First slide

Capacitance

Question

The energy stored in the capacitor as shown in the figure (a) is 4.5 x 10^-4J. If the battery is replaced by another capacitor of 900 pF as shown in figure (b), then the total energy of system is

Moderate

A

4.5 x 10^-6J
B

2.25 x 10^-6J
C

Zero
D

9 x 10^-6 J

Solution

$E_{i} = \frac{1 Q^{2}}{2 C} = 4 .5 \times 10^{- 6} J$
$(∵ Q = CV = 900 \times 10^{- 12} \times 100 = 9 \times 10^{- 8} C) In fig (b) C_{eq.} = 2 C = 2 \times 900 pF ∴ E_{f} = \frac{1}{2} \frac{Q^{2}}{C_{Q}} = \frac{Q^{2}}{2 (x)} = \frac{1}{2} \times 4 .5 \times 10^{- 6} = 2 .25 \times 10^{- 6} J$

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