UEE 2201 Electromagnetics I

3

Spring, 2015

Static Electric Fields 1. Building a theoretical EM model by deduction (see textbook §1.2) (a) physical quantities to be studied (source and field quantities) (b) rules of operation – vector analysis (c) fundamental postulates – Maxwell’s equations 2. Quatities in EM model (functions of position and time) (a) source quantities - electric charge: * point charge q, Q electron: e = 1.6 × 10−19 (Coulomb) * distribution of charge and charge density lim ∆q (C/m3 ) volume charge density: ρ = ∆v→0 ∆v lim ∆q (C/m2 ) surface charge density: ρs = ∆s→0 ∆s lim ∆q (C/m) line charge density: ρ` = ∆`→0 ∆` * Conservation of electric charge must be satisfied at all times and under any circumstance. - current (charge in motion; flowing through a finite area, not a point function; scalar) I=

dq dt

(C/s or A)

current density (vector) lim ∆I a J = ∆s→0 ∆s v

(A/m2 )

surface current density (vector) lim ∆I a Js = ∆`→0 ∆` v

(A/m)

c

Yi Chiu, ECE Dept., NCTU

3-1

UEE 2201 Electromagnetics I

Spring, 2015

(b) field quantities: E: electric field intensity (V/m) D: electric flux density (electric displacement) (C/m2 ) B: magnetic flux density (Tesla) H: magnetic field intensity (A/m) 3. SI units and universal constant (see textbook §1.3) Meter - SI (MKSA):

(length) - Universal constants

Kilogram -

(mass)

Ampere

Second -

(time)

-

(current)

(a) c: velocity of EM waves in vacuum c ∼ 3 × 108 m/s (b) 0 : permittivity of free space (介電係數)(D = 0 E) 1 × 10−9 (F/m) 0 = 8.854 × 10−12 ∼ 36π (c) µ0 : permeability of free space (導磁係數) (H = µ10 B) µ0 = 4π × 10−7 (H/m)

3.1

Introduction

1. Electrostatics - source (charge q, Q, ρ) and field (E) do not change with time - no magnetic field 2. Coulomb’s law F12 = aR12 k

q1 q2 2 , where R12

F12 = force on q2 excerted by q1 , aR12 = unit vector from q1 to q2 , k = constant depending on medium and units, q1 , q2 = same sign → repulsive force, or = opposite sign → attractive force c

Yi Chiu, ECE Dept., NCTU

3-2

UEE 2201 Electromagnetics I

3.2

Spring, 2015

Postulates of Electrostatics in Free Space

1. Definition of electric field intensity E (= force per unit charge)

- To measure E or F, q must be small enough not to disturb the charge disribution of the source Total electrostatic force on a finite charge q: F = qE 2. Postulates of electrostatics in free space - Differential form (independent of coordinate system)  ρ   ∇ · E = 0 , where ρ = volume charge density (C/m3 )       0 = permittivity of free space   (charge is the source of electric flux)      ∇ × E = 0 (static electric field is irrotational/conservative)      (static electric field has no vortex/circulation source) - Integral form (a) From Eq. 3-4,

(Total outward flux of the electrostatic field intensity over any closed surface S is proportional to the total charge enclosed in the surface.) (b) From Eq. 3-5,

(Circulation of electrostatic field intensity around any closed path is zero.) c

Yi Chiu, ECE Dept., NCTU

3-3

UEE 2201 Electromagnetics I

Spring, 2015

3. Consider two points P1 and P2 in free space and two arbitrary paths C1 and C2 connecting the two points. From Eq. 3-8,

⇒ Scalar line integral of the electrostatic field E (work done by the E field to move a unit charge from P1 to P2 ) is: – independent of paths (C1 or C2 ) – dependent on the end points (P1 and P2 ) only ⇒ Eq. 3-11 is a statement of conservation of energy in an electrostatic field.

3.3

Coulomb’s Law

1. Consider a point charge q in a boundless free space. Due to the spherical symmetry, E = aR ER is in the aR direction and ER = ER (R) is a function of R only. From Eq. 3-7,

- For a positive charge q, the electric field is in the outward radial direction. - |E| ∝ q, |E| ∝ 1/R2 2. Please check ∇ × E = 0 3. If the charge q is not at the origin, E(P ) =

q(R − R0 ) 4π0 |R − R0 |3 c

Yi Chiu, ECE Dept., NCTU

3-4

UEE 2201 Electromagnetics I

Spring, 2015

4. Force F12 experienced by q2 in the field E12 generated by q1 at R12 F12 = q2 E12 = aR12

q1 q2 2 4π0 R12

(Coulomb’s law)

