A parallel-plate capacitor features a plate area of 100 cm ^2 and a plate separation of 1.0 cm . A glass plate (dielectric constant 6.0 ) with a thickness of 6.0 mm and an ebonite plate (dielectric constant 4.0 ) are placed one over the other to completely fill the space between the capacitor plates. Determine the new capacitance.

Step-by-step solution

Given plate area A = 100 cm ^2 = 10⁻² m ^2 and separation d = 1.0 cm = 10⁻² m . The space between the plates is completely filled with two dielectric slabs. The thickness of the glass plate is t₁ = 6.0 mm = 6 10⁻³ m with dielectric constant K₁ = 6.0 . The thickness of the ebonite plate is t₂ = d - t₁ = 1.0 cm - 0.6 cm = 0.4 cm = 4 10⁻³ m with dielectric constant K₂ = 4.0 . The equivalent capacitance of the parallel-plate capacitor with two dielectric slabs in series is given by: C = ₀ A t₁ K₁ + t₂ K₂ Substituting t