A 4 µF capacitor is charged by a 200 V supply.

Solution:

Capacitance of a charged capacitor, $C_{1}=4 \mu \mathrm{F}=4 \times 0^{-6} \mathrm{~F}$

Supply voltage, V1 = 200 V

Electrostatic energy stored in C1 is given by,

$E_{1}=\frac{1}{2} C_{1} V_{1}^{2}$

$=\frac{1}{2} \times 4 \times 10^{-6} \times(200)^{2}$

$=8 \times 10^{-2} \mathrm{~J}$

Capacitance of an uncharged capacitor, $C_{2}=2 \mu \mathrm{F}=2 \times 0^{-6} \mathrm{~F}$

When C2 is connected to the circuit, the potential acquired by it is V2.

According to the conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.

$\therefore V_{2}\left(C_{1}+C_{2}\right)=C_{1} V_{1}$

$V_{2} \times(4+2) \times 10^{-6}=4 \times 10^{-6} \times 200$

$V_{2}=\frac{400}{3} \mathrm{~V}$

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

$E_{2}=\frac{1}{2}\left(C_{1}+C_{2}\right) V_{2}^{2}$

$=\frac{1}{2}(2+4) \times 10^{-6} \times\left(\frac{400}{3}\right)^{2}$

$=5.33 \times 10^{-2} \mathrm{~J}$

Hence, amount of electrostatic energy lost by capacitor C1

= E1 − E2

= 0.08 − 0.0533 = 0.0267

 

= 2.67 × 10−2 J