A parallel-plate capacitor has plate area 4 cm² and separation d = 1.77 mm. The space is filled by two dielectric layers of equal thickness d/2 with dielectric constants 5 and 3, as shown in the original figure. This capacitor is connected in parallel with a 7.5 pF capacitor. Given ε₀ = 8.85×10⁻¹² F/m, find the total equivalent capacitance in pF.

Question

G1 Physics · Accepted Answer

The two dielectric layers are stacked along the electric field, so they form two capacitors in series. Plate area: A = 4 cm² = 4×10⁻⁴ m². Each layer thickness: d/2 = 1.77×10⁻³/2 m. For dielectric K=5: C₁ = Kε₀A/(d/2) = 5(8.85×10⁻¹²)(4×10⁻⁴)/(1.77×10⁻³/2) = 20 pF. For dielectric K=3: C₂ = 3(8.85×10⁻¹²)(4×10⁻⁴)/(1.77×10⁻³/2) = 12 pF. Series equivalent: C s C_s C s ​ = C₁C₂/(C₁+C₂) = (20×12)/(20+12) = 240/32 = 7.5 pF. This is in parallel with another 7.5 pF capacitor: C t o t a l C_total C t ​ o t…