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NEET PHYSICSGeneralMedium

Question

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}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICScapacitorcapacitancechargedbatteryfigure

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