• # question_answer A parallel plate capacitor has plates with area A and separation between it is d. A battery charges the plates to a potential difference${{V}_{0}}$The battery is then disconnected and a dielectric slab of thickness d is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced is A)  K                                 B)  $\frac{1}{K}$ C)  $\frac{A}{{{d}^{2}}K}$               D)  $\frac{{{d}^{2}}K}{A}$

$\Sigma mvr=\,({{l}_{system}})\omega$ Now, $\Rightarrow$ As battery is disconnected, Q remains unaltered. So, $mv\frac{l}{2}=\frac{(2m)\,{{(2l)}^{2}}}{12}\omega =\frac{2m(4{{l}^{2}})}{12}\omega$ $\Rightarrow$            $\omega =\frac{3v}{4l}$