Skip to content
# Example 6-7: Displacement CurrentDensityThe conduction current flowing through a wire with conductivity σ = 2 × 107 S/m and relative permittivity ‘r = 1 is givenby Ic = 2 sin ωt (mA). If ω = 109 rad/s, find the displacementcurrent.Solution: The conduction current Ic = J A = σEA, where Ais the cross section of the wire. Hence,E = IcσA = 2 × 10−3 sin ωt2 × 107A = 1 × 10−10Asin ωt (V/m).Application of Eq. (6.44), with D = ‘E, leads toId = JdA = ‘A∂E∂t = ‘A∂∂t!1 × 10−10Asin ωt”= ‘ω × 10−10 cos ωt = 0.885 × 10−12 cos ωt (A),where we used ω = 109 rad/s and ‘ = ‘0 = 8.85×10−12 F/m.Note that Ic and Id are in phase quadrature (90◦ phase shiftbetween them). Also, Id is about nine orders of magnitudesmaller than Ic, which is why the displacement current usuallyis ignored in good conductors.

%d bloggers like this: