Q. A parallel plate capacitor is of area $6 \; cm^2$ and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1, = 10, K_2 = 12 $ and $K_3 = 14$. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be :

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Solution:

Let dielectric constant of material used be K.
$\therefore \frac{10\in_{0} A/3}{d} + \frac{12\in_{0} A/3}{d} + \frac{14\in_{0} A/3}{d} =\frac{K\in_{0} A}{d} $
$ \Rightarrow K = 12 $

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