A parallel plate capacitor has charge $$5 \times 10^{-6}$$ 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 $$4 \times 10^{-6}$$ C then the dielectric constant of the slab is _____.

Let the free charge on the plates of the capacitor be $$Q_f = 5 \times 10^{-6}\, \text{C}$$.

When a dielectric slab is inserted, the molecules of the slab polarise and create two layers of induced (bound) charge. The magnitude of the induced charge on either face of the slab is given as $$Q_{\text{ind}} = 4 \times 10^{-6}\, \text{C}$$.

For a parallel-plate capacitor completely (or almost completely) filled with a dielectric of relative permittivity $$K$$, the relation between the free surface charge density $$\sigma_f$$ and the bound surface charge density $$\sigma_b$$ is derived as follows:

• The electric displacement vector $$\mathbf{D}$$ inside the dielectric is $$\mathbf{D} = \sigma_f \hat{n}$$ (since $$\mathbf{D}$$ originates from free charge only).
• Electric field inside the dielectric: $$E = \frac{\sigma_f}{K \varepsilon_0}$$.
• Polarisation $$P$$ of the dielectric: $$P = \varepsilon_0 \chi_e E = \varepsilon_0 (K-1) E$$, where $$\chi_e = K-1$$.
• Surface bound charge density: $$\sigma_b = P = \varepsilon_0 (K-1) \frac{\sigma_f}{K \varepsilon_0} = \frac{K-1}{K} \, \sigma_f$$.

Therefore the ratio of the magnitudes of bound to free charge is

$$\frac{Q_{\text{ind}}}{Q_f} = \frac{\sigma_b}{\sigma_f} = \frac{K-1}{K}\;.$$

Substitute the given values:

$$\frac{4 \times 10^{-6}}{5 \times 10^{-6}} = \frac{K-1}{K}$$

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

Cross-multiplying:

$$5(K-1) = 4K \quad\Rightarrow\quad 5K - 5 = 4K$$

$$K = 5$$

Hence, the dielectric constant of the slab is $$5$$.