A parallel plate capacitor is filled with a dielectric of dielectric constant (relative permittivity) 5 between its plates and is charged to acquire an energy E. Then it is isolated (disconnected from the battery) and the dielectric is replaced by another dielectric of dielectric constant (relative permittivity) 2. The new energy stored in the capacitor

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2.5E

10E

answer is B.

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Detailed Solution

When battery in disconnected
$\begin{array}{l} \Rightarrow \frac{E}{E^{'}} = \frac{Q^{2} / 2 kc}{Q^{2} / 2 k} = \frac{K^{1}}{K} = \frac{2}{5} \\ E^{'} = \frac{5}{2} E; E^{'} = 2 .5 E \end{array}$

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