A capacitor having a capacitance of 200 µF is charged to a potential difference of 20V. The charging battery is disconnected and the capacitor is connected to another battery of emf 10V with the positive plate of the capacitor joined with the positive terminal of the battery.
Text Solution
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[ 4000 µC, 2000 µC
2000 µC
work is done on the battery 20 mJ
30 mJ
(e) 10 mJ]
Sol. Charge on capacitor before connection
Q 1 = CV 1 = 4000 µC
Charge on capacitor after connection
Q 2 = CV 2 = 200 × 10 = 2000 µC
Charge flown through the 10V battery = 4000 – 2000 = 2000 µC
Work is done on the battery.
Work done = Q 2 × V 2 = 2000 × 10 = 20mJ.
The decrease in electrostatic field energy
= U i – U f = 
CV 1 2 – 
CV 2 2 = 
× 200 × (20) 2 – 
× 200 (10) 2
= 30 mJ
(e) W = Δ U + Δ H
- 20 mJ = – 30 mJ + Δ H
[ Δ H = 10 mJ]
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