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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery B as shown in Figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in Figure (b). The electrostatic energy stored by the system (b) is:

A

1.5 × 10⁻⁶ J

B

4.5 × 10⁻⁶ J

C

3.25 × 10⁻⁶ J

D

2.25 × 10⁻⁶ J

Step-by-Step Solution

  1. Initial State (Figure a):
  • Capacitance C1=900 pF=9×10−10 FC_1 = 900 \text{ pF} = 9 \times 10^{-10} \text{ F}.
  • Voltage V1=100 VV_1 = 100 \text{ V}.
  • Initial Energy stored: Ui=12C1V12=12(9×10−10)(100)2=4.5×10−6 JU_i = \frac{1}{2}C_1 V_1^2 = \frac{1}{2}(9 \times 10^{-10})(100)^2 = 4.5 \times 10^{-6} \text{ J}.
  • Charge stored: Q=C1V1=(900×10−12)(100)=9×10−8 CQ = C_1 V_1 = (900 \times 10^{-12})(100) = 9 \times 10^{-8} \text{ C}.
  1. Final State (Figure b):
  • The charged capacitor C1C_1 is disconnected and connected in parallel to an identical uncharged capacitor C2=900 pFC_2 = 900 \text{ pF}.
  • Charge Redistribution: The total charge QQ remains conserved but distributes equally between the two identical capacitors. Thus, charge on each is Q′=Q/2Q' = Q/2.
  • System Capacitance: The two capacitors are in parallel, so equivalent capacitance Ceq=C1+C2=2C=1800 pFC_{eq} = C_1 + C_2 = 2C = 1800 \text{ pF}.
  • Final Energy: The energy stored in the system is Uf=Q22CeqU_f = \frac{Q^2}{2C_{eq}}.
  • Substituting Ceq=2C1C_{eq} = 2C_1: Uf=Q22(2C1)=12(Q22C1)=12UiU_f = \frac{Q^2}{2(2C_1)} = \frac{1}{2} \left( \frac{Q^2}{2C_1} \right) = \frac{1}{2} U_i.
  1. Calculation:
  • Uf=12×(4.5×10−6 J)=2.25×10−6 JU_f = \frac{1}{2} \times (4.5 \times 10^{-6} \text{ J}) = 2.25 \times 10^{-6} \text{ J}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. 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.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEcapacitorcapacitancechargedbatteryfigure

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