A capacitor having a capacitance of 100 µF is charged to a potential difference of 24 V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12 V with the positive plate of the capacitor joined with the positive terminal of the battery. (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the 12 V battery. (c) Is work done by the battery or is it done on the battery? Find its magnitude. (d) Find the decrease in electrostatic field energy. (e) Find the heat developed during the flow of charge after reconnection.
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(a) Charge on the capacitor before and after the reconnection = 12V
(b) Charge flown through the 12 V battery = 1200 V
(c) Work done = 14.4mJ
(d) Decrease in electrostatic field energy = 21.6 mJ
(e) heat developed during the flow of charge = 7.2 mJ
Explanation:
As given in the question C = 100 µF, V = 24 V
a) Charge (Q) = CV = 2400 µC
After Connection C = 100 µF
⇒ (Q) = CV = 1200 µC
b) C = 100, V = 12V
Charge flown through the 12 V battery (∴ q) = CV = 1200v
c) We know V = W/q
Work done on the battery (W) = vq
⇒12 × 1200 = 14400J = 14.4mJ
d) Initial electrostatic field energy Ui = (1/2) CV1 2
Final Electrostatic field energy Uf = (1/2) CV22
⇒ Decrease in Electrostatic Field energy = (1/2) CV1 2 – (1/2) CV22
⇒ (1/2) C (V12 – V22) = (1/2) × 100(576 –144) = 21600J
∴ Energy = 21600j = 21.6mJ
e) The energy appeared = (1/2) CV2 = (1/2) × 100 × 144 = 7200J = 7.2mJ
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