Question

A series $R C$ circuit is formed using a resistance $R$, a capacitor without dielectric having a capacitance $C=2\, F$ and a battery of emf $E=3\, V$. 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

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