A parallel plate capacitor with a dielectric slab $(ε_{r} = 3)$ filling the space between the plates is charged to potential V and isolated. Then the dielectric slab is drawn out and another dielectric of equal thickness but $ε_{r} = 2$ is introduced between the plates. The ratio of the energy stored in the capacitor later to that initially is

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$9 : 4$

$2 : 3$

$3 : 2$

$4 : 9$

answer is B.

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Detailed Solution

$V_{i} = \frac{q^{2}}{2 k_{1} c}$

$V_{f} = \frac{q^{2}}{2 k_{2} c} \Rightarrow \frac{V_{f}}{V_{i}} = \frac{K_{1}}{K_{2}}$

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