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A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of same area but thickness d/2 is inserted between the plates as shown in figure having dielectric constant K(= 4). The ratio of new capacitance to its original capacitance will be,

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2 : 1

6 : 5

8 : 5

4 : 1

answer is C.

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Detailed Solution

$C_{0} = \frac{ε_{0} A}{d}$

After inserting dielectric

$\begin{aligned} C = \frac{ε_{0} A}{(d - t) + \frac{t}{k}} \\ = \frac{ε_{0} A}{\frac{d}{2} + \frac{d}{8}} \\ = \frac{8 ε_{0} A}{5 d} \\ = \frac{8}{5} C_{0} \\ So, \frac{C}{C_{0}} = \frac{8}{5} \end{aligned}$

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