Displacement current

In the case of a very simple dielectric material the constitutive relation holds:

$${\displaystyle \mathbf {D} =\varepsilon \,\mathbf {E} ~,}$$

where the permittivity ${\displaystyle \varepsilon =\varepsilon _{0}\,\varepsilon _{\mathrm {r} }}$ is the product of:

• ε₀, the permittivity of free space, or the electric constant; and
• ε_r, the relative permittivity of the dielectric.

In the equation above, the use of ε accounts for the polarization (if any) of the dielectric material.

The scalar value of displacement current may also be expressed in terms of electric flux:

$${\displaystyle I_{\mathrm {D} }=\varepsilon \,{\frac {\,\partial \Phi _{\mathrm {E} }\,}{\partial t}}~.}$$

The forms in terms of scalar ε are correct only for linear isotropic materials. For linear non-isotropic materials, ε becomes a matrix; even more generally, ε may be replaced by a tensor, which may depend upon the electric field itself, or may exhibit frequency dependence (hence dispersion).

For a linear isotropic dielectric, the polarization P is given by:

$${\displaystyle \mathbf {P} =\varepsilon _{0}\chi _{\mathrm {e} }\,\mathbf {E} =\varepsilon _{0}(\varepsilon _{\mathrm {r} }-1)\,\mathbf {E} ~,}$$

where χ_e is known as the susceptibility of the dielectric to electric fields. Note that

$${\displaystyle \varepsilon =\varepsilon _{\mathrm {r} }\,\varepsilon _{0}=\left(1+\chi _{\mathrm {e} }\right)\,\varepsilon _{0}~.}$$

An example illustrating the need for the displacement current arises in connection with capacitors with no medium between the plates. Consider the charging capacitor in the figure. The capacitor is in a circuit that causes equal and opposite charges to appear on the left plate and the right plate, charging the capacitor and increasing the electric field between its plates. No actual charge is transported through the vacuum between its plates. Nonetheless, a magnetic field exists between the plates as though a current were present there as well. One explanation is that a displacement current I_D "flows" in the vacuum, and this current produces the magnetic field in the region between the plates according to Ampère's law:[3][4]

An electrically charging capacitor with an imaginary cylindrical surface surrounding the left-hand plate. Right-hand surface R lies in the space between the plates and left-hand surface L lies to the left of the left plate. No conduction current enters cylinder surface R, while current I leaves through surface L. Consistency of Ampère's law requires a displacement current I_D = I to flow across surface R.

$${\displaystyle \oint _{C}\mathbf {B} \cdot \operatorname {d} \!{\boldsymbol {\ell }}=\mu _{0}I_{\mathrm {D} }~,}$$

where

• ${\displaystyle \oint _{C}}$ is the closed line integral around some closed curve C;
• ${\displaystyle \mathbf {B} }$ is the magnetic field measured in teslas;
• ${\displaystyle \operatorname {\cdot } ~}$ is the vector dot product;
• ${\displaystyle \mathrm {d} {\boldsymbol {\ell }}}$ is an infinitesimal line element along the curve C, that is, a vector with magnitude equal to the length element of C, and direction given by the tangent to the curve C;
• ${\displaystyle \mu _{0}\,}$ is the magnetic constant, also called the permeability of free space; and
• ${\displaystyle I_{\mathrm {D} }\,}$ is the net displacement current that passes through a small surface bounded by the curve C.

The magnetic field between the plates is the same as that outside the plates, so the displacement current must be the same as the conduction current in the wires, that is,

$${\displaystyle I_{\mathrm {D} }=I\,,}$$

which extends the notion of current beyond a mere transport of charge.

Next, this displacement current is related to the charging of the capacitor. Consider the current in the imaginary cylindrical surface shown surrounding the left plate. A current, say I, passes outward through the left surface L of the cylinder, but no conduction current (no transport of real charges) crosses the right surface R. Notice that the electric field E between the plates increases as the capacitor charges. That is, in a manner described by Gauss's law, assuming no dielectric between the plates:

$${\displaystyle Q(t)=\varepsilon _{0}\oint _{S}\mathbf {E} (t)\cdot \operatorname {d} \!\mathbf {S} \,,}$$

where S refers to the imaginary cylindrical surface. Assuming a parallel plate capacitor with uniform electric field, and neglecting fringing effects around the edges of the plates, according to charge conservation equation

$${\displaystyle I=-{\frac {\mathrm {d} Q}{\mathrm {d} t}}=-\varepsilon _{0}\oint _{S}{\frac {\partial \mathbf {E} }{\partial t}}\cdot \operatorname {d} \!\mathbf {S} =S\,\varepsilon _{0}{\Biggl .}{\frac {\partial \mathbf {E} }{\partial t}}{\Biggr |}_{R}~,}$$

where the first term has a negative sign because charge leaves surface L (the charge is decreasing), the last term has a positive sign because unit vector of surface R is from left to right while the direction of electric field is from right to left, S is the area of the surface R. The electric field at surface L is zero because surface L is in the outside of the capacitor. Under the assumption of a uniform electric field distribution inside the capacitor, the displacement current density J_D is found by dividing by the area of the surface:

$${\displaystyle \mathbf {J} _{\mathrm {D} }={\frac {\mathbf {I} _{\mathrm {D} }}{S}}={\frac {\mathbf {I} }{S}}=\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}={\frac {\partial \mathbf {D} }{\partial t}}~,}$$

where I is the current leaving the cylindrical surface (which must equal I_D) and J_D is the flow of charge per unit area into the cylindrical surface through the face R.

Combining these results, the magnetic field is found using the integral form of Ampère's law with an arbitrary choice of contour provided the displacement current density term is added to the conduction current density (the Ampère-Maxwell equation):[5]

$${\displaystyle \oint _{\partial S}\mathbf {B} \cdot \operatorname {d} \!{\boldsymbol {\ell }}=\mu _{0}\int _{S}\left(\mathbf {J} +\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\right)\cdot \operatorname {d} \!\mathbf {S} \,.}$$

This equation says that the integral of the magnetic field B around the edge ${\displaystyle \partial S}$ of a surface S is equal to the integrated current J through any surface with the same edge, plus the displacement current term ${\displaystyle \varepsilon _{0}\partial \mathbf {E} /\partial t}$ through whichever surface.

Example showing two surfaces S₁ and S₂ that share the same bounding contour ∂S. However, S₁ is pierced by conduction current, while S₂ is pierced by displacement current. Surface S₂ is closed under the capacitor plate.

As depicted in the figure to the right, the current crossing surface S₁ is entirely conduction current. Applying the Ampère-Maxwell equation to surface S₁ yields:

$${\displaystyle B={\frac {\mu _{0}I}{2\pi r}}~.}$$

However, the current crossing surface S₂ is entirely displacement current. Applying this law to surface S₂, which is bounded by exactly the same curve ${\displaystyle \partial S}$, but lies between the plates, produces:

$${\displaystyle B={\frac {\mu _{0}I_{\mathrm {D} }}{2\pi r}}~.}$$

Any surface S₁ that intersects the wire has current I passing through it so Ampère's law gives the correct magnetic field. However a second surface S₂ bounded by the same edge ${\displaystyle \partial S}$ could be drawn passing between the capacitor plates, therefore having no current passing through it. Without the displacement current term Ampere's law would give zero magnetic field for this surface. Therefore, without the displacement current term Ampere's law gives inconsistent results, the magnetic field would depend on the surface chosen for integration. Thus the displacement current term ${\displaystyle \varepsilon _{0}\partial \mathbf {E} /\partial t}$ is necessary as a second source term which gives the correct magnetic field when the surface of integration passes between the capacitor plates. Because the current is increasing the charge on the capacitor's plates, the electric field between the plates is increasing, and the rate of change of electric field gives the correct value for the field B found above.

