Question

A 0.6 pF capacitor is charged to 27 V. After charging, the capacitor is disconnected from the voltage source and is connected to another uncharged capacitor. The final voltage is 9 V. (a) What is the capacitance of the other capacitor? x PF Enter a number. (b) How much energy was lost when the connection was made? e non se conserva una

Answer 1

(a) INITIALLy total charge , Q = C V = 0.6 x 27

Qi = 16.2 uC  

finally, Qf = Q1 + Q2 = (0.6 x 9) + (9 C)

= 5.4 + 9 C

Applying charge conservation,

16.2 = 5.4 + 9 C

C = 1.2 uF .........Ans

(B) Ui = (0.6 x 10^-6)(27^2)/2

Ui = 2.187 x 10^-4 J

UF = (0.6 x 10^-6 + 1.2 x 10^-6) (9^2) / 2

Uf = 7.29 x 10^-5 J  

Energy lost = Ui - Uf = 1.46 x 10^-4 J

= 146 uJ