# 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(C) = 4 μF = 4 10-6F

Voltage supplied to the capacitor (V) = 200 V

Electrostatic energy of the capacitor (E) = CV2 = = 8 10-2 J

Capacitance of an uncharged capacitor (C’) = 2 μF = 2 10-6F

When both the capacitors are connected in a circuit, then initial charge on charged capacitor is equal to the final charge on both the capacitors in circuit. (According to law of conservation of charges)

Since, Charge = Voltage Capacitence

Therefore, C V = (C + C’) V’, where V’ is the voltage in the circuit when both capacitors are connected.

4 10-6F 200V = (4 10-6F + 2 10-6F) V’ V = V

Now, Electrostatic energy for the combination of two capacitors = 1/2 (C + C’)V’2 U = 5.33 10-2 J

Thus, the electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation = 8 10-2 J-5.33 10-2 J = 6.67 10-2J

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