question_answer 27)

A 4µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer:

Energy stored,                                     = 8 × 10?2J                 On connection to an uncharged capacitor,                 Total charge,                 q = C1V1 + C2V2                 = 4 x 10?6 x 200 + 0                 = 8 x 10?4C                 Total capacitance, C = C1 + C2                 = 4 + 2 = 6 µF                 Common potential,                                 Energy stored,                                 = 5.33 × 10?2J                 Energy stored, U1 = U2                 = 8 x 10?6 ? 5.33 × 10?2                 = 2.67 x 10?2 J.