A rectangular dielectric slab has dimensions of 20 cm 20 cm 1.0 mm and a dielectric constant of 4.0 . If its two square faces are metal-coated, determine the capacitance between these coated surfaces.

Step-by-step solution

The capacitance of a parallel plate capacitor with a dielectric is given by C = K ₀ A d Given: Dielectric constant, K = 4.0 Area of the square faces, A = 20 cm 20 cm = 400 cm ^2 = 4 10⁻² m ^2 Distance between the plates (thickness), d = 1.0 mm = 10⁻³ m Permittivity of free space, ₀ = 8.85 10⁻¹² F/m Substituting the values into the formula: C = 4.0 8.85 10⁻¹² 4 10⁻² 10⁻³ C = 141.6 10⁻¹¹ F C = 1.416 10⁻⁹ F 1.42 nF