Question

Question 2.27: A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-90

Answer 1

$C_1=4\mu F , V_1=200V ,C_2=2\mu F , V_2=0$
so, common potential difference across the two capacitors after connection is
$V=\frac{C_1V_1+C_2V_2}{C_1+C_2}\\\\\frac{4\times10^{-6}\times200+0}{(4+2)\times10^{-6}}\\\\=133.33V$

initially, total energy stored in capacitors before connection is $U_i=\frac{1}{2}C_1V_1^2$
$so,U_i=\frac{1}{2}\times4\times10^{-6}\times200^2=0.08J$

and total energy stored in capacitors after connection is $U_f=\frac{1}{2}(C_1+C_2)V^2$
$so,U_f=\frac{1}{2}(4+2)\times10^{-6}\times133.33^2=0.053J$

so, energy lost due to connection is
$\boxed{\Delta{U}=U_f-U_i}$
∆U = 0.053 - 0.08 = -0.027 J