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. The electric field between the plates of a parallel plate capacitor is E = (sigma/ε₀) , where sigma = (Q/A) is the surface charge density, Q is the charge, and A is the overlapping area. If the area A is reduced while Q is constant, sigma increases ( sigma = (Q/A) ), but in the region of overlap, E = (sigma/ε₀) depends on sigma