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Academic session 2026-27, Chapter 2 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Physics - Chapter Electrostatic Potential and Capacitance. This page offers a step-by-step solution to the specific question from Exercise 1, Question 27:
A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Capacitance of a charged capacitor, C1=4µF = 4 x 10-6 F
Supply voltage, V1 = 200 V
Electrostatic energy stored in C1 is given by,

Capacitance of an uncharged capacitor, C2=2µF = 2 x 10-6 F
When C2 is connected to the circuit, the potential acquired by it is V2.
According to the conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.

Electrostatic energy for the combination of two capacitors is given by,

Hence, amount of electrostatic energy lost by capacitor C1
= E1 - E2
= 0.08 - 0.0533 = 0.0267
=2.67 × 10 - 2 J
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 27: A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is....
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nice one
Nice but must be more elaborate in ur topic