The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C₀ and W₀. 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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$\frac{C_{o}}{5}, 5 W_{o}$

$5 C_{o}, \frac{W_{0}}{5}$

$5 C_{o}, 5 W_{o}$

$\frac{C_{o}}{5}, \frac{W_{o}}{5}$

answer is B.

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Detailed Solution

When a dielectric K is introduced in a parallel plate condenser its capacity becomes K times. Hence $C' = 5 C_{0}$ . Energy stored $W_{0} = \frac{q^{2}}{2 C_{0}}$

$∴ W^{'} = \frac{q^{2}}{2 C^{'}} = \frac{q^{2}}{2 \times 5 C_{0}} \Rightarrow W^{'} = \frac{W_{0}}{5}$

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