A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 300 pF capacitor. Calculate how much electrostatic energy is lost in the process. What is the source of energy loss?
CBSE OUTSIDE DELHI - SET 2 CHENNAI 2015
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We have, energy stored $\frac{1}{2}\frac{\text{Q}}{\text{c}^{2}}$
and Equivalent Capacitance $= C_1 +C_2$
$=(600+300)\ pF$
Charge on the capacitor $= Q = 600 \times 200 \times 10^{-12}$
$= 12 \times 10^{-8} C$
Initial Energy $ = \frac{1}{2}\frac{\text{Q}^{2}}{\text{c}} = \frac{1}{2}\frac{\text{Q}^{2}}{600\times10^{-12}}$
Final Energy $= \frac{1}{2}=\frac{\text{Q}^{2}}{900\times10^{-12}}$
Loss in Energy $ =\frac{1}{2}\frac{144\times10^{-16}}{10^{-12}}\bigg[\frac{1}{600} - \frac{1}{900}\bigg]$
$ = 4 \times10^{-6}\text{J}$
The source of energy loss is the energy converted into heat due to sharing of charge between the two capacitors. (Also accept: heat produced).
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