A slab of material of dielectric constant k has the area (1/4)th as that of the plates of a parallel plate capacitor and has thickness 3/4 d where d is the separation of plates. How is the capacitance changed when the slab is inserted between the plates?

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$L e t i n i t i a l c a p a c i \tan c e o f p a r a l l e l p l a t e c a p a c i t o r b e C = ε A / d, w h e r e A = a r e a o f p l a t e d = d i s \tan c e b e t w e e n t w o p l a t e s a n d w h e n d i l e c t r i c i s i n s e r t e d t h e n e w c a p a c i \tan c e b e C_{1} = \frac{ε A}{(d - t + \frac{t}{K})}, w h e r e k i s t h e d i l e c t r i c c o n s \tan t o f t h e s l a b . A r e a o f d i l e c t r i c = \frac{A}{4}, t h i c k n e s s = \frac{3}{4} d C_{1} = \frac{ε \frac{A}{4}}{(d - \frac{3}{4} d + \frac{3 d}{4 K}) = \frac{ε A}{4}} \times \frac{4 K}{(K d + 3 d)} = \frac{ε A K}{(K d + 3 d)} \frac{C_{1}}{C} = \frac{ε A K}{(K d + 3 d)} \times \frac{d}{ε A} = \frac{K}{K + 3} C_{1} = (\frac{K}{K + 3}) C r e g a r d s$

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