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Question

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $$3d/4$$, where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?


Solution

$$C = \cfrac{A \epsilon_0}{d}$$

$$C' = \cfrac{A \epsilon_0}{d - t + \cfrac{t}{k}}$$

Put $$t = \cfrac{3d}{4}$$ ;

$$C' = \cfrac{4k}{3 + k} . \cfrac{A \epsilon_0}{d}$$

$$C' = \cfrac{4k}{3 + k} . C$$


Physics
NCERT
Standard XII

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