A combination of parallel plate capacitors is maintained at a certain potential difference. When a $3$ mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by $2.4$ mm. Find the dielectric constant of the slab.
Answer: (D) $5$
Each capacitor must keep its capacitance (and hence the same charge at the same potential difference). A slab of thickness $t$ in a gap $d$ gives
$$C = \frac{\varepsilon_0A}{d - t + t/K}$$
so the effective gap shrinks by $t\left(1 - \dfrac1K\right)$. Restoring it needs the plates to move apart by the same amount:
$$3\left(1 - \frac1K\right) = 2.4 \;\Rightarrow\; \frac1K = 0.2 \;\Rightarrow\; K = 5$$
Solution by Sreeraj P, M.Sc Physics