The two plates of a parallel plate capacitor are 4 mm apart. A slab...

Here, distance between parallel plates d = 4 mm = 0.004 m, K = 3, thickness t = 3 mm = 0.003 m and ${d}_{1}$ =?

$∴ C = \dfrac{{ε}_{0} A}{d}$ and ${C}_{1} = \dfrac{{ε}_{0} A}{{d}_{1} - t( 1 - \dfrac{1}{K} )}$

since ${C}_{1} = \dfrac{2}{3} C$ (given)
$$\begin{array}{ll}∴ \dfrac{{ε}_{0} A}{{d}_{1} - t( 1 - \dfrac{1}{K} )} = \dfrac{2}{3} \dfrac{{ε}_{0} A}{d} \\ \dfrac{1}{{d}_{1} - t( 1 - \dfrac{1}{K} )} = \dfrac{2}{3d} \\ \dfrac{1}{{d}_{1} - 0.003( 1 - \dfrac{1}{3} )} = \dfrac{2}{3 \times 0.004} \\ \dfrac{1}{{d}_{1} - 0.003 \times \dfrac{2}{3}} = \dfrac{1}{0.006} \\ \dfrac{1}{{d}_{1} - 0.002} = \dfrac{1}{0.006} \\ {d}_{1} - 0.002 = 0.006 \\ {d}_{1} = 0.006 + 0.002 = 0.008\text{m} = 8\text{mm}\text{.}\end{array}$$