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 dissipated in the form of heat and electromagnetic radiation?

• A. $2.67\times 10^{-2}J$
• B. $2.67\times 10^{-4}J$
• C. $3.67\times 10^{-2}J$
• D. $3.67\times 10^{-4}J$

Answer 1

Answer: (A)

Here, $C_1=4\mu F=4\times 10^{-6}F,V_1=200V$
Initial electrostatic energy stored in $C_1$ is
$U_1=\frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}\times 4\times 10^{-6}\times 200\times 200$
$U_1=8\times 10^{-2}J$
When $4\mu F$ capacitor is connected to uncharged capacitor of $2\mu F$, charge flows and both acquire a common potential which is given by
$V=\frac{Totalcharge}{Totalcapacity};V=\frac{C_1{V}_1}{C_1+C_2}$
$V=\frac{4\times 10^{-6}\times 200}{\left(4+2\right)\times 10^{-6}}=\frac{800}{6}V$
$\therefore$ Final electrostatic energy of both capacitors
$U_2=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}\times 6\times 10^{-6}\times \frac{800}{6}\times \frac{800}{6}$
$=5.33\times 10^{-2}J$
$\therefore$ Energy dissipated in the form of heat and electromagnetic radiation is
$U_1-U_2=\left(8\times 10^{-2}-5.33\times 10^{-2}\right)J=2.67\times 10^{-2}J$