A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is [figure]
- 4.5 10-6 J
- 3.25 10-6 J
- 2.25 10-6 J
- 1.5 10-6 J
Answer
(3) 2.25 10-6 J
Solution
Charge on the capacitor in (a):
Q = CV = 900 10-12 100 = 9 10-8 C
In (b) the two equal capacitors are in parallel and share this charge (charge is conserved):
Ceq = 2C = 1800 pF, common potential V' = Q2C = 50 V
U = Q22Ceq
U = (9 10-8)22 1800 10-12
U = 81 10-1636 10-10
U = 2.25 10-6 J
Answer: (3)