will be 1.008*10^-6 - 2.52*10^-7 = 7.56*10^-7 C
Since the capacitors are in parallel, the equivalent capacitance will be 70 + 280 pF = 350 pF
Remember, q = CV, then V = q/C
V = 7.56*10^-7 C / 350*10^12 F
V = 2160 V
b) The energy before is 1/2 C*v²
E = 1/2 * 70*10^-12 * 3600² + 1/2 * 280*10^-12 * 3600²
E = 4.536*10^-4 J + 1.814*10^-3 J
E = 2.268 kJ
The energy After is 1/2 Cv²
E = 1/2 * 70*10^-12 * 2160² + 1/2 * 280*10^-12 * 2160²
E = 3.266*10^-4 J + 1.306*10^-3 J
E = 1.632 kJ
so the loss is 2.268 - 1.632 = 0.636 kJ