In a more mathematical vein, the same results can be obtained from the underlying differential equations. Consider for simplicity a non-magnetic medium where the relative magnetic permeability is unity, and the complication of magnetization current (bound current) is absent, so that ${\displaystyle \mathbf {M} =0}$ and ${\displaystyle \mathbf {J} =\mathbf {J} _{\mathrm {f} }}$. The current leaving a volume must equal the rate of decrease of charge in a volume. In differential form this continuity equation becomes:

$${\displaystyle \nabla \cdot \mathbf {J} _{\mathrm {f} }=-{\frac {\partial \rho _{\mathrm {f} }}{\partial t}}\,,}$$

where the left side is the divergence of the free current density and the right side is the rate of decrease of the free charge density. However, Ampère's law in its original form states:

$${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} _{\mathrm {f} }\,,}$$

which implies that the divergence of the current term vanishes, contradicting the continuity equation. (Vanishing of the divergence is a result of the mathematical identity that states the divergence of a curl is always zero.) This conflict is removed by addition of the displacement current, as then:[6][7]

$${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\left(\mathbf {J} +\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\right)=\mu _{0}\left(\mathbf {J} _{\mathrm {f} }+{\frac {\partial \mathbf {D} }{\partial t}}\right)\,,}$$

and

$${\displaystyle \nabla \cdot \left(\nabla \times \mathbf {B} \right)=0=\mu _{0}\left(\nabla \cdot \mathbf {J} _{\mathrm {f} }+{\frac {\partial }{\partial t}}\nabla \cdot \mathbf {D} \right)\,,}$$

which is in agreement with the continuity equation because of Gauss's law:

$${\displaystyle \nabla \cdot \mathbf {D} =\rho _{\mathrm {f} }\,.}$$

The added displacement current also leads to wave propagation by taking the curl of the equation for magnetic field.[8]

$${\displaystyle \mathbf {J} _{\mathrm {D} }=\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\,.}$$

Substituting this form for J into Ampère's law, and assuming there is no bound or free current density contributing to J:

$${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} _{\mathrm {D} }\,,}$$

with the result:

$${\displaystyle \nabla \times \left(\nabla \times \mathbf {B} \right)=\mu _{0}\epsilon _{0}{\frac {\partial }{\partial t}}\nabla \times \mathbf {E} \,.}$$

However,

$${\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial }{\partial t}}\mathbf {B} \,,}$$

leading to the wave equation:[9]

$${\displaystyle -\nabla \times \left(\nabla \times \mathbf {B} \right)=\nabla ^{2}\mathbf {B} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}}{\partial t^{2}}}\mathbf {B} ={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\mathbf {B} \,,}$$

where use is made of the vector identity that holds for any vector field V(r, t):

$${\displaystyle \nabla \times \left(\nabla \times \mathbf {V} \right)=\nabla \left(\nabla \cdot \mathbf {V} \right)-\nabla ^{2}\mathbf {V} \,,}$$

and the fact that the divergence of the magnetic field is zero. An identical wave equation can be found for the electric field by taking the curl:

$${\displaystyle \nabla \times \left(\nabla \times \mathbf {E} \right)=-{\frac {\partial }{\partial t}}\nabla \times \mathbf {B} =-\mu _{0}{\frac {\partial }{\partial t}}\left(\mathbf {J} +\epsilon _{0}{\frac {\partial }{\partial t}}\mathbf {E} \right)\,.}$$

If J, P, and ρ are zero, the result is:

$${\displaystyle \nabla ^{2}\mathbf {E} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}}{\partial t^{2}}}\mathbf {E} ={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\mathbf {E} \,.}$$

The electric field can be expressed in the general form:

$${\displaystyle \mathbf {E} =-\nabla \varphi -{\frac {\partial \mathbf {A} }{\partial t}}\,,}$$

where φ is the electric potential (which can be chosen to satisfy Poisson's equation) and A is a vector potential (i.e. magnetic vector potential, not to be confused with surface area, as A is denoted elsewhere). The ∇φ component on the right hand side is the Gauss's law component, and this is the component that is relevant to the conservation of charge argument above. The second term on the right-hand side is the one relevant to the electromagnetic wave equation, because it is the term that contributes to the curl of E. Because of the vector identity that says the curl of a gradient is zero, ∇φ does not contribute to ∇×E.