A capacitor has air as dielectric medium and two conducting plates of area $12\ \text{cm}^2$ and they are $0.6\ \text{cm}$ apart. When a slab of dielectric having area $12\ \text{cm}^2$ and $0.6\ \text{cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2\ \text{cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is: (Given $\epsilon_0 = 8.834\times10^{-12}\ \text{F/m}$)
Answer: (D) 1.50
The slab raises the capacitance, so the plates are moved apart to a separation $d' = 0.8\ \text{cm}$. For equal capacitance, the effective air gap must stay $0.6\ \text{cm}$:
$$d' - t + \frac tK = d \;\Rightarrow\; 0.8 - 0.6 + \frac{0.6}{K} = 0.6 \;\Rightarrow\; K = \frac{0.6}{0.4} = 1.5$$
Solution by Sreeraj P, M.Sc Physics