In a parallel plate air capacitor of plate separation '$d$', a dielectric slab of thickness '$t$' is introduced between the plates. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

The capacitance of a parallel plate air capacitor is given by:

$C_0 = \dfrac{\epsilon_0 A}{d}$

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates, the new capacitance becomes:

$C = \dfrac{\epsilon_0 A}{d - t + \dfrac{t}{K}}$

According to the given condition, the new capacitance is one-third of the original value:

$C = \dfrac{C_0}{3}$

Substituting the expressions for $C$ and $C_0$:

$\dfrac{\epsilon_0 A}{d - t + \dfrac{t}{K}} = \dfrac{\epsilon_0 A}{3d}$

Equating the denominators:

$d - t + \dfrac{t}{K} = 3d$

Rearranging the terms to solve for $K$:

$\dfrac{t}{K} = 2d + t$

$K = \dfrac{t}{2d + t}$

Answer: $\dfrac{t}{2d + t}$