A dielectric slab of thickness $6 c m$ is placed between the plates of a parallel plate capacitor. If the distance between the plates is reduced by $4 c m$ , the capacity of the capacitor remains same. Find the dielectric constant of the medium.

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Solution

The correct option is B $3$
$\begin{matrix} T h e c a p a c i tan c e i s g i v e n b y C^{'} = \frac{\in_{0} A}{d} w h e r e A i s p l a t e a r e a a n d d i s i n i t i a l s e p a r a t i o n b e t w e e n t h e m, c a p a c i tan c e w h e n a d i e l e c t r i c s l a b o f t h i c k n e s s t i s i n t r o d u c c e d i s t h e n C^{'} = \frac{\in_{0} A}{d} - t + \frac{t}{K} N o w, a s a f t e r i n t r o d u c i n g t h e s l a b C w i l l i n c r e a s e a n d t o e q u a l i z e C a n d C^{'} t h e s p a r a t i o n w a s i n c r e a s e d t o 4 c m S o, \frac{\in_{0} A}{d} = \frac{\in_{0} A}{{d + 4 - t + (\frac{t}{K})}} \Rightarrow d = d + 4 - 6 + \frac{6}{K} \Rightarrow K = 3 \end{matrix}$
Hence,
option $(B)$ is correct answer.

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