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

818-721

  • A

    $\frac{{2{ \in _r}}}{{1\, + \, \in _r^2}}$

  • B

    ${ \in _r}$

  • C

    $\frac{1}{{{ \in _r}}}$

  • D

    $\frac{{{ \in _r}}}{{1\, + \, \in _r}}$

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  • [AIIMS 2009]

Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be