# 6.7 Displacement Current

Ampère’s law in differential form is given by

∇​×H=J+∂D∂t (A m p è r e's l a w).(6.41)

Integrating both sides of Eq. (6.41) over an arbitrary open surface S with contour C, we have

∫S(∇×H)⋅d s=∫S J⋅d s+∫S∂D∂t⋅d s.(6.42)

The surface integral of J equals the conduction current I c flowing through S, and the surface integral of ∇×H can be converted into a line integral of H over the contour C bounding C by invoking Stokes’s theorem. That is,

∫S J⋅d s=I c,∫S(∇​×H)⋅d s=∮C H⋅d l,

and

The second term on the right-hand side of Eq. (6.43) of course has the same unit (amperes) as the current I c, and because it is proportional to the time derivative of the electric flux density D, which is also called the electric displacement, it is called the displacement current I d. That is,

where J d=∂D/∂t represents a displacement current density. In view of Eq. (6.44),

∮C H⋅d l=I c+I d=I,(6.45)

where I is the total current. In electrostatics, ∂D/∂t=0; therefore, I d=0 and I=I c. The concept of displacement current was first introduced in 1873 by James Clerk Maxwell when he formulated the unified theory of electricity and magnetism under time-varying conditions.

The parallel-plate capacitor is commonly used as an example with which to illustrate the physical meaning of the displacement current I d. The simple circuit shown in Fig. 6-13 consists of a capacitor and an ac source with voltage V s(t) given by

V s(t)=V 0 cos ω t (V).(6.46)

According to Eq. (6.45), the total current flowing through any surface consists, in general, of a conduction current I c and a displacement current I d. Let us find I c and I d through each of the following two imaginary surfaces: (1) the cross section of the conducting wire, S 1, and (2) the cross section of the capacitor S 2 (Fig. 6-13). We denote the conduction and displacement currents in the wire as I 1 c and I 1 d and those through the capacitor as I 2 c and I 2 d.

# Figure 6-13 The displacement current I 2 d in the insulating material of the capacitor is equal to the conducting current I 1 c in the wire.

Figure 6 13 Full Alternative Text

In the perfectly conducting wire, D=E=0; hence, Eq. (6.44) gives I 1 d=0. As for I 1 c, we know from circuit theory that it is related to the voltage across the capacitor V C by

I 1 c=C d V C d t=C d d t(V 0 cos ω t)=−C V 0 ω sin ω t,(6.47)

where we used the fact that V C=V s(t). With I 1 d=0, the total current in the wire is simply I 1=I 1 c=−C V 0 ω sin ω t.

In the perfect dielectric with permittivity ε between the capacitor plates, σ=0. Hence, I 2 c=0 because no conduction current exists there. To determine I 2 d, we need to apply Eq. (6.44). From Example 4-14, the electric field E in the dielectric spacing is related to the voltage V c across its plates by

E=y ˆ V c d=y ˆ V 0 d cos ω t,(6.48)

where d is the spacing between the plates and y ˆ is the direction from the higher-potential plate toward the lower-potential plate at t=0. The displacement current I 2 d is obtained by applying Eq. (6.44) with d s=y ˆ d s:

I 2 d=∫S∂D∂t⋅d s=∫A∂∂t y ˆ ε V 0 d cos ω t⋅(y ˆ d s)=−ε A d V 0 ω sin ω t=−C V 0 ω sin ω t,(6.49)

where we used the relation C=ε A/d for the capacitance of the parallel-plate capacitor with plate area A. The expression for I 2 d in the dielectric region between the conducting plates is identical with that given by Eq. (6.47) for the conduction current I 1 c in the wire. The fact that these two currents are equal ensures the continuity of total current flow through the circuit.

In the capacitor example, we treated the wire as a perfect conductor, and we assumed that the space between the capacitor plates was filled with a perfect dielectric. If the wire has a finite conductivity σ w, then D in the wire would not be zero; therefore, the current I 1 would consist of a conduction current I 1 c as well as a displacement current I 1 d; that is, I 1=I 1 c+I 1 d. By the same token, if the dielectric spacing material has a nonzero conductivity σ d, then free charges would flow between the two plates, and I 2 c would not be zero. In that case, the total current flowing through the capacitor would be I 2=I 2 c+I 2 d. No matter the circumstances, the total capacitor current remains equal to the total current in the wire. That is, I 1=I 2.

# Example 6-7: Displacement Current Density

The conduction current flowing through a wire with conductivity σ=2×10 7 S/m and relative permittivity ε r=1 is given by I c=2 sin ω t (mA). If ω=10 9 rad/s, find the displacement current.

Solution: The conduction current I c=J A=σ E A, where A is the cross section of the wire. Hence,

E=I c σ A=2×10−3 sin ω t 2×10 7 A=1×10−10 A sin ω t (V/m).

Application of Eq. (6.44), with D=ε E, leads to

I d=J d A=ε A∂E∂t=ε A∂∂t 1×10−10 A sin ω t=ε ω×10−10 cos ω t=0.885×10−12 cos ω t(A),

where we used ω=10 9 rad/s and ε=ε 0=8.85×10−12 F/m. Note that I c and I d are in phase quadrature (90∘ phase shift between them). Also, I d is about nine orders of magnitude smaller than I c, which is why the displacement current usually is ignored in good conductors.