Question

A $4\mu F$ capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged $2\mu F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer 1

Capacitance of a charged capacitor, $C_1=4\mu F=4\times {10}^{-6}F$

Supply voltage, $V_1=200V$

Electrostatic energy stored in C1 is given by,

$E_1=\frac{1}{2}{C}_1{V}_1^2$

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

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

Capacitance of an uncharged capacitor, $C_2=2\mu F=2\times {10}^{-6}F$

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 $C_2$

$\therefore V_2(C_1+C_2)=C_1{V}_1$

$V_2\times (4+2)\times {10}^{-6}=4\times {10}^{-6}\times 200$

$V_2=\frac{400}{3}V$

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

$E_2=\frac{1}{2}(C_1+C_2)V_2^2$

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

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

Hence, amount of electrostatic energy lost by capacitor $C_1$

$=E_1-E_2$

$=0.08-0.0533=0.0267$

$=2.67\times {10}^{-2}J$