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?
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?
Capacitance of a charged capacitor, $c_{1}=4\, \mu \,F=4 \times 10^{-6}\, F$
Supply voltage, $V _{1}=200\, V$ Electrostatic energy stored in $C _{1}$ 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 $C _{2}$ $\therefore V_{2}\left(C_{1}+C_{2}\right)=C_{1} V_{1}$
$V_{2} \times(4+2) \times 10^{-6}=4 \times 10^{-6} \times 200$
$V_{2}=\frac{400}{3} \,V$
Electrostatic energy for the combination of two capacitors is given by, $E_{2}=\frac{1}{2}\left(C_{1}+C_{2}\right) V_{2}^{2}$
$=\frac{1}{2}(2+4) \times 10^{-6} \times\left(\frac{400}{3}\right)^{2}$
$=5.33 \times 10^{-2} \,J$
Hence, amount of electrostatic energy lost by capacitor
$C _{1}= E _{1}- E _{2}=0.08-0.0533=0.0267$ $=2.67 \times 10^{-2} \;J$
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