1.2.2 Capacitors—Electrical Energy Storage

In circuit theory, when the power supply charges the capacitor to V0, the energy provided is:

W We 1/2C V02 ( J )

where C is the capacitance of the capacitor.

In the electromagnetic field theory, the electrical energy stored between the parallel plate capacitors is related not only with parallel plate capacitor plate area S and board spacing d, but also with the dielectric ε between the two plates, i.e.,

We V02ε S/2d (J)

From the above analysis, the capacitance of the parallel plate capacitor is

C 2 We/ V02 ε S/d (F)

It is consistent with the capacitance of the parallel plate capacitor in circuit theory. This example shows that the power stored in the system can be used to find its capacitance. At the same voltage, the greater the power stored in the system, the greater the capacitance. If there is no electrical energy storage, the capacitance is zero. Thus, the capacitance of the system is a representation of the ability of the

system to store electrical energy.

1.2.3 Inductor—Magnetic Energy Storage

In the circuit theory, the energy supplied by a current source to a single-turn coil is

W Wm 1/2L I02 (J)

where L is the amount of inductance.

In the electromagnetic field theory, the magnetic energy stored in single-turn coil is related to the circumference of the coil l × d, width of the coil w, and the medium filled in the coil.

Wm 1/2(μ l d/w)I02 (J)

where μ is the magnetic conductivity.

From the analysis above, we can get the inductance of this coil as

L 2Wm /I02 μ l d/w ( H)

16 1 Electromagnetic Field and Wave

Table 1.1 Recommended maximum linear dimension of circuits and systems using DC or LF methods

which is consistent with the formula for the inductor inductance of a single-turn coil in circuit theory.

It can be seen from this example that the inductance of the system can be calculated from magnetic energy, which is a representation of the ability of the system to store magnetic energy.

1.2.4 Examples of Device Properties Analysis

Example 1.3 There is a capacitor with 10 nH self-inductance, and the capacitance measured at 2 MHz is 10 pF. What is the capacitance of the capacitor at 450 MHz?

Solution

f 0.45 GHz

X L 2π fG H z Ln H 28.3

XC 1/(2π fH z CF ) 159/( fG H z C pF ) 35.2

j XT j X L − j XC − j7 − j159/( fG H z C pF )

C pF −159/( fG H z XT ) 50 pF

where XT is the total reactance.

When the frequency is less than the resonant frequency, the capacitor plays the major role. The closer to the resonant frequency, the more obvious the effect of self-inductance. At the resonant frequency, the capacitor is no longer a capacitor. When the frequency is greater than the resonant frequency, the capacitor is actually equivalent to an inductor.

Table 1.1 lists the wavelength corresponding to the frequency from 100 kHz to 10 GHz and the maximum linear dimension of circuits and systems that will not cause intolerable error when using DC or low frequency to analyze the characteristics of the circuit or system.

It can be seen from the above analysis and discussion that in an energy storage component, if the energy is stored in the form of electrical energy, the component will be capacitive to the outside; if the energy is stored in the form of magnetic energy, the component will exhibit inductivity to the outside. When the energy is stored as

both electrical energy and magnetic energy, then when We > Wm , it is externally capacitive; when We < Wm , it is externally inductive. Usually, in an alternating field, the properties of the components are not single. Therefore, the influx of so-called lead inductance and distributed capacitance is often considered in circuit theory. Essentially, the nature of the component depends on the form in which it is stored.

1.3 The Reflection of Electromagnetic Wave

The electromagnetic shielding used in EMC usually utilizes the concept of electric wall in principle. The surface where the tangential component of electric field is zero is called the electric wall, which can shield the electric field.

1.3.1Boundary Conditions of the Electromagnetic Field on the Ideal Conductor Surface

The boundary refers to the discontinuous interface in the electromagnetic field. If the electromagnetic properties of the two regions are different, there should be a change in the electromagnetic field in the crossed interface.

The boundary condition is the relationship of electromagnetic fields between the two sides of the interface in which such electromagnetic properties are discontinuous.

According to Faraday’s law of electromagnetic induction from Maxwell’s equations

E · ds − d μ0 H · da

C dt S

and the modified Ampere’s circuital law

the tangential boundary conditions of the electric and magnetic fields can be concluded as

i n × ( E1 − E2) 0, i n × ( H1 − H2) K

where in is the normal direction of the boundary; E1 and E2 represent the electric field intensity in area 1 and area 2; H1 and H2 represent the magnetic field intensity in region 1 and region 2, respectively.

It can be seen from i n × ( E1 − E2) 0 that the tangential component of the electric field is continuous within the boundary. From in × ( H1 − H2) K , we

see that the boundary component of the magnetic field intensity on both sides of the tangential direction at the boundary is discontinuous when current exits on the boundary surface, and the mold of the difference is equal to the mold of the surface current density.

According to Gauss’s law of electric field from Maxwell’s equations

ε0 E · da ρdV Qnet

S V

and Gauss’s law for magnetism

μ0 H · da 0

S

the normal boundary conditions of electric and magnetic fields will be known as

i n · (ε0 E1 − ε0 E2) η, i n · (μ0 H1 − μ0 H2) 0

where i n · (ε0 E1 − ε0 E2) η shows that the normal component of the electric flux vector on both sides of the boundary is discontinuous when there is a surface charge. i n · (μ0 H1 − μ0 H2) 0 shows that the normal component of the flux vector on both sides of the boundary is continuous.

We will explain the above principles with three examples: (1) A closed metal shell has a shielding effect from the electric field generated by a charged system with zero charge (Example 1.4); (2) a closed metal shell does not have shielding effect when the net charge is not equal to zero (Example 1.5); (3) a closed and grounded metal shell shields the electric field generated by a charged system whether the charge is equal to zero (Example 1.6).

Example 1.4 There is a hollow conductor spherical shell with inner radius a and outer radius b. An electric dipole is placed at the center of the sphere. What is the spatial potential distribution?

Solution The problem can be solved with the electrostatic field separation variable method.

First, divide the space into three areas: area I(0 < rS < a), area II(a < rS < b), and area III(rS > b).

The potentials in the three regions are Φ1(r), Φ2(r), Φ3(r) respectively, which all satisfy the Laplace equation. The four boundary conditions are

rS a, Φ1(r) Φ2(r) C (Constant)

rS b, Φ2(r) Φ3(r) C (Constant)

rS → ∞, Φ3(r) → 0

p

rS → 0, Φ1(r) → 4 π ε0rS2 cos θ

Then, the solved spatial distribution of the potential is [1]

It is easy to see that the potential outside the metal spherical shell is zero, which indicates that the metal spherical shell has shielded the electric field generated by the electric dipole.

Example 1.5 There is a hollow metal spherical shell with inner radius a and outer radius b, with a point charge placed at the center of the sphere. Calculate the spatial potential distribution.

Φ1(r), Φ2(r), Φ3(r), respectively, all of which satisfy the Laplace equation. The four boundary conditions are

rS a, Φ1(r) Φ2(r) C (Constant)

rS b, Φ2(r) Φ3(r) C (Constant)