A capacitor of capacitance 5uF is charged to a potential of 100 V. Another capacitor of capacitance 10uF is charged to a potential of 150 V. The two capacitors are then joined with plates of like charges connected together.
(a) Find the total energy stored in the system before and after the capacitors are joined.
(b) What happens to the difference in energy in the two cases?
Answers
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(a) Initial energy, E = 0.1375 J
Final energy, É = 0.1332 J
(b) Difference in energy, ΔE = 0.0043 J
Explanation:
Capacitance C1 = 5 μF = 5 * 10^-6 F
Potential V1 = 100 V
Capacitance C2 = 10 μF = 10 * 10^-6 F
Potential V1 = 150 V
The total energy of the capacitors before being joined =
E = E1 + E2
= 1/2.C1²V1² + 1/2.C2²V2²
= 1/2 [(5 * 10^-6) * 100² + 10 * 10^-6 * 150)² ]
= 1/2 [ (50 * 10^-3) + (225 * 10^-3) ]
= 275/2 * 10^-3
= 137.5 * 10^-3 J
E = 0.1375 J --------------------(1)
Charge of first capacitor Q1 = C1V1 = 5 * 10^-6 * 100 = 500 * 10^-6 F
Charge of second capacitor Q2 = C2V2 = 10 * 10^-6 * 150 = 1500 * 10^-6 F
Now common potential = total charge / total capacitance
V = (500 + 1500) * 10^-6 / (5+10) * 10^-6
= 2000 / 15
= 133.3 V
After the capacitors are joined, total energy É = 1/2 (C1+C2) V²
= 1/2 (5+10) * 10^-6 * 133.3²
É = 0.1332 J ---------------(2)
ΔE = E - É = 0.1375 - 0.1332 = 0.0043 J --------------(3)