- Concept of force and field force: effect acting on one object (q2 ) by another object (q1 ), with or without medium in the space between the two objects; simultaneous action field: property of space created by one object (q1 ) and experienced by another object (q2 ); no direct interaction between two objects; propagation of field variation (wave) can be discussed Ex 3-2: Total charge Q is deposited uniformly on a thin shell of radius b. Find E field inside the shell. Sol:

1. From spherical symmetry, E = aR ER (R) For a spherical Gaussian surface S inside the shell,

2. The uniform surface charge density is ρs = Q/4πb2 From Coulomb’s law,

From geometrical similarity of the two cones,

note: Eq. 3-19: Coulomb’s inverse-square law, experimental Eq. 3-20’: geometric relation, exact ⇒ dE = 0 only when Coulomb’s law is an exact inverse square law ⇒ We can use the vanishing electric field inside a charged shell to verify the accuracy of the inverse square law. c

Yi Chiu, ECE Dept., NCTU

3-5

UEE 2201 Electromagnetics I

Spring, 2015

Ex 3-3: Electrostaitc deflection in a cathode ray tube (CRT) (Please see textbook) - d0 ∝ E d - two pairs of orthogonal deflection plates Ex: Ink jet printer

- q = n∆q, Ed = constant, d0 ∝ q ∝ n 5. System of discrete point charges (linear superposition) E field due to n discrete point charges E(R) =

n 1 X qk (R − R0k ) 4π0 k=1 |R − R0k |3

6. Electric dipole (n = 2)

c

Yi Chiu, ECE Dept., NCTU

3-6

UEE 2201 Electromagnetics I

Spring, 2015

Compared to Coulomb’s law (E ∝ q/R2 ): - p : source of dipole field - 1/R3 dependence due to cancellation of fields of +q and −q charges 7. Continuous distribution of charge - volume charge distribution (superposition principle)

Z

1 E= dE = 4π0 V0

Z

ρ 1 aR 2 dv 0 = 4π0 R V0

Z V0

ρR 0 dv R3

(ρ, R, R3 are functions of positions of P and dv 0 ) - surface charge distribution Z Z Z 1 1 ρs 0 ρs R 0 dE = aR 2 ds = ds E= 4π0 S 0 R 4π0 S 0 R3 S0 - linear charge distribution Z Z Z 1 1 ρ` 0 ρ` R 0 E= dE = aR 2 d` = d` 4π0 L0 R 4π0 S 0 R3 L0 Ex 3-4: Find the E field due to an infinitely long straight line with uniform charge density ρ` in air. Sol: - primed coordinate for source points → on the line - unprimed coordinate for field points → in the air - cylindrical symmetry → independent of φ and z - similarly, dEz will be canceled and only dEr needs to be considered c

Yi Chiu, ECE Dept., NCTU

3-7

UEE 2201 Electromagnetics I

c

Yi Chiu, ECE Dept., NCTU

Spring, 2015

3-8

UEE 2201 Electromagnetics I

3.4

Spring, 2015

Gauss’s Law and Applications

- total outward flux ∝ total charge enclosed - can be used to find E in a symmetric system, e.g. E due to a point charge - S: any real or hypothetical (mathematical) closed surface (Gaussian surface) Ex 3-5: Use Gauss’s law to solve Ex 3-4. Sol: Due to cylindrical symmetry,    E ⊥ line charge   ⇒ E in the radial direction     E = a E (r), E is independent of φ and z r r r

Ex 3-6: E due to an infinite uniform planar charge distribution ρs (e.g. charge on a conductor surface) Sol: Due to symmetry, E k az   a E (z), z > 0 z z E=  −a E (z), z < 0 z z ! I Z Z Z E · ds = + + E · ds = Ez A + Ez A + 0 S top bottom side =

c

Yi Chiu, ECE Dept., NCTU

3-9

UEE 2201 Electromagnetics I

- surface source

Spring, 2015

E ∝ R0

line source

E ∝ R−1

point source

E ∝ R−2

dipole source

E ∝ R−3

Ex 3-7: E due  to a spherical charge distribution, e.g. electron cloud,  ρ(R), 0 ≤ R ≤ b, ρ=  0, R≥b Sol: (Gaussian surfaces: concentric spherical surfaces) (a) Si inside the charge distribution, Ri ≤ b

(b) So outside the charge distribution, Ro ≥ b

c

Yi Chiu, ECE Dept., NCTU

3-10

UEE 2201 Electromagnetics I

3.5

Spring, 2015

Electric Potential

1. Since electrostatic field E is irrotatioinal (∇ × E = 0), a scalar field can be defined such that E = −∇V . V is called electric potential. 2. E field is defined as the force on a unit charge. Therefore, work must be done against the field to move a charge q from P1 to P2 : Z

P2

Z

P2

qE · d`

(−F) · d` = −

W =

P1

P1

(independent of path since ∇ × E = 0; depends only on P1 and P2 ) ⇒ For a unit charge, a scalar function V (r) can be defined such that

- V = electric potential energy per unit charge stored in the electrostatic field - V = potential difference between P1 and P2 , not absolute potential → a reference is needed (usually V at ground or infinity) - minus sign: potential increases if going against E field - E = −∇V is perpendicular to equipotential (constant-V ) surfaces 3. Electric potential due to a point charge

Choose P1 = ∞, d` in aR direction, ⇒ 4. Potential difference between two points P1 and P2 Z P20 Z P2 ! V21 = V2 − V1 = − + E · d` P1 P20   q 1 1 = − 4π0 R2 R1 c

Yi Chiu, ECE Dept., NCTU

3-11

UEE 2201 Electromagnetics I

Spring, 2015

5. Electric potential due to n discrete point charges    V (R) = Vq1 (R) + Vq2 (R) + · · · + Vqn (R)    n  1 X qi = (scalar sum) 4π0 i=1 |R − R0i |       E = −∇V 6. Electric dipole (n = 2)

- Equipotential lines in an electric dipole field → find R(R, θ, φ) such that V (R, θ, φ) = constant ⇒ - E field lines → find R such that d` k E

Fig. 3-15 Is the calculation correct for all positions in the space? c

Yi Chiu, ECE Dept., NCTU

3-12

UEE 2201 Electromagnetics I

Spring, 2015

7. Electric potential due to continuous distribution of charge: Z

1 dV = volume charge: V = 4π0 V0 Z

1 surface charge: V = dV = 4π0 S0 Z

line charge:

1 V = dV = 4π0 C0

Z V0

ρ 0 dv R

S0

ρs 0 ds R

Z

Z C0

ρ` 0 d` R

Ex 3-9: E on the axis of a circular disk (r = b) with a uniform charge density ρs Sol: (Can Gauss’s law be applied to simplify calculation?)

- Can we calculate Ex and Ey from V ? - On the axis, Ex =?, Ey =? - It can be shown that for |z|  b, (see textbook) Q Ez ≈ ±az = E of a point charge Q = πb2 ρs 4π0 z 2 Ex 3-10 (see textbook)

c

Yi Chiu, ECE Dept., NCTU

3-13

UEE 2201 Electromagnetics I

3.6

Spring, 2015

Conductors in Static Electric Field

1. Classification of materials conductor . . . . . . . . . . . semiconductor . . . . . . . . . . . insulator (dielectrics) many free electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . few free electrons 2. Inside a conductor,

⇒ Inside a conductor under static condition,

- consistant with Gauss’s law - charge distributed only on the surface (ρs ) 3. On the surface of the conductor, if Et 6= 0 ⇒ surface current is induced ⇒ not static ⇒ Et = 0 for static condition ⇒ E is normal to the conductor surface ⇒ conductor surface is an equipotential surface 4. For an internal point P , the potential difference between P and the surface S is ⇒ entire conductor is an equipotential body 5. Boundary conditions Conditions or constraints that must be satisfied by field quantities across the boundary of two different regions.

c

Yi Chiu, ECE Dept., NCTU

3-14

UEE 2201 Electromagnetics I

Spring, 2015

(a) Tangential components Et

(b) Normal components En

Ex 3-11 A point charge Q at the center of a conducting shell Due to spherical symmetry, E = aR ER (R) (a) R > Ro

(b) Ri ≤ R ≤ Ro (inside the conductor)

(c) R ≤ Ri

- V is continuous, E can be discontinuous c

Yi Chiu, ECE Dept., NCTU

3-15

UEE 2201 Electromagnetics I

3.7

Spring, 2015

Dielectrics in Static Elecrtric Field

1. Dielectric material: no free charge; bound charge has limited moving range and does not have screening effect

2. Polarized dielectrics

For the induced dipole moment, we can define a polarization vector P, P(R) = volume density of induced electric dipole moment =

Electrostatic potential dV due to dp in dv 0 is

Total electrostatic potential due to the polaized dielectric Z Z 1 P · aR 0 V = dV = dv 4π0 V 0 R2

1 But in general, V = 4π0

Z

dq R

⇒ Potential due to polarized dielectrics can be written as I  Z 1 ρps ds0 ρp dv 0 + , where V = 4π0 S 0 R R V0 ρps = P · an 0 = equivalent polarization surface charge density ρp = −∇0 · P = equivalent polarization volume charge density c

Yi Chiu, ECE Dept., NCTU

3-16

UEE 2201 Electromagnetics I

Spring, 2015

- Charge neutrality: please show that the total induced charge Q in a polarized dielectrics is zero.

3.8

Electric Flux Density and Dielectric Constant

1. In free space with free charge density ρ,

In dielectric materials with induced dipole moment,

We can define D , 0 E + P = electric flux density (electric displacement) [C/m2 ], ⇒

2. For linear and isotropic dielectric material,

 0 = relative permittivity (or dielectric constant)

r = 1 + χe =

= measurement of polarizability

c

Yi Chiu, ECE Dept., NCTU

3-17

UEE 2201 Electromagnetics I

Spring, 2015

For example, r ∼ 1.00059 (air) 2.3 (oil) 4 - 10 (glass) 80 (water) 3. Simple media: linear, isotropic, and homogeneous (r = scalar constant) ⇒ Ex 3-12: A point charge Q is placed at the center of a spherical dielectric shell. Find E, V, D, P as functions of R. Sol: Spherical symmetry → all vector fields in aR direction. Gauss’s law can be applied with concentric Gaussian surfaces to simplify the calculation. (a) R > Ro net charge in the dielectric shell is zero ⇒ same as a point charge Q at center Q ⇒ ER1 = 4π0 R2 Q V1 = 4π R 0

(b) Ri < R < Ro

c

Yi Chiu, ECE Dept., NCTU

3-18

UEE 2201 Electromagnetics I

Spring, 2015

(c) R < Ri (free space with Q at center) Q 4π0 R2 Q DR3 = 4πR2 PR3 = 0

⇒ ER3 =

* Please find the equivalent surface and volume charge density in the dielectric shell. Ex 3-13 Lightning rod For two spherical conductors at same potential,

⇒ E field is larger near sharper points ⇒ breakdown of air will occur near the tip of lightning rod (see ”dielectric strength” and Table 3-1) Ex: Field emission microscopy

c

Yi Chiu, ECE Dept., NCTU

3-19

UEE 2201 Electromagnetics I

3.9

Spring, 2015

General Boundary Conditions for Electrostatic Fields

1. Tangential components Et and Dt

⇒ Tangential component of E is continuous across the boundry 2. Normal components En and Dn

⇒ Normal component of D is discontinuous if free surface charge exists on the boundary - If medium 2 is a condcutor, D2 = 0. ⇒ D1n = ρs = 1 E1n - If both media are dielectrics with no free charge on the surface (ρs = 0),   D =D 1n 2n ⇒   E = E 1 1n 2 2n Ex 3-14: Dielectric sheet in uniform electric field (see textbook)

c

Yi Chiu, ECE Dept., NCTU

3-20

UEE 2201 Electromagnetics I

3.10

Spring, 2015

Capacitance and Capacitor

1. Surface charge on an isolated conductor  I   ρs ds0  Q = 0 S I 1 ρs ds0   V =  4π0 S 0 R if ρs 0 (R0 ) = kρs (R0 ), (same spatical distribution)   Q0 = kQ  V 0 = kV ⇒ 2. Capacitor composed of two conductors

3. How to calculate C? (a) Q → E → V (integration of field) (b) V → Q (boundary value problem; solving differential equation) Ex 3-7: Parallel-plate capacitor Sol: (a) Deposit ±Q charge on the two plates, assume uniform charge distribution, and neglect fringing field. On the upper plate,

(b) Apply a potential difference V12 between two plates

From B.C. of the conductor/air interface (Eq. 3-122),

c

Yi Chiu, ECE Dept., NCTU

3-21

UEE 2201 Electromagnetics I

Spring, 2015

Ex 3-8: Cylindrical capacitor Sol: Assume ±Q in the two cylindrical surfaces, apply Gauss’s law and neglect fringing field,

Ex 3-9: Spherical capacitor (see textbook)

4. Electrostatic shielding

c

Yi Chiu, ECE Dept., NCTU

3-22

UEE 2201 Electromagnetics I

3.11

Spring, 2015

Electrostatic Energy and Force (see pp. 172-173 of Ulaby)

1. In a parallel plate capacitor,

Electrostatic energy stored in the capacitor is

where V = volume of the capacitor = volume of space with electric field ⇒ an electrostatic energy density can be defined

- Eq. 4.123 is valid for any electric field distribution in dielectric media - we (r) is a function of position for a non-uniform field distribution E(r) or a non-uniform dielectric medium (r) Z Z 1 - Total electrostatic energy stored in V is We = we dv = E 2 dv 2 V V Ex 3-25: capacitance of a cylindrical capacitor (see textbook) Sol:

c

Yi Chiu, ECE Dept., NCTU

3-23

UEE 2201 Electromagnetics I

Spring, 2015

2. Electrostatic force

If Q on the conductor is kept constant, work done by the electrostatic force dW = change of electrostatic energy is dWe = In an isolated system (Q= constant), dW + dWe = 0, ⇒ 3. For a parallel plate capacitor,

c

Yi Chiu, ECE Dept., NCTU

3-24