The capacity and the energy stored in a charged parallel plate condenser with air between its plates are respectively $C 0$ and $W 0.$ If the air is replaced by glass (dielectric constant = 5) between the plates, the capacity of the plates and the energy stored in it will respectively be $:$

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Solution

$C = K C_{0} = 5 C_{0}$
$U = \frac{Q^{2}}{2 C} = \frac{Q^{2}}{2 K C_{0}}$
$= \frac{W_{0}}{K} = \frac{W_{0}}{5} .$

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The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_{0} a n d W_{0}$ .If the air is replaced by the glass (dielectric constant = $5$ ) between the plates, the capacity of the plates and the energy stored in it will respectively be: