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Question
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 capacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
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Solution
Capacitance of a charged capacitor, `"C"_1 = mu"F" = 4 xx 10^-6 "F"`
Supply voltage, V1 = 200 V
Electrostatic energy stored in C1 is given by,
`"E"_1 = 1/2"C"_1"V"_1^2`
= `1/2 xx 4 xx 10^-6 xx (200)^2`
= `8 xx 10^-2 "J"`
Capacitance of an uncharged capacitor, `"C"_2 = 2mu"F" = 2 xx 10^-6 "F"`
When C2 is connected to the circuit, the potential acquired by it is V2.
According to the conservation of charge, the initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.
∴ `"V"_2("C"_1 + "C"_2) = "C"_1"V"_1`
`"V"_2 xx (4 + 2) xx 10^-6 = 4 xx 10^-6 xx 200`
`"V"_2 = 400/3` V
Electrostatic energy for the combination of two capacitors is given by,
`"E"_2 = 1/2("C"_1 + "C"_2)"V"_2^2`
= `1/2(2 + 4)xx 10^-6 xx (400/3)^2`
= `5.33 xx 10^-2` J
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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