A $600\; pF$ capacitor is charged by a $200\; V$ supply. It is then disconnected from the supply and is connected to another uncharged $600\; pF$ capacitor. How much electrostatic energy is lost in the process?
A $600\; pF$ capacitor is charged by a $200\; V$ supply. It is then disconnected from the supply and is connected to another uncharged $600\; pF$ capacitor. How much electrostatic energy is lost in the process?
Capacitance of the capacitor, $C =600\, pF$
Potential difference, $V =200 \,V$
Electrostatic energy stored in the capacitor is given by, $E_{1}=\frac{1}{2} C V^{2}$
$=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2} \,J$
$=1.2 \times 10^{-5} \,J$
If supply is disconnected from the capacitor and another capacitor of capacitance $C=600\, pF$ is connected to it, then equivalent capacitance ( $C_{ eq }$ ) of the combination is given by,
$\Rightarrow \frac{1}{C_{e q}}=\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300}$
$\Rightarrow C_{e q}=300\, pF$
New electrostatic energy can be calculated as $E_{2}=\frac{1}{2} C_{e q} V^{2}$
$=\frac{1}{2} \times 300 \times(200)^{2} \,J$
$=0.6 \times 10^{-5} \,J$
Loss in electrostatic energy $= E _{1}- E _{2}$ $=1.2 \times 10^{-5}-0.6 \times 10^{-5} \,J$
$=0.6 \times 10^{-5} \,J$
$=6 \times 10^{-6}\, J$
Therefore, the electrostatic energy lost in the process is $6 \times 10^{-6}\; J$
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