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