A parallel plate capacitor has plate area 50 ~cm ^2 and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4 . The capacitance becomes ( ₀= permittivity of free space)

Step-by-step solution

The capacitance of a parallel plate capacitor with a dielectric slab of thickness t and dielectric constant K inserted between plates separated by distance d is given by: C = ₀ A d - t + t K Substituting the given values A = 50 10⁻⁴ m^2 , d = 3 10⁻³ m , t = 1 10⁻³ m , and K = 4 : C = ₀ (50 10⁻⁴) (3 10⁻³) - (1 10⁻³) + (1 10⁻³) 4 Evaluating the denominator: d - t = 2 10⁻³ m t K = 0.25 10⁻³ m Total denominator = 2.25 10⁻³ m The capacitance simplifies to: C = 50 10⁻⁴ 2.25 10⁻³ ₀ = 5 2.25 ₀ = 20 9 ₀ This result correspo