To find the dielectric constant $ K $ of the slab, we use the formula for the induced charge $ Q_{\text{in}} $ in a parallel plate capacitor:

$ Q_{\text{in}} = Q \left(1 - \frac{1}{K}\right) $

Given:

Total charge $ Q = 5 \times 10^{-6} \, \text{C} $

Induced charge $ Q_{\text{in}} = 4 \times 10^{-6} \, \text{C} $

We substitute the values into the formula:

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

To solve for the dielectric constant $ K $, simplify the equation:

Divide both sides by $ 5 \times 10^{-6} $:

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

Rearrange to solve for $ \frac{1}{K} $:

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

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

Invert both sides to find $ K $:

$ K = 5 $

Therefore, the dielectric constant of the slab is $ K = 5 $.