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

Answer 1

Answer: (A)

Here, $C_{1} = 4\,\mu F = 4 × 10^{-6 } \,F, \,V_{1} = 200 \,V$
Initial electrostatic energy stored in $C_{1}$ is
$U_{1} = \frac{1}{2}C_{1}V_{1}^{2} = \frac{1}{2}×4×10^{-6}×200×200$
$U_{1} = 8 × 10^{-2} \,J$
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
$V = \frac{Total\,charge}{Total\,capacity} ; V = \frac{C_{1}V_{1}}{C_{1}+C_{2}}$
$V = \frac{4×10^{-6}×200}{\left(4+2\right)×10^{-6}} = \frac{800}{6}\,V$
$\therefore \quad$ Final electrostatic energy of both capacitors
$U_{2} = \frac{1}{2}\left(C_{1}+C_{2}\right)V^{2} = \frac{1}{2}×6×10^{-6}× \frac{800}{6}× \frac{800}{6}$
$= 5.33 × 10^{-2} \,J$
$\therefore \quad$ Energy dissipated in the form of heat and electromagnetic radiation is
$U_{1} - U_{2} = \left(8 × 10^{-2} - 5.33 ×10^{-2}\right) \,J = 2.67 × 10^{-2} \,J$