The distance between the plates of a parallel plate capacitor is $d$ . A metal plate of thickness $d / 2$ is placed between the plates. The capacitance would be then be

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Solution

The correct option is B Initial
$\begin{matrix} C a s e I : (A i r b e t w e e n p l a t e s) C a p a c i tan c e, C = \frac{E_{0} A}{d} C a s e I I : (M e t a l p l a t e i s i n s e r t e d b e t w e e n t h e p l a t e s) c a p a c i tan e, C^{'} = \frac{E_{0} A}{d - t + \frac{t}{k}} W h e r e t = t h i c k n e s s o f m e t a l p l a t e = \frac{d}{2} k = D i e l e c t r i c c o n s tan t o f m e t a l p l a t e = \infty] C^{''} = \frac{E_{0} A}{d - t + \frac{\frac{d}{2}}{\infty}} = \frac{E_{0} A}{\frac{d}{2}} = \frac{2 E_{0} A}{d} = 2 C H e n c e, i t c a n b e c o n c u l e d t h a t t h e n e w c a p a c i tan c e b e c o m e s d o u b l e t h a t o f i n i t i a l c a p a c i tan e . \end{matrix}$

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The distance between the plates of a parallel plate capacitor is d. A metal plate of thickness d/2 is placed between the plates. The capacitance would be then be

The distance between the plates of a parallel plate capacitor is $d$ . A metal plate of thickness $d / 2$ is placed between the plates. The capacitance would be then be