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?

Answer

Capacitance of a charged capacitor, *C*_{1}=4µF = 4 x 10^{-6} F

Supply voltage,* V*_{1} = 200 V

Electrostatic energy stored in *C*_{1} is given by,

Capacitance of an uncharged capacitor, *C*_{2}=2µF = 2 x 10^{-6} F

When *C*_{2} is connected to the circuit, the potential acquired by it is *V*_{2}.

According to the conservation of charge, initial charge on capacitor *C*_{1} is equal to the final charge on capacitors, *C*_{1} and *C*_{2}.

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

Hence, amount of electrostatic energy lost by capacitor *C*_{1}

= *E*_{1} - *E*_{2}

= 0.08 - 0.0533 = 0.0267

=2.67 × 10 ^{- 2} J

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Answer carefully:

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The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.

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sandeep choudhary
2018-07-19 09:20:26

nice one

Nadhu
2018-07-04 07:08:31

Nice but must be more elaborate in ur topic

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