A slab of dielectric constant 4.5 is placed in a field of 7.2 × 10⁴ V/m, at right angles to its…

Step 1. The field inside. $E = \dfrac{E_0}{K} = \dfrac{72\ \text{kV/m}}{4.5} = 16\ \text{kV/m}$.

Step 2. The field of the faces. It acts against the applied field, so $E_p = E_0 - E = 72\ \text{kV/m} - 16\ \text{kV/m} = 56\ \text{kV/m}$.

Step 3. The bound charge. $E_p = \dfrac{\sigma_b}{\varepsilon_0}$, so $\sigma_b = \varepsilon_0 E_p = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times 5.6 \times 10^{4}\ \text{V/m} = 4.96 \times 10^{-7}\ \text{C/m}^2$, that is $0.496\ \mu\text{C/m}^2$.