In the figure shown P₁and P₂ are two conducting plates having charges of equal magnitude and opposite sign. Two dielectrics of dielectric constant K₁ and K₂ fill the space between the plates as shown in the figure. The ratio of electrical energy in 1^st dielectric to that in the 2^nd dielectric is :

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K₂ : K₁

K₁ : K₂

K₂² : K₁²

1 : 1

answer is C.

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Detailed Solution

Let $σ$ be the charge density of conducting plate and V be the volume of either dielectric

$∴ \frac{U_{1}}{U_{2}} = \frac{(\frac{1}{2} k_{1} \in_{0} E_{1}^{2}) V}{(\frac{1}{2} k_{2} \in_{0} E_{2}^{2}) V} = \frac{k_{1}}{k_{2}} \frac{{(\frac{σ}{k_{1} \in_{0}})}^{2}}{{(\frac{σ}{k_{2} E_{0}})}^{2}} = \frac{k_{2}}{k_{1}}$

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