Practice question

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. When a dielectric ( K > 1 ) is inserted into a disconnected capacitor, the charge Q remains constant. The dielectric polarizes, creating an induced field opposing the applied field. The effective field inside the dielectric becomes E = (E₀/K) , where E₀ = (sigma/ε₀) is the field without the dielectric ( sigma = Q/A ). Since K > 1 , the field decreases