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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?

$\begin{array}{1 1} 2.7 \times 10^{-2}\;J \\2.67 \times 10^{-2}\;J \\ 17 \times 10^{-2}\;J \\ 2.5 \times 10^{-2}\;J \end{array} $

1 Answer

Solution :
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,
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 C1 is equal to the final charge on capacitors, $C_1$ and $C_2$.
$V_2(C_1+C_2) =C_1V_1$
$V_2 \times (4+2) \times 10^{-6} \times \bigg( \large\frac{400}{3}\bigg)^2$
$\qquad= 5.33 \times 10^{-2}\;J$
Hence, amount of electrostatic energy lost by capacitor $C_1$
$\qquad= E_1-E_2$
$\qquad= 0.08 -0.0533=0.0267$
$\qquad= 2.67 \times 10^{-2}\;J$
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