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Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is
$\frac{{2{ \in _r}}}{{1\, + \, \in _r^2}}$
${ \in _r}$
$\frac{1}{{{ \in _r}}}$
$\frac{{{ \in _r}}}{{1\, + \, \in _r}}$
Solution
$ \mathrm{u}_{1} =\frac{1}{2} \times 2 \mathrm{Cv}^{2}=\mathrm{Cv}^{2} $
${{\rm{u}}_2} = \frac{1}{2}{\rm{C}}{ \in _{\rm{r}}}{{\rm{v}}^2} + \frac{{{{\rm{C}}^2}{{\rm{v}}^2}}}{{2{_{\rm{r}}}{\rm{C}}}}$
$ = \frac{{{ \in _1}{\rm{C}}{{\rm{v}}^2}}}{2} + \frac{{{\rm{C}}{{\rm{v}}^2}}}{{2 \in {\rm{v}}}}$
${{\rm{u}}_2} = \frac{{{\rm{C}}{{\rm{v}}^2}\left[ { \in _{\rm{r}}^2 + 1} \right]}}{{2{ \in _{\rm{r}}}}} \Rightarrow \frac{{{{\rm{u}}_1}}}{{{{\rm{u}}_2}}} = \frac{{2{ \in _{\rm{r}}}}}{{1 + \in _1^2}}$
