1. Given data

• Charge on capacitor plates:
  $Q = 5 \times 10^{-6}\,\text{C}$
• Induced charge on one face of dielectric slab:
  $q' = 4 \times 10^{-6}\,\text{C}$

We need to find the dielectric constant $K$.

2. Relation between free charge and induced charge

When a dielectric slab completely (or almost completely) fills the space between the plates of a charged capacitor, the induced surface charge on the dielectric is related to the free charge on the plates by

$q' = Q\left(1 - \frac{1}{K}\right)$

This comes from polarization of the dielectric.

3. Substitute the values

$4 \times 10^{-6} = 5 \times 10^{-6}\left(1 - \frac{1}{K}\right)$

Cancel $10^{-6}$ from both sides:

$4 = 5\left(1 - \frac{1}{K}\right)$

$\frac{4}{5} = 1 - \frac{1}{K}$

$\frac{1}{K} = 1 - \frac{4}{5} = \frac{1}{5}$

Hence,

$K = 5$

4. Final answer

The dielectric constant of the slab is

$\boxed{5}$