A $5.0 \mathrm{\mu F}$ capacitor having a charge of $20 \mathrm{\mu C}$ is discharged through a wire of resistance  $5.0 \mathrm{\Omega }$. Find the heat ( in $\mathrm{\mu J}$ ) dissipated in the wire between 25 $\mathrm{\mu s}$ to 50 $\mathrm{\mu s}$ after the connections are made.

The charge on the capacitor at time $t$ after the connections are made is 

$Q=Q_0{e}^{-t/RC} \Rightarrow i=\frac{dQ}{dt}=-(Q_0/RC)e^{-t/RC}$

Heat dissipated during the time $t_1$ to $t_2$ is $U=\underset{t_1}{\overset{t_2}{\int }}{i}^{2}Rdt$

$=\underset{t_1}{\overset{t_2}{\int }}\frac{Q_0}{R{C}^2}{e}^{-2t/RC}dt=\frac{Q_0^2}{2C}(e^{-\frac{2t_1}{RC}}-e^{-\frac{2t_2}{RC}})...............\left(1\right)$

The time constant RC is $5\mathrm{\Omega }\times 5.0\mathrm{\mu F}=25\mathrm{\mu s}$

Putting ${\mathrm{t}}_1=25\mathrm{\mu s},{\mathrm{t}}_2=50\mathrm{\mu s}$ and other values in (i), 

$U=\frac{{\left(20\mu C\right)}^2}{2\times 5.0\mu F}(e^{-2}-e^{-4})=4.7\mathrm{\mu J}$