# CBSE Class 12-science Answered

A 4 μF 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 heat and electromagnetic radiation?

Capacitance of a charged capacitor, C_{1} = 4 F = 4 × 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 μF = 2 × 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}.

therefore, V_{2 }(C_{1} + C_{2}) = C_{1}V_{1}

V_{2 }×_{ }(4 + 2) × 10^{-6} × 200

V_{2} = V

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 x 10^{-2} J