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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system

A

increases by a factor of 4

B

decreases by a factor of 2

C

remains the same

D

increases by a factor of 2

Step-by-Step Solution

  1. Initial State: Let the capacitance be CC and the potential be VV. The initial charge is Q=CVQ = CV. The initial energy stored is Ui=12CV2=Q22CU_i = \frac{1}{2}CV^2 = \frac{Q^2}{2C}.
  2. Disconnection and Reconnection: When the battery is removed, the charge QQ is conserved. Connecting an identical uncharged capacitor (CC) in parallel creates a system with equivalent capacitance Ceq=C+C=2CC_{eq} = C + C = 2C.
  3. Final State: The total charge QQ is now distributed across the equivalent capacitance. The final energy is Uf=Q22CeqU_f = \frac{Q^2}{2C_{eq}}.
  4. Calculation: Substituting Ceq=2CC_{eq} = 2C, we get Uf=Q22(2C)=12(Q22C)=Ui2U_f = \frac{Q^2}{2(2C)} = \frac{1}{2} \left( \frac{Q^2}{2C} \right) = \frac{U_i}{2}.
  5. Conclusion: The total electrostatic energy decreases by a factor of 2. The missing energy is dissipated as heat and electromagnetic radiation during the redistribution of charge (as mentioned in NCERT Example 2.10).

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 CAPACITANCEcapacitorchargedbatterybatteryremoved

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