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A capacitor of capacitance C=900 pFC = 900 \text{ pF} is charged fully by 100 V100 \text{ V} battery BB as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C=900 pFC = 900 \text{ pF} as shown in figure (b). The electrostatic energy stored by the system (b) is

A

4.5×10−6 J4.5 \times 10^{-6} \text{ J}

B

3.25×10−6 J3.25 \times 10^{-6} \text{ J}

C

2.25×10−6 J2.25 \times 10^{-6} \text{ J}

D

1.5×10−6 J1.5 \times 10^{-6} \text{ J}

Step-by-Step Solution

Initial charge q1=CV=900×10−12×100=9×10−8 Cq_1 = CV = 900 \times 10^{-12} \times 100 = 9 \times 10^{-8} \text{ C}. When connected to an identical uncharged capacitor, the common potential is V′=C1V1+C2V2C1+C2=9×10−8+01800×10−12=50 VV' = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} = \frac{9 \times 10^{-8} + 0}{1800 \times 10^{-12}} = 50 \text{ V}. The total energy stored is U=12(C1+C2)(V′)2=12×1800×10−12×(50)2=900×10−12×2500=2.25×10−6 JU = \frac{1}{2}(C_1 + C_2)(V')^2 = \frac{1}{2} \times 1800 \times 10^{-12} \times (50)^2 = 900 \times 10^{-12} \times 2500 = 2.25 \times 10^{-6} \text{ J}.

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