Question

A series $RC$ circuit is formed using a resistance $R$, a capacitor without dielectric having a capacitance $C=2F$ and a battery of emf $E=3V$. The circuit is completed and it is allowed to attain the steady state. After this, at $t=0$, half the thickness of the capacitor is filled with a dielectric of constant $K=2$ as shown in the figure. The system is again allowed to attain a steady state. What will be the heat generated (in joule) in the capacitor between $t=0$ and $t=\infty$ ?

Answer 1

Answer: 3

Initial charge on capacitor $=CE$
Initial potential energy of capacitor $=C{E}^2/2$
Now, $C={\epsilon }_{0}A/d$
New capacitance: $C^{\prime }=\frac{{\epsilon }_{0}A}{d/2}$ in series with $\frac{{\epsilon }_{0}KA}{d/2}$
$\Rightarrow C^{\prime }=\frac{2{\epsilon }_{0}A}{d}$ in series with $\frac{4{\epsilon }_{0}A}{d}$
$=\frac{{\epsilon }_{0}A}{d}\left(\frac{2\times 4}{2+4}\right)=\frac{4{\epsilon }_{0}A}{3d}=\frac{4}{3}C$
$\Rightarrow$ New charge $=\frac{4}{3}CE$
New energy $=\frac{1}{2}\times \frac{4}{3}C{E}^2=\frac{2}{3}C{E}^2$.
Now, $W_{\text{battery }}=\Delta H+\Delta U$
$\Rightarrow E\left(\frac{4}{3}CE-CE\right)$
$=\Delta H+\frac{2}{3}C{E}^2-\frac{1}{2}C{E}^2$
$\Rightarrow \Delta H=\frac{1}{6}C{E}^2=3J$