A parallel plate capacitor has charge $5 \times 10^{-6} \mathrm{C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \mathrm{C}$ then the dielectric constant of the slab is _______________.

<p>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:</p> <p>$ Q_{\text{in}} = Q \left(1 - \frac{1}{K}\right) $</p> <p>Given:</p> <p><p>Total charge $ Q = 5 \times 10^{-6} \, \text{C} $</p></p> <p><p>Induced charge $ Q_{\text{in}} = 4 \times 10^{-6} \, \text{C} $</p></p> <p>We substitute the values into the formula:</p> <p>$ 4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right) $</p> <p>To solve for the dielectric constant $ K $, simplify the equation:</p> <p><p>Divide both sides by $ 5 \times 10^{-6} $:</p> <p>$ \frac{4}{5} = 1 - \frac{1}{K} $</p></p> <p><p>Rearrange to solve for $ \frac{1}{K} $:</p> <p>$ 1 - \frac{4}{5} = \frac{1}{K} $</p> <p>$ \frac{1}{5} = \frac{1}{K} $</p></p> <p><p>Invert both sides to find $ K $:</p> <p>$ K = 5 $</p></p> <p>Therefore, the dielectric constant of the slab is $ K = 5 $.</p